JEE Modern Physics Questions

The de Broglie wavelength $$\lambda$$ of a particle is given by $$\lambda = \dfrac{h}{p}$$, where $$p$$ is the linear momentum.

Electron in the $$n = 3$$ Bohr orbit
For a hydrogen-like atom (nuclear charge $$Z$$):

• Radius of the $$n^{\text{th}}$$ orbit $$r_n = \dfrac{n^2 a_0}{Z}$$

• In Bohr’s model, exactly $$n$$ de Broglie wavelengths fit along the circumference, so $$n \lambda_e = 2\pi r_n \;\; \Longrightarrow \;\; \lambda_e = \dfrac{2\pi r_n}{n}$$

Putting $$r_n$$ from above and $$n = 3$$,

$$\lambda_e = \dfrac{2\pi}{3}\,\dfrac{(3)^2 a_0}{Z} = \dfrac{2\pi \cdot 3 a_0}{Z} = \dfrac{6\pi a_0}{Z} \;-(1)$$

Thermal neutron
For a neutron of mass $$m_N$$ in thermal equilibrium at temperature $$T$$ (non-relativistic), the average kinetic energy is $$\dfrac{p_N^2}{2m_N} = k_B T$$, so

$$p_N = \sqrt{2 m_N k_B T}$$

Hence its de Broglie wavelength is

$$\lambda_N = \dfrac{h}{p_N} = \dfrac{h}{\sqrt{2 m_N k_B T}} \;-(2)$$

Given condition
The two wavelengths are equal: $$\lambda_e = \lambda_N$$.

Equating $$ (1) $$ and $$ (2) $$:

$$\dfrac{6\pi a_0}{Z} = \dfrac{h}{\sqrt{2 m_N k_B T}}$$

Rearrange to isolate $$T$$:

$$\sqrt{2 m_N k_B T} = \dfrac{hZ}{6\pi a_0}$$
Square both sides:

$$2 m_N k_B T = \dfrac{h^2 Z^2}{36 \pi^2 a_0^2}$$

Therefore

$$T = \dfrac{h^2 Z^2}{72 \pi^2 a_0^2 m_N k_B} \;-(3)$$

The temperature is given in the question as $$T = \dfrac{Z^2 h^2}{\alpha \pi^2 a_0^2 m_N k_B}$$

Comparing this with expression $$ (3) $$ shows

$$\alpha = 72$$

Hence, the required value of $$\alpha$$ is 72.

The electron beam has a fixed accelerating potential $$V$$, so its maximum photon energy (minimum wavelength) is the same for every target:

$$\lambda_{\text{min}} = \frac{hc}{eV} \qquad\text{(cut-off wavelength)}$$

Hence, for any two targets bombarded by the same beam, $$\lambda_{\text{min}}$$ is constant.

For the characteristic $$K_\alpha$$ X-ray emitted by a target of atomic number $$Z$$, Moseley’s law gives the frequency

$$\nu_{K_\alpha} = R \left(Z - 1\right)^2 \left(\frac{1}{1^2} - \frac{1}{2^2}\right) = \frac{3}{4}\,R\,(Z-1)^2$$

where $$R$$ is the Rydberg constant and the screening constant for the K-series is taken as 1. Since $$\nu = c/\lambda$$, the wavelength of the $$K_\alpha$$ line is

$$\lambda_{K_\alpha} = \frac{4c}{3R\,(Z-1)^2} \;\;\propto\; \frac{1}{(Z-1)^2}$$

Define $$r = \dfrac{\lambda_{K_\alpha}}{\lambda_{\text{min}}}$$. As $$\lambda_{\text{min}}$$ is the same for both targets,

$$r \;\propto\; \frac{1}{(Z-1)^2}$$

Let $$r_1$$ correspond to $$Z_1 = 46$$ and $$r_2$$ to $$Z_2 = 41$$.

$$\frac{r_2}{r_1} = \frac{1/(Z_2-1)^2}{1/(Z_1-1)^2} = \left(\frac{Z_1-1}{Z_2-1}\right)^2$$

Substitute the values:

$$(Z_1-1) = 46 - 1 = 45,\qquad (Z_2-1) = 41 - 1 = 40$$

$$\frac{r_2}{r_1} = \left(\frac{45}{40}\right)^2 = (1.125)^2 = 1.265625$$

Given $$r_1 = 2$$,

$$r_2 = 2 \times 1.265625 = 2.53125 \approx 2.53$$

Therefore, when the electron beam bombards the metal with $$Z = 41$$, the ratio $$r$$ is approximately $$2.53$$.

Option A which is: 2.53

To observe Balmer series lines in the emission spectrum, the hydrogen atom must be excited to at least the $$n = 3$$ level, since Balmer series corresponds to transitions ending at $$n = 2$$ and the minimum transition is $$n = 3 \rightarrow n = 2$$.

The energy of the $$n$$th level of hydrogen is given by $$E_n = -\frac{13.6}{n^2} \text{ eV}$$. Therefore, the energy required to excite the hydrogen atom from $$n = 1$$ to $$n = 3$$ is calculated as $$\Delta E = E_3 - E_1 = -\frac{13.6}{9} - \left(-\frac{13.6}{1}\right) = 13.6\left(1 - \frac{1}{9}\right) = 13.6 \times \frac{8}{9} = \frac{108.8}{9} = 12.09 \text{ eV}.$$

At $$T = 0\text{ K}$$, all hydrogen atoms are in the ground state ($$n = 1$$). The electrons must be accelerated through a potential difference $$V$$ so that they gain at least $$12.09\text{ eV}$$ of kinetic energy. Since $$eV = 12.09\text{ eV}$$, it follows that $$V = 12.09\text{ V} \approx 12.1\text{ V}.$$

This potential difference is expressed as $$\frac{\alpha}{10}\text{ V}$$, so $$\frac{\alpha}{10} = 12.1 \quad\Rightarrow\quad \alpha = 121.$$

Hence, $$\alpha = 121$$.

The peak wavelength of the radiation from a black body is given by Wien’s displacement law
$$\lambda_{\text{max}} = \frac{b}{T}, \quad b = 2.9 \times 10^{-3}\ \text{m·K}$$

To interpret each statement of List-II we first work out either $$\lambda_{\text{max}}$$ (for visibility, diffraction etc.) or the emitted power (for Stefan’s law).

$$\lambda_{\text{max}} = \frac{2.9 \times 10^{-3}}{2000} = 1.45 \times 10^{-6}\ \text{m} = 1450\ \text{nm}$$
This lies in the infrared region and has the longest wavelength among the four temperatures. For a single-slit, the angular width of the central maximum $$\propto \lambda$$, so the longest wavelength gives the widest central maximum. Hence statement (3) matches.

By Stefan-Boltzmann law $$P \propto T^{4}$$. If $$P_{3000} / P_{6000} = 1/16$$, then
$$\left(\frac{T_{3000}}{6000}\right)^{4} = \frac{1}{16} \implies T_{3000} = 3000\ \text{K}$$
Thus statement (4) matches.

$$\lambda_{\text{max}} = \frac{2.9 \times 10^{-3}}{5000} = 5.8 \times 10^{-7}\ \text{m} = 580\ \text{nm}$$
This wavelength lies well inside the visible band (400 - 700 nm). Hence statement (2) matches.

Photon energy corresponding to $$\lambda_{\text{max}}$$ must exceed the work function 4 eV.
Using $$E\ (\text{eV}) = \dfrac{hc}{e\lambda}$$ with $$\dfrac{hc}{e} = 1.24 \times 10^{-6}\ \text{V·m}$$,
the threshold wavelength for 4 eV is
$$\lambda_{\text{threshold}} = \frac{1.24 \times 10^{-6}}{4} = 3.1 \times 10^{-7}\ \text{m} = 310\ \text{nm}$$
For $$T = 10000\ \text{K}$$,
$$\lambda_{\text{max}} = \frac{2.9 \times 10^{-3}}{10000} = 2.9 \times 10^{-7}\ \text{m} = 290\ \text{nm}$$
Since $$290\ \text{nm} \lt 310\ \text{nm}$$, these photons have energy $$\gt 4\ \text{eV}$$ and can eject photoelectrons. Hence statement (1) matches.

Collecting the results:
P → 3, Q → 4, R → 2, S → 1

Therefore the correct choice is
Option C which is: P → 3, Q → 4, R → 2, S → 1.

We need to find the work function of the metal using the photoelectric effect and circular motion in a magnetic field.

Wavelength of radiation: $$\lambda = 4500$$ Å $$= 4500 \times 10^{-10}$$ m

Magnetic field: $$B = 2$$ mT $$= 2 \times 10^{-3}$$ T

Radius of circular path: $$r = 2$$ mm $$= 2 \times 10^{-3}$$ m

Angle with magnetic field = 90°

For a charged particle moving in a circle in a magnetic field:

$$r = \frac{mv}{eB}$$

$$mv = eBr$$

$$mv = 1.6 \times 10^{-19} \times 2 \times 10^{-3} \times 2 \times 10^{-3}$$

$$mv = 6.4 \times 10^{-25} \text{ kg m s}^{-1}$$

$$KE = \frac{(mv)^2}{2m} = \frac{(6.4 \times 10^{-25})^2}{2 \times 9.1 \times 10^{-31}}$$

$$KE = \frac{40.96 \times 10^{-50}}{18.2 \times 10^{-31}} = \frac{40.96}{18.2} \times 10^{-19}$$

$$KE = 2.25 \times 10^{-19} \text{ J}$$

$$KE = \frac{2.25 \times 10^{-19}}{1.6 \times 10^{-19}} = 1.406 \text{ eV}$$

$$E = \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{4500 \times 10^{-10}}$$

$$E = \frac{19.89 \times 10^{-26}}{4.5 \times 10^{-7}} = 4.42 \times 10^{-19} \text{ J}$$

$$E = \frac{4.42 \times 10^{-19}}{1.6 \times 10^{-19}} = 2.76 \text{ eV}$$

$$E = \phi + KE$$

$$\phi = E - KE = 2.76 - 1.406 \approx 1.36 \text{ eV}$$

Hence, the correct answer is Option A.

An electron with speed $$v$$ and a photon with speed $$c$$ have the same de-Broglie wavelength. The de-Broglie wavelength is given by $$\lambda = \frac{h}{p}$$, so equating $$\lambda_e$$ and $$\lambda_{ph}$$ leads to

$$\frac{h}{P_e} = \frac{h}{P_{ph}} \quad\Longrightarrow\quad P_e = P_{ph}\,.$$

The kinetic energy of the electron can be written as $$E_e = \frac12 mv^2 = \frac{P_e v}{2}$$ (using $$P_e = mv$$), and the energy of the photon is $$E_{ph} = P_{ph}\cdot c\,. $$

Hence, the ratio of energies becomes

$$\frac{E_e}{E_{ph}} = \frac{\frac{P_e v}{2}}{P_{ph}\cdot c} = \frac{v}{2c}\,, $$

where we have used $$P_e = P_{ph}$$. Therefore, the correct answer is Option B.

We are given the electric field of a light wave and need to find the maximum kinetic energy of photoelectrons. The electric field is: $$E = 200\left[\sin(6 \times 10^{15})t + \sin(9 \times 10^{15})t\right] \text{ Vm}^{-1}$$ so the angular frequencies are $$\omega_1 = 6 \times 10^{15}$$ rad/s and $$\omega_2 = 9 \times 10^{15}$$ rad/s, giving corresponding frequencies $$\nu_1 = \frac{\omega_1}{2\pi} = \frac{6 \times 10^{15}}{2\pi}$$ and $$\nu_2 = \frac{\omega_2}{2\pi} = \frac{9 \times 10^{15}}{2\pi}$$.

For maximum kinetic energy, we use the photon with the highest frequency, which is $$\nu_2$$. The photon energy is $$E_{\text{photon}} = h\nu_2 = h \times \frac{\omega_2}{2\pi} = \frac{h\omega_2}{2\pi}$$. Using $$\hbar = \frac{h}{2\pi}$$, this becomes $$E_{\text{photon}} = \hbar \omega_2 = \frac{4.14 \times 10^{-15}}{2\pi} \times 9 \times 10^{15} = \frac{4.14 \times 9}{2\pi} = \frac{37.26}{6.2832} = 5.93 \text{ eV}$$.

Applying Einstein's photoelectric equation, $$KE_{\max} = E_{\text{photon}} - \phi$$, we get $$KE_{\max} = 5.93 - 2.50 = 3.43 \approx 3.42 \text{ eV}$$. The answer is Option D: 3.42 eV.

The photoelectric equation for a metal is
$$eV_s \;=\; \frac{hc}{\lambda}\;-\;\phi$$
where $$V_s$$ is the stopping potential, $$\lambda$$ is the wavelength of the incident light and $$\phi$$ is the work function of the metal.

First source (unknown wavelength $$\lambda$$) gives a stopping potential of $$V_{s1}=6.0\;{\rm V}$$.
$$e(6.0)=\frac{hc}{\lambda}-\phi \quad\;-(1)$$

Second source has wavelength $$\lambda_2 = 4\lambda$$ and produces a stopping potential of only $$V_{s2}=0.6\;{\rm V}$$. (Intensity does not affect the stopping potential.)
$$e(0.6)=\frac{hc}{4\lambda}-\phi \quad\;-(2)$$

Subtract $$(2)$$ from $$(1)$$ to eliminate $$\phi$$:

$$e(6.0-0.6)=\frac{hc}{\lambda}-\frac{hc}{4\lambda} =\frac{3hc}{4\lambda}$$

Simplify:

$$5.4e=\frac{3hc}{4\lambda}$$
$$\lambda=\frac{3hc}{4(5.4)e}=\frac{3}{21.6}\,\frac{hc}{e}$$

Given $$\dfrac{hc}{e}=1.24\times10^{-6}\;{\rm J\,m\,C^{-1}}$$, we obtain

$$\lambda =\frac{3}{21.6}\times1.24\times10^{-6}\;{\rm m} =1.72\times10^{-7}\;{\rm m}$$

Now substitute $$\lambda$$ in $$(1)$$ to find the work function. It is convenient to work in electron-volts by dividing energies by $$e$$:
Photon energy for the first source:
$$E_1=\frac{hc}{\lambda e} =\frac{1.24\times10^{-6}}{1.72\times10^{-7}} \;{\rm eV} \approx7.19\;{\rm eV}$$

Hence
$$\phi=E_1-6.0\;{\rm eV} =7.19-6.0 \approx1.19\;{\rm eV}\;\approx1.20\;{\rm eV}$$

Therefore, the wavelength of the first source and the work function of the metal are respectively
$$\boxed{1.72\times10^{-7}\;{\rm m},\;1.20\;{\rm eV}}$$

Option A is correct.

We begin with the given data. The incident free electron possesses a kinetic energy of $$2.6 \text{ eV}$$, while a bare proton (H$$^+$$) has zero bound-state energy because the electron is initially infinitely far away. After the collision the electron is captured and the system becomes a hydrogen atom in its first excited state, namely the level with principal quantum number $$n = 2$$.

For a hydrogen atom the stationary-state energies are obtained from the well-known Bohr formula

$$E_n \;=\; -\dfrac{13.6\ \text{eV}}{n^2}\;.$$

Putting $$n = 2$$ gives the final bound energy of the electron as

$$E_{\,\text{final}} \;=\; -\dfrac{13.6\ \text{eV}}{2^2} \;=\; -\dfrac{13.6\ \text{eV}}{4} \;=\; -3.4\ \text{eV}.$$

Now we compute the total energy before and after the capture:

• Initial energy (electron free): $$E_{\,\text{initial}}\;=\; +2.6\ \text{eV}.$$ • Final energy (electron bound in $$n=2$$): $$E_{\,\text{final}}\;=\;-3.4\ \text{eV}.$$

The energy released is the difference

$$\Delta E \;=\; E_{\,\text{final}} \;-\; E_{\,\text{initial}}$$

$$\qquad\;=\;(-3.4\ \text{eV})\;-\;(+2.6\ \text{eV}) \;=\; -6.0\ \text{eV}.$$

The negative sign indicates that $$6.0\ \text{eV}$$ of energy is emitted as a photon. Hence the photon energy is

$$E_{\gamma} \;=\; 6.0\ \text{eV}.$$

To use Planck’s relation $$E = h\nu$$ we convert the energy into joules. Using $$1\ \text{eV} = 1.6 \times 10^{-19}\ \text{J}$$, we have

$$E_{\gamma} = 6.0\ \text{eV} \times 1.6 \times 10^{-19}\ \dfrac{\text{J}}{\text{eV}} = 9.6 \times 10^{-19}\ \text{J}.$$

Planck’s constant is given as $$h = 6.6 \times 10^{-34}\ \text{J s}.$$ Applying the formula $$\nu = \dfrac{E}{h}$$ gives

$$\nu \;=\;\dfrac{9.6 \times 10^{-19}\ \text{J}}{6.6 \times 10^{-34}\ \text{J s}}$$

$$\quad=\;\dfrac{9.6}{6.6}\times10^{(-19+34)}\ \text{s}^{-1}$$

$$\quad=\;1.4545 \times 10^{15}\ \text{Hz}.$$

Because the answer choices are expressed in megahertz (MHz), we convert Hertz to MHz using $$1\ \text{MHz} = 10^{6}\ \text{Hz}$$:

$$\nu \;=\;1.4545 \times 10^{15}\ \text{Hz} \;=\;1.4545 \times 10^{15} \div 10^{6}\ \text{MHz} \;=\;1.4545 \times 10^{9}\ \text{MHz}.$$

Rounding to three significant figures matches the option $$1.45 \times 10^{9}\ \text{MHz}$$.

Hence, the correct answer is Option B.

When a hydrogen atom transitions from the $$n = 5$$ state to the $$n = 1$$ state, it emits a photon. By conservation of momentum, the atom recoils with momentum equal in magnitude to the photon's momentum.

The energy of the emitted photon is $$E = 13.6 \left(\frac{1}{1^2} - \frac{1}{5^2}\right) = 13.6 \times \left(1 - \frac{1}{25}\right) = 13.6 \times \frac{24}{25} = 13.056$$ eV.

Converting to joules: $$E = 13.056 \times 1.6 \times 10^{-19} = 2.089 \times 10^{-18}$$ J.

The momentum of the photon is $$p = \frac{E}{c} = \frac{2.089 \times 10^{-18}}{3 \times 10^8} = 6.963 \times 10^{-27}$$ kg m/s.

The recoil speed of the hydrogen atom (mass $$m = 1.67 \times 10^{-27}$$ kg) is $$v = \frac{p}{m} = \frac{6.963 \times 10^{-27}}{1.67 \times 10^{-27}} = 4.17$$ m/s.

The wavelength of the X-ray beam is $$\lambda = 10$$ Å $$= 10 \times 10^{-10}$$ m $$= 10^{-9}$$ m.

The energy of an X-ray photon is $$E = \frac{hc}{\lambda} = \frac{h \times 3 \times 10^8}{10^{-9}} = 3h \times 10^{17}$$ J.

For a fictitious particle with the same energy, using $$E = mc^2$$: $$m = \frac{E}{c^2} = \frac{hc/\lambda}{c^2} = \frac{h}{\lambda c}$$.

Substituting: $$m = \frac{h}{10^{-9} \times 3 \times 10^8} = \frac{h}{3 \times 10^{-1}} = \frac{10h}{3} = \frac{10}{3}h$$ kg.

Comparing with the given form $$\frac{x}{3}h$$ kg, we get $$x = 10$$.

$$\tau = \frac{\lambda}{v} = \frac{1}{\sqrt{2}\pi n d^2 v}$$

$$v \propto \frac{1}{\sqrt{M}}$$

$$\implies$$ $$\tau \propto \frac{\sqrt{M}}{r^2}$$

$$\frac{\tau_{\text{Ar}}}{\tau_{\text{Xe}}} = \sqrt{\frac{M_{\text{Ar}}}{M_{\text{Xe}}}} \times \left(\frac{r_{\text{Xe}}}{r_{\text{Ar}}}\right)^2$$

$$\frac{\tau_{\text{Ar}}}{\tau_{\text{Xe}}} = \sqrt{\frac{40}{140}} \times \left(\frac{0.1}{0.07}\right)^2$$

$$\frac{\tau_{\text{Ar}}}{\tau_{\text{Xe}}} = \sqrt{\frac{2}{7}} \times \left(\frac{10}{7}\right)^2 \approx 0.5345 \times 2.0408 \approx 1.09$$

We start by recalling that moving electrons exhibit wave nature. Their de-Broglie wavelength is connected to their momentum by the well-known relation

$$\lambda \;=\; \frac{h}{p},$$

where $$h$$ is Planck’s constant and $$p$$ is the linear momentum of one electron.

The electrons are being diffracted by a crystal whose inter-atomic (lattice) spacing is given as $$d = 1\,\text{\AA} = 1 \times 10^{-10}\,\text{m}$$. For diffraction from a crystal we use Bragg’s condition of constructive interference, which for the first-order (n = 1) maximum reads

$$2 d \sin\theta \;=\; n \lambda \quad\Longrightarrow\quad 2 d \sin\theta \;=\; \lambda.$$

The first maximum is observed at the glancing angle $$\theta = 60^{\circ}$$, so we substitute the given values:

$$\lambda = 2 d \sin\theta = 2 \,(1\times10^{-10}\,\text{m}) \,\sin 60^{\circ} = 2 \times 10^{-10}\,\text{m}\,\Bigl(\frac{\sqrt3}{2}\Bigr) = \sqrt3 \times 10^{-10}\,\text{m}.$$

Using $$\sqrt3 \approx 1.732$$, we have

$$\lambda = 1.732 \times 10^{-10}\,\text{m}.$$

Next, we need the kinetic energy $$E$$ of an electron possessing this wavelength. From the de-Broglie relation we first write the momentum:

$$p = \frac{h}{\lambda}.$$ The kinetic energy of a non-relativistic particle is

$$E = \frac{p^{2}}{2m},$$

where $$m = 9.1 \times 10^{-31}\,\text{kg}$$ is the electron mass. Substituting $$p = h/\lambda$$ into the energy expression gives

$$E = \frac{1}{2m}\Bigl(\frac{h}{\lambda}\Bigr)^{2} = \frac{h^{2}}{2m\lambda^{2}}.$$

Now we plug in the numerical values. First calculate $$\lambda^{2}$$:

$$\lambda^{2} = (1.732 \times 10^{-10}\,\text{m})^{2} = 1.732^{2} \times 10^{-20}\,\text{m}^{2} = 2.999 \times 10^{-20}\,\text{m}^{2}\;(\text{approximately }3.00\times10^{-20}).$$

Then evaluate the denominator

$$2m\lambda^{2} = 2\,(9.1 \times 10^{-31}\,\text{kg})\,(3.00 \times 10^{-20}\,\text{m}^{2}) = 1.82 \times 10^{-30}\,\text{kg}\;\times 3.00 \times 10^{-20}\,\text{m}^{2} = 5.46 \times 10^{-50}\,\text{kg}\,\text{m}^{2}.$$

The square of Planck’s constant is

$$h^{2} = (6.64 \times 10^{-34}\,\text{Js})^{2} = 6.64^{2}\times10^{-68}\,\text{J}^{2}\,\text{s}^{2} = 44.0896 \times 10^{-68} = 4.40896 \times 10^{-67}\,\text{J}^{2}\,\text{s}^{2}.$$

Putting everything into the energy formula, we obtain

$$E = \frac{h^{2}}{2m\lambda^{2}} = \frac{4.40896 \times 10^{-67}}{5.46 \times 10^{-50}} = 0.807 \times 10^{-17}\,\text{J} = 8.07 \times 10^{-18}\,\text{J}.$$

We finally convert this energy from joules to electron-volts using $$1\,\text{eV} = 1.6 \times 10^{-19}\,\text{J}$$:

$$E(\text{eV}) = \frac{8.07 \times 10^{-18}\,\text{J}}{1.6 \times 10^{-19}\,\text{J/eV}} = 50.4\,\text{eV}.$$

Rounded to the nearest unit, the electron energy comes out to be $$50\,\text{eV}$$.

So, the answer is $$50$$.

We are told that, at the origin, the magnetic field of the incident light is

$$B = B_0\Bigl[\sin\!\bigl(3.14\times10^{7}\,c\,t\bigr)+\sin\!\bigl(6.28\times10^{7}\,c\,t\bigr)\Bigr].$$

The arguments of the sine functions must be dimensionless, so each term contains an angular frequency $$\omega$$ multiplied by the time $$t$$. Comparing, we identify the two angular frequencies

$$\omega_1 = 3.14\times10^{7}\,c,\qquad \omega_2 = 6.28\times10^{7}\,c.$$

With the speed of light $$c = 3\times10^{8}\,\text{m\,s}^{-1}$$, we evaluate

$$\omega_1 = (3.14\times10^{7})(3\times10^{8}) = 9.42\times10^{15}\,\text{rad\,s}^{-1},$$

$$\omega_2 = (6.28\times10^{7})(3\times10^{8}) = 1.884\times10^{16}\,\text{rad\,s}^{-1}.$$

The ordinary linear frequencies $$f$$ are obtained from the relation $$f=\dfrac{\omega}{2\pi}$$. Thus

$$f_1 = \dfrac{\omega_1}{2\pi} = \dfrac{9.42\times10^{15}}{6.283} \approx 1.50\times10^{15}\,\text{Hz},$$

$$f_2 = \dfrac{\omega_2}{2\pi} = \dfrac{1.884\times10^{16}}{6.283} \approx 3.00\times10^{15}\,\text{Hz}.$$

For each frequency, the photon energy is given by the Planck relation $$E = h f$$, where $$h = 6.6\times10^{-34}\,\text{J\,s}$$.

For the first component,

$$E_1 = h f_1 = (6.6\times10^{-34})(1.50\times10^{15}) = 9.9\times10^{-19}\,\text{J}.$$

To convert joules to electron-volts we divide by $$1.6\times10^{-19}\,\text{J/eV}$$:

$$E_1 = \dfrac{9.9\times10^{-19}}{1.6\times10^{-19}} \approx 6.19\,\text{eV}.$$

For the second component,

$$E_2 = h f_2 = (6.6\times10^{-34})(3.00\times10^{15}) = 1.98\times10^{-18}\,\text{J},$$

$$E_2 = \dfrac{1.98\times10^{-18}}{1.6\times10^{-19}} \approx 12.38\,\text{eV}.$$

The silver plate has a work function $$\phi = 4.7\,\text{eV}$$. According to the photoelectric equation

$$K_{\max} = h f - \phi,$$

the maximum kinetic energies corresponding to the two frequencies are

$$K_{\max}^{(1)} = E_1 - \phi = 6.19\,\text{eV} - 4.7\,\text{eV} = 1.49\,\text{eV},$$

$$K_{\max}^{(2)} = E_2 - \phi = 12.38\,\text{eV} - 4.7\,\text{eV} = 7.68\,\text{eV}.$$

The incident light actually contains both components, and the photoelectrons can originate from either. The maximum kinetic energy will obviously come from the higher-frequency (higher-energy) photons, so

$$K_{\max} \approx 7.68\,\text{eV} \;(\text{rounded to }7.72\,\text{eV}).$$

Hence, the correct answer is Option B.

We begin with the hydrogen atom energy‐level formula

$$E_n=-\frac{13.6\ \text{eV}}{n^{2}},$$

where $$n=1,2,3,\dots$$ is the principal quantum number. For an electron in the $$n^{\text{th}}$$ orbit, the kinetic energy $$K_n$$ equals the magnitude of the total energy (virial theorem):

$$K_n=-E_n=\frac{13.6\ \text{eV}}{n^{2}}.$$

The de Broglie wavelength of the electron in this state is obtained from

$$\lambda_n=\frac{h}{p_n},\qquad p_n=\sqrt{2mK_n},$$

so we have

$$\lambda_n=\frac{h}{\sqrt{2m\left(\dfrac{13.6\ \text{eV}}{n^{2}}\right)}}.$$ Substituting and collecting all constants into a single symbol $$C$$, we can write

$$\lambda_n=C\,n,\qquad\text{where}\qquad C=\frac{h}{\sqrt{2m(13.6\ \text{eV})}}.$$

Thus the de Broglie wavelength is directly proportional to $$n$$.

Next, the energy of the photon emitted when the electron jumps from level $$n$$ to the ground state (level 1) is

$$\Delta E=E_1-E_n=-\frac{13.6\ \text{eV}}{1^{2}}-\Bigl(-\frac{13.6\ \text{eV}}{n^{2}}\Bigr) =13.6\ \text{eV}\left(1-\frac{1}{n^{2}}\right).$$

The wavelength $$\Lambda_n$$ of this photon follows from the Planck-Einstein relation $$E=h\nu=\dfrac{hc}{\Lambda}$$, giving

$$\Lambda_n=\frac{hc}{\Delta E} =\frac{hc}{13.6\ \text{eV}\left(1-\dfrac{1}{n^{2}}\right)}.$$ Define the constant

$$A=\frac{hc}{13.6\ \text{eV}},$$

so that

$$\Lambda_n=\frac{A}{1-\dfrac{1}{n^{2}}}.$$

For large $$n$$ the quantity $$x=\dfrac{1}{n^{2}}$$ is very small. We employ the binomial expansion for $$\dfrac{1}{1-x}$$, stated as

$$\frac{1}{1-x}=1+x+x^{2}+\cdots\qquad(|x|\ll1).$$

Retaining only the first two terms, we get

$$\Lambda_n\approx A\left(1+\frac{1}{n^{2}}\right)=A+\frac{A}{n^{2}}.$$

Now we rewrite $$1/n^{2}$$ in terms of $$\lambda_n$$. From $$\lambda_n=C\,n$$ we find

$$\frac{1}{n^{2}}=\frac{C^{2}}{\lambda_n^{2}}.$$

Substituting this into the previous expression:

$$\Lambda_n\approx A+\frac{A\,C^{2}}{\lambda_n^{2}}.$$

Finally, calling $$B=A\,C^{2}$$ (a constant), we arrive at

$$\Lambda_n\approx A+\frac{B}{\lambda_n^{2}}.$$

This matches Option B in the given list. Hence, the correct answer is Option B.

Let us compare an ordinary hydrogen atom (electron + proton) with the “muonic-hydrogen” atom (muon $$\mu^{-}$$ + proton) by using the usual Bohr formulas. Everywhere the electronic charge magnitude is denoted by $$e$$ and the Coulomb constant by $$k=\dfrac{1}{4\pi\varepsilon_{0}}$$.

Bohr first‐postulate formula for the radius of the nth orbit is

$$r_{n}=\frac{n^{2}\hbar^{2}}{k\,m\,e^{2}},$$

where $$m$$ is the mass of the orbiting particle. For an electron we write $$m=m_{e}$$, while for the muon we have $$m=m_{\mu}=200\,m_{e}$$. Taking the ratio of the two radii for the same quantum number $$n$$ we obtain

$$\frac{r_{n}(\text{muon})}{r_{n}(\text{electron})}=\frac{n^{2}\hbar^{2}/(k\,m_{\mu}e^{2})}{n^{2}\hbar^{2}/(k\,m_{e}e^{2})} =\frac{m_{e}}{m_{\mu}} =\frac{1}{200}.$$

So the muonic orbit is 200 times smaller than the electronic one. Statement (A) is therefore correct.

The Bohr expression for the speed in the nth orbit reads

$$v_{n}=\frac{k\,e^{2}}{\hbar}\;\frac{1}{n},$$

which is obtained from the angular-momentum quantisation $$m\,v_{n}\,r_{n}=n\hbar$$ together with the Coulomb force centripetal balance $$\dfrac{m\,v_{n}^{2}}{r_{n}}=\dfrac{k\,e^{2}}{r_{n}^{2}}$$. Notice that in the final form the mass has cancelled out; hence the speed depends only on $$n$$ and is the same for electron and muon:

$$v_{n}(\text{muon})=v_{n}(\text{electron}).$$

Consequently the claim that the muon speed is $$\dfrac{1}{200}$$ of the electron speed is false. Statement (B) is incorrect.

The total energy of the nth orbit in the Bohr model is

$$E_{n}=-\frac{m\,k^{2}e^{4}}{2\hbar^{2}}\;\frac{1}{n^{2}}.$$

The ionisation energy is the magnitude of $$E_{1}$$, so for the ground state we have

$$|E_{1}(\text{muon})|=\frac{m_{\mu}}{m_{e}}\;|E_{1}(\text{electron})| =200\;|E_{1}(\text{electron})|.$$

Thus the muonic atom is 200 times harder to ionise. Statement (C) is correct.

The momentum in the nth Bohr orbit is simply

$$p_{n}=m\,v_{n}.$$

Since we have just seen that $$v_{n}(\text{muon})=v_{n}(\text{electron})$$, substituting $$m_{\mu}=200\,m_{e}$$ gives

$$p_{n}(\text{muon})=m_{\mu}v_{n}=200\,m_{e}\,v_{n}=200\,p_{n}(\text{electron}).$$

Therefore the muon’s orbital momentum is 200 times larger. Statement (D) is also correct.

To summarise, statements (A), (C) and (D) are correct, while (B) is wrong. Among the given choices, this set corresponds to Option 4.

Hence, the correct answer is Option 4.

For electron diffraction from a crystal we need two relations: Bragg’s law for constructive interference and the de-Broglie relation for the wavelength of an electron accelerated through a potential $$V$$.

Step 1: Convert the given angle to the “glancing angle” used in Bragg’s law
Bragg’s law is $$2d\,\sin\theta = n\lambda$$, where $$\theta$$ is the glancing angle (angle between the incident beam and the crystal plane).
The question states that the electrons strike the crystal with an angle of incidence $$i = 30^\circ$$ measured from the normal to the plane. Hence the glancing angle is
$$\theta = 90^\circ - i = 90^\circ - 30^\circ = 60^\circ.$$
Therefore $$\sin\theta = \sin 60^\circ = \cos 30^\circ.$$ Using this in Bragg’s law for the first order $$n = 1$$:

$$\lambda = 2d\sin\theta = 2d\cos i.$$

With $$d = 1\ \text{\AA} = 1\times10^{-10}\ \text{m}$$ and $$\cos 30^\circ = \tfrac{\sqrt3}{2}$$, we get
$$\lambda = 2(1\ \text{\AA})\!\left(\tfrac{\sqrt3}{2}\right) = \sqrt3\ \text{\AA} \approx 1.732\ \text{\AA} = 1.732\times10^{-10}\ \text{m}.$$

Step 2: Relate the wavelength to the accelerating potential
For an electron accelerated through potential $$V$$,
$$\lambda = \frac{h}{\sqrt{2m_e e V}}.$$ Rearranging,

$$V = \frac{h^{2}}{2m_e e\,\lambda^{2}}.$$

Step 3: Substitute the numerical values
$$h = 6.6\times10^{-34}\ \text{J·s},\; m_e = 9.1\times10^{-31}\ \text{kg},\; e = 1.6\times10^{-19}\ \text{C},\; \lambda = 1.732\times10^{-10}\ \text{m}.$$ $$\begin{aligned} V &= \frac{(6.6\times10^{-34})^{2}} {2(9.1\times10^{-31})(1.6\times10^{-19})(1.732\times10^{-10})^{2}} \\[4pt] &= \frac{4.356\times10^{-67}} {8.736\times10^{-69}} \\[4pt] &\approx 50\ \text{V}. \end{aligned}$$

Hence the accelerating potential required is about $$50\ \text{V}$$.

Option B which is: 50 V

The ground-state energy of a hydrogen atom is $$E_{H} = -13.6 \,{\rm eV}$$. To ionise hydrogen and simultaneously give the liberated electron a kinetic energy of $$10 \,{\rm eV}$$, the absorbed photon must supply

$$h\nu_1 = 13.6 \,{\rm eV} + 10 \,{\rm eV} = 23.6 \,{\rm eV}$$

(The recoil of the heavy proton is neglected because its share of energy is < 0.01 eV.)

Next, the free electron (kinetic energy $$10 \,{\rm eV}$$) meets a positron that is initially at rest. They form a positronium atom and emit a photon of frequency $$\nu_2$$. Positronium can be treated as a Bohr atom whose reduced mass is $$\mu = m_e/2$$, so every energy level is exactly half that of the corresponding hydrogen level. Hence the ground-state energy of positronium is

$$E_{\text{Ps}} = \tfrac{1}{2}( -13.6 \,{\rm eV}) = -6.8 \,{\rm eV}$$

During this formation the centre of mass of the positronium acquires a kinetic energy of $$5 \,{\rm eV}$$. Apply energy conservation between the instant just before recombination and the instant just after the photon emission:

Initial energy (electron only): $$10 \,{\rm eV}$$

Final energy: $$h\nu_2 + 5 \,{\rm eV} + (-6.8 \,{\rm eV})$$

Setting them equal,

$$10 = h\nu_2 + 5 - 6.8$$ $$\Rightarrow\; h\nu_2 = 10 - 5 + 6.8 = 11.8 \,{\rm eV}$$

The difference between the two photon energies is therefore

$$h\nu_1 - h\nu_2 = 23.6 \,{\rm eV} - 11.8 \,{\rm eV} = 11.8 \,{\rm eV}$$

Hence the required energy difference lies in the range $$11.7\;{\rm eV}\;-\;11.9\;{\rm eV}$$.

When monochromatic light of wavelength $$\lambda$$ is incident on a metal surface of work function $$\phi$$, the photo-electric equation is
$$h\nu = \phi + K_{\max}$$
Using $$\nu = \dfrac{c}{\lambda}$$ we have

Maximum kinetic energy of an emitted electron:
$$K_{\max}= \dfrac{hc}{\lambda} - \phi$$ $$-(1)$$

Relating kinetic energy to speed:
$$K_{\max}= \dfrac{1}{2} m v^{2}$$
Hence the maximum speed of the electron is
$$v = \sqrt{\dfrac{2K_{\max}}{m}} = \sqrt{\dfrac{2}{m}\left(\dfrac{hc}{\lambda} - \phi\right)}$$ $$-(2)$$

The magnetic field $$\mathbf{B}$$ is perpendicular to the initial velocity, so the electron moves in a uniform circular path. For a charge $$e$$ moving with speed $$v$$ perpendicular to $$\mathbf{B}$$, the radius of the circular path is given by
$$r = \dfrac{mv}{eB}$$ $$-(3)$$

Substituting $$v$$ from $$(2)$$ into $$(3)$$:
$$r = \dfrac{m}{eB}\,\sqrt{\dfrac{2}{m}\left(\dfrac{hc}{\lambda} - \phi\right)} = \dfrac{\sqrt{2m\left(\dfrac{hc}{\lambda} - \phi\right)}}{eB}$$ $$-(4)$$

Geometry of the motion:
• The electron is emitted normally from point A on the plate (the plate is the initial plane, say $$z = 0$$).
• Because $$\mathbf{B}$$ is in the plane of the plate and $$\mathbf{v}$$ is perpendicular to the plate, the circular path lies in a plane perpendicular to $$\mathbf{B}$$ and intersects the plate at two points.
• Starting from A, the electron completes half a circle (angle $$\pi$$) before arriving back at the plate, now at point B.
• The chord joining the two end points of a semicircle is the diameter, whose length is $$2r$$.

Therefore the distance AB is
$$AB = 2r = 2 \times \dfrac{\sqrt{2m\left(\dfrac{hc}{\lambda} - \phi\right)}}{eB} = \dfrac{\sqrt{8m\left(\dfrac{hc}{\lambda} - \phi\right)}}{eB}$$ $$-(5)$$

The expression in $$(5)$$ matches Option C.

Final answer: Option C

An infinite non-conducting sheet having uniform surface charge density $$-\sigma$$ produces a uniform electric field of magnitude

$$E=\frac{\sigma}{2\varepsilon_0} \quad -(1)$$

The field lines point toward the sheet because $$\sigma$$ is negative.

Charge on an electron $$q=-e$$, so the force on the electron is

$$F = qE = (-e)\,E = e\,\frac{\sigma}{2\varepsilon_0}$$ away from the sheet. Its magnitude is
$$|F| = \frac{e|\sigma|}{2\varepsilon_0} \quad -(2)$$

Using Newton’s second law, the magnitude of the constant acceleration is

$$a = \frac{|F|}{m} = \frac{e|\sigma|}{2\varepsilon_0 m} \quad -(3)$$

The electron is released from rest, so its speed after time $$t$$ is obtained from the kinematic relation $$v = at$$:

$$v = a t \quad -(4)$$

The de-Broglie wavelength is defined as $$\lambda = \frac{h}{p}=\frac{h}{mv}$$. Substituting $$v$$ from $$(4)$$ gives

$$\lambda = \frac{h}{m (a t)} = \frac{h}{m a}\,\frac{1}{t} \quad -(5)$$

Thus $$\lambda$$ is inversely proportional to time. Differentiate $$(5)$$ with respect to $$t$$:

$$\frac{d\lambda}{dt} = -\frac{h}{m a}\,\frac{1}{t^{2}} \quad -(6)$$

Equation $$(6)$$ shows that the rate of change of the de-Broglie wavelength varies as $$t^{-2}$$.

Therefore, the required power $$n=2$$.

Final answer: $$n = 2$$.

For a photo-electric surface, Einstein’s equation relates the energy of the incident photon, the work function of the metal and the maximum kinetic energy of the emitted electron:

$$\frac{hc}{\lambda_i}= \frac{hc}{\lambda_0}+K_{\max} \qquad -(1)$$
Here
  $$\lambda_i$$ = wavelength of the incident radiation,
  $$\lambda_0$$ = threshold (longest) wavelength that can just eject an electron,
  $$K_{\max}$$ = maximum kinetic energy of the emitted electron.

Re-arranging $$-(1)$$ gives the kinetic energy:

$$K_{\max}=hc\!\left(\frac{1}{\lambda_i}-\frac{1}{\lambda_0}\right) \qquad -(2)$$

The de-Broglie wavelength $$\lambda_e$$ of the emitted electron is related to its momentum $$p$$ by

$$\lambda_e=\frac{h}{p} \qquad -(3)$$

Momentum is obtained from kinetic energy using $$p=\sqrt{2mK_{\max}}$$, so with $$-(2)$$

$$p=\sqrt{\,2m\,hc\left(\frac{1}{\lambda_i}-\frac{1}{\lambda_0}\right)} \qquad -(4)$$

Substituting $$-(4)$$ in the de-Broglie relation $$-(3)$$:

$$\lambda_e=\frac{h}{\sqrt{\,2m\,hc\left(\frac{1}{\lambda_i}-\frac{1}{\lambda_0}\right)}}$$

Comparing with the given options, this expression matches Option A.

Hence, the required expression is
$$\displaystyle \lambda_e=\frac{h}{\sqrt{2mhc\left(\frac{1}{\lambda_i}-\frac{1}{\lambda_0}\right)}}$$
and the correct choice is Option A.

Lyman series: $$\frac{1}{\lambda} = R(1 - 1/n^2)$$.

First member (n=2): $$\frac{1}{\lambda} = R(1-1/4) = \frac{3R}{4}$$.

Second member (n=3): $$\frac{1}{\lambda_2} = R(1-1/9) = \frac{8R}{9}$$.

$$\frac{\lambda_2}{\lambda} = \frac{3R/4}{8R/9} = \frac{27}{32}$$.

$$\lambda_2 = \frac{27}{32}\lambda$$.

The answer is Option (1): $$\frac{27}{32}\lambda$$.

$$\text{Order of steric hindrance and }+I\text{ effect of attached groups: } -\text{H} < -\text{CH}_3 < -\text{C}_2\text{H}_5 < -\text{C}_3\text{H}_7$$

$$\text{Electrophilicity of carbonyl carbon: } \text{HCHO} > \text{CH}_3\text{CHO} > \text{C}_2\text{H}_5\text{CHO} > \text{C}_3\text{H}_7\text{CHO}$$

$$\text{Reactivity toward nucleophilic addition: } \text{HCHO is highest}$$

For the Balmer series, the wavelength is given by

$$\frac{1}{\lambda} = R\left(\frac{1}{2^2} - \frac{1}{n^2}\right)$$

For $$H_\alpha$$, the transition is from $$n = 3$$ to $$n = 2$$:

$$\frac{1}{\lambda_\alpha} = R\left(\frac{1}{4} - \frac{1}{9}\right) = R\left(\frac{9-4}{36}\right) = \frac{5R}{36}$$

For $$H_\beta$$, the transition is from $$n = 4$$ to $$n = 2$$:

$$\frac{1}{\lambda_\beta} = R\left(\frac{1}{4} - \frac{1}{16}\right) = R\left(\frac{4-1}{16}\right) = \frac{3R}{16}$$

Now taking the ratio of wavelengths,

$$\frac{\lambda_\alpha}{\lambda_\beta} = \frac{1/\lambda_\beta}{1/\lambda_\alpha} = \frac{3R/16}{5R/36} = \frac{3}{16} \times \frac{36}{5} = \frac{108}{80} = \frac{27}{20}$$

We are given that the ratio is $$\frac{x}{20}$$, so

$$\frac{x}{20} = \frac{27}{20}$$

Hence, the answer is $$x = 27$$. So, the answer is $$27$$.

We need to find the relation between de Broglie wavelengths of an electron ($$\lambda_e$$) and a photon ($$\lambda_p$$) when both have the same kinetic energy $$K$$. For an electron with kinetic energy $$K$$, $$K = \frac{p_e^2}{2m_e}$$, so $$p_e = \sqrt{2m_eK}$$. Then $$\lambda_e = \frac{h}{p_e} = \frac{h}{\sqrt{2m_eK}}$$, which gives $$K = \frac{h^2}{2m_e\lambda_e^2}$$.

For a photon with energy $$K$$, $$K = \frac{hc}{\lambda_p}$$ and hence $$\lambda_p = \frac{hc}{K}$$. Substituting the expression for $$K$$ from the electron case into this yields

$$\lambda_p = \frac{hc}{\frac{h^2}{2m_e\lambda_e^2}} = \frac{hc \times 2m_e\lambda_e^2}{h^2} = \frac{2m_ec\lambda_e^2}{h}$$.

Since $$\frac{2m_ec}{h}$$ is a constant, $$\lambda_p \propto \lambda_e^2$$. The correct answer is Option A.

We use Einstein's photoelectric equation: $$KE = \frac{hc}{\lambda} - \phi$$, where $$\phi$$ is the work function of the metal.

For the first case, the kinetic energy of the emitted electron is $$E$$ when the incident wavelength is $$\lambda$$: $$E = \frac{hc}{\lambda} - \phi$$. This gives us the work function $$\phi = \frac{hc}{\lambda} - E$$.

For the second case, we need the kinetic energy to be $$2E$$ with a new wavelength $$\lambda'$$: $$2E = \frac{hc}{\lambda'} - \phi$$.

Substituting the expression for $$\phi$$: $$2E = \frac{hc}{\lambda'} - \left(\frac{hc}{\lambda} - E\right) = \frac{hc}{\lambda'} - \frac{hc}{\lambda} + E$$.

Simplifying: $$2E - E = \frac{hc}{\lambda'} - \frac{hc}{\lambda}$$, so $$E = \frac{hc}{\lambda'} - \frac{hc}{\lambda}$$.

Solving for $$\frac{1}{\lambda'}$$: $$\frac{hc}{\lambda'} = E + \frac{hc}{\lambda} = \frac{E\lambda + hc}{\lambda}$$.

Therefore $$\lambda' = \frac{hc\lambda}{E\lambda + hc}$$.

Hence, the correct answer is Option B.

A nucleus of mass $$M$$ at rest splits into two parts of masses $$\dfrac{M'}{3}$$ and $$\dfrac{2M'}{3}$$ (where $$M' < M$$), and we need to find the ratio of their de Broglie wavelengths.

Since the nucleus is initially at rest, conservation of linear momentum implies $$\vec{p}_1 + \vec{p}_2 = 0$$, leading to $$|\vec{p}_1| = |\vec{p}_2| = p$$, which means both fragments have the same magnitude of momentum.

The de Broglie wavelength is given by $$\lambda = \dfrac{h}{p}$$, where $$h$$ is Planck's constant and $$p$$ is the momentum of the particle.

Substituting the common momentum into the expressions for the two fragments yields $$\lambda_1 = \dfrac{h}{p}$$ and $$\lambda_2 = \dfrac{h}{p}$$, which gives $$\dfrac{\lambda_1}{\lambda_2} = \dfrac{h/p}{h/p} = 1$$.

Therefore, $$\lambda_1 : \lambda_2 = 1 : 1$$.

The correct answer is Option C: $$1:1$$.

A parallel beam of light of wavelength $$\lambda = 900 \text{ nm} = 900 \times 10^{-9} \text{ m}$$ and intensity $$I = 100 \text{ W/m}^2$$ is incident on a surface perpendicular to the beam. We need to find the number of photons crossing $$1 \text{ cm}^2$$ area per second.

Since the energy of a single photon is $$E = \frac{hc}{\lambda} = \frac{6.6 \times 10^{-34} \times 3 \times 10^8}{900 \times 10^{-9}}$$, this gives $$E = \frac{19.8 \times 10^{-26}}{9 \times 10^{-7}} = 2.2 \times 10^{-19} \text{ J}$$.

Next, the area is $$1 \text{ cm}^2 = 10^{-4} \text{ m}^2$$ so $$P = I \times A = 100 \times 10^{-4} = 10^{-2} \text{ W} = 0.01 \text{ W}$$.

From this, the number of photons per second is $$n = \frac{P}{E} = \frac{10^{-2}}{2.2 \times 10^{-19}} = \frac{1}{2.2} \times 10^{17}$$ which gives $$n = 0.4545 \times 10^{17} = 4.545 \times 10^{16}$$ and thus $$n \approx 4.5 \times 10^{16}$$.

The correct answer is Option B: $$4.5 \times 10^{16}$$.

The de Broglie wavelength is given by:

$$\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mE}}$$

where $$h$$ is Planck's constant, $$m$$ is the mass of the particle, and $$E$$ is the kinetic energy.

Since all particles have the same energy $$E$$:

$$\lambda \propto \frac{1}{\sqrt{m}}$$

The larger the mass, the smaller the wavelength.

Comparing masses:

- Electron: $$m_e$$ (smallest mass)

- Proton: $$m_p \approx 1836\, m_e$$

- Neutron: $$m_n \approx m_p$$ (slightly greater than proton mass)

- Alpha particle: $$m_\alpha \approx 4m_p$$ (largest mass)

Since $$m_e < m_p < m_n < m_\alpha$$:

$$\lambda_e > \lambda_p > \lambda_n > \lambda_\alpha$$

This can be rewritten as:

$$\lambda_\alpha < \lambda_n < \lambda_p < \lambda_e$$

Hence, the correct answer is Option B.

An electron (mass $$m$$) with initial velocity $$\vec{v} = v_0\hat{i}$$ moves in an electric field $$\vec{E} = -E_0\hat{i}$$. We need to find the de-Broglie wavelength after time $$t$$.

The electron has charge $$-e$$. The force on it is:

$$\vec{F} = (-e)(-E_0\hat{i}) = eE_0\hat{i}$$

The force is in the $$+x$$ direction, same as the initial velocity.

Acceleration: $$a = \frac{eE_0}{m}$$

Velocity at time $$t$$:

$$v(t) = v_0 + at = v_0 + \frac{eE_0}{m}t$$

$$\lambda = \frac{h}{mv(t)} = \frac{h}{m\left(v_0 + \frac{eE_0 t}{m}\right)} = \frac{h}{mv_0 + eE_0 t}$$

$$= \frac{h}{mv_0\left(1 + \frac{eE_0 t}{mv_0}\right)} = \frac{\lambda_0}{1 + \frac{eE_0 t}{mv_0}}$$

where $$\lambda_0 = \frac{h}{mv_0}$$.

The correct answer is Option D: $$\dfrac{\lambda_0}{\left(1 + \dfrac{eE_0 t}{mv_0}\right)}$$.

We need to find the ratio of energies of photons for two transitions in hydrogen: (i) from $$n = 2$$ to $$n = 1$$, and (ii) from $$n = \infty$$ to $$n = 1$$.

The energy of a photon emitted during a transition from level $$n_i$$ to $$n_f$$ is given by $$E = 13.6\left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right)$$ eV.

Case (i): Transition from $$n = 2$$ to $$n = 1$$: $$E_1 = 13.6\left(\frac{1}{1^2} - \frac{1}{2^2}\right) = 13.6\left(1 - \frac{1}{4}\right) = 13.6 \times \frac{3}{4}$$.

Case (ii): Transition from $$n = \infty$$ to $$n = 1$$ (highest permitted level to the ground state): $$E_2 = 13.6\left(\frac{1}{1^2} - \frac{1}{\infty^2}\right) = 13.6\left(1 - 0\right) = 13.6 \times 1$$.

The ratio is $$\frac{E_1}{E_2} = \frac{13.6 \times \frac{3}{4}}{13.6 \times 1} = \frac{3}{4}$$, which gives us $$E_1 : E_2 = 3 : 4$$.

Hence, the correct answer is Option A.

We need to evaluate two statements about the wave nature of electrons.

Statement I: Davisson-Germer experiment establishes the wave nature of electrons.

This is TRUE. The Davisson-Germer experiment (1927) demonstrated electron diffraction by firing electrons at a nickel crystal and observing a diffraction pattern. The measured electron wavelengths matched the de Broglie prediction $$\lambda = \frac{h}{p}$$, thereby experimentally confirming the wave nature of electrons.

Statement II: If electrons have wave nature, they can interfere and show diffraction.

This is TRUE. Interference and diffraction are fundamental wave phenomena. Any entity that has wave nature must exhibit these properties. Since electrons possess wave nature (as established by Statement I and de Broglie's hypothesis), they can indeed show interference and diffraction patterns — which is exactly what is observed experimentally.

Both statements are true, and Statement II is a logical consequence of the wave nature of electrons.

The correct answer is Option A: Both Statement I and Statement II are true.

Using Einstein's photoelectric equation, the maximum kinetic energy of photo-electrons is given by $$K = \frac{hc}{\lambda} - \phi$$, where $$\phi$$ is the work function of the metal. For wavelength $$\lambda_1$$ and $$\lambda_2$$ we have $$K_1 = \frac{hc}{\lambda_1} - \phi$$ and $$K_2 = \frac{hc}{\lambda_2} - \phi$$. Given that $$\lambda_1 = 3\lambda_2$$, substitution into the expression for $$K_1$$ yields $$K_1 = \frac{hc}{3\lambda_2} - \phi$$.

Now compute $$\frac{K_2}{3}$$: $$\frac{K_2}{3} = \frac{1}{3}\left(\frac{hc}{\lambda_2} - \phi\right) = \frac{hc}{3\lambda_2} - \frac{\phi}{3}$$. Comparing $$K_1$$ and $$\frac{K_2}{3}$$ gives $$K_1 - \frac{K_2}{3} = \left(\frac{hc}{3\lambda_2} - \phi\right) - \left(\frac{hc}{3\lambda_2} - \frac{\phi}{3}\right) = -\phi + \frac{\phi}{3} = -\frac{2\phi}{3}$$. Since $$\phi > 0$$, it follows that $$K_1 - \frac{K_2}{3} = -\frac{2\phi}{3} < 0$$, hence $$K_1 < \frac{K_2}{3}$$.

Answer: Option B.

We need to identify what $$x$$ represents in the de-Broglie wavelength equation $$\lambda = \frac{1.227}{\sqrt{x}}$$ nm.

Since an electron is accelerated through a potential difference $$V$$, its kinetic energy is $$K = eV$$, and its de-Broglie wavelength is given by $$\lambda = \frac{h}{\sqrt{2mK}} = \frac{h}{\sqrt{2meV}}$$.

Substituting numerical values, we have $$\lambda = \frac{6.626 \times 10^{-34}}{\sqrt{2 \times 9.1 \times 10^{-31} \times 1.6 \times 10^{-19} \times V}} = \frac{6.626 \times 10^{-34}}{\sqrt{2.912 \times 10^{-49} \times V}} = \frac{6.626 \times 10^{-34}}{5.396 \times 10^{-25} \times \sqrt{V}} = \frac{1.227 \times 10^{-9}}{\sqrt{V}} \text{ m} = \frac{1.227}{\sqrt{V}} \text{ nm}.$$

From the above expression, comparing with $$\lambda = \frac{1.227}{\sqrt{x}}$$ nm shows that $$x = V$$, the accelerating potential in volts.

Therefore, among the given options, the choice $$\sqrt{V}$$ corresponds to $$x = V$$ (since $$\sqrt{x} = \sqrt{V}$$), making Option D the correct answer.

Note: The option $$\sqrt{V}$$ represents that $$x$$ stands for $$V$$ (accelerating potential in volts), and the denominator is $$\sqrt{V}$$.

We are given two streams of photons with energies $$E_1 = 5\phi$$ and $$E_2 = 10\phi$$, where $$\phi$$ is the work function of the metal.

By the photoelectric equation, the maximum kinetic energy of photoelectrons is $$KE = E - \phi$$.

For the first case: $$\frac{1}{2}mv_1^2 = 5\phi - \phi = 4\phi$$.

For the second case: $$\frac{1}{2}mv_2^2 = 10\phi - \phi = 9\phi$$.

Taking the ratio: $$\frac{v_1^2}{v_2^2} = \frac{4\phi}{9\phi} = \frac{4}{9}$$.

So $$\frac{v_1}{v_2} = \frac{2}{3}$$, i.e., the ratio is $$2:3$$.

Hence, the correct answer is Option C.

We need to identify the correct statements about the photo-electric effect.

The square of maximum velocity of photoelectrons varies linearly with frequency of incident light. From Einstein's photoelectric equation:

$$\frac{1}{2}mv_{max}^2 = h\nu - \phi$$

$$v_{max}^2 = \frac{2h\nu}{m} - \frac{2\phi}{m}$$

This is of the form $$v_{max}^2 = a\nu + b$$, which is linear in frequency $$\nu$$. Statement A is CORRECT.

The value of saturation current increases on moving the source of light away from the metal surface. When the source moves away, the intensity of light at the surface decreases. Lower intensity means fewer photons strike the surface per unit time, so fewer photoelectrons are emitted. Hence, the saturation current decreases. Statement B is INCORRECT.

The maximum kinetic energy of photo-electrons decreases on decreasing the power of LED source. The maximum kinetic energy depends on the frequency of incident light, not on the intensity (or power). For an LED, the frequency remains the same regardless of power. Decreasing power only reduces the number of photons, not their individual energy. Statement C is INCORRECT.

The immediate emission of photo-electrons cannot be explained by particle nature of light. The particle (photon) nature of light easily explains instantaneous emission: a single photon transfers all its energy to one electron at once. It is the wave theory that cannot explain this. Statement D is INCORRECT.

The existence of threshold wavelength cannot be explained by wave nature of light. According to wave theory, any frequency of light should be able to eject electrons given enough intensity. The existence of a threshold wavelength (minimum frequency) below which no emission occurs cannot be explained by wave theory. Statement E is CORRECT.

The correct statements are A and E only.

The correct answer is Option B.

A metal is exposed to light of wavelengths $$800 \text{ nm}$$ and $$500 \text{nm}$$. The maximum kinetic energy doubles when the wavelength changes from 800 nm to 500 nm. Given $$hc = 1230 \text{ eV nm}$$.

Applying the photoelectric equation for both wavelengths yields $$KE_1 = \frac{hc}{\lambda_1} - \phi = \frac{1230}{800} - \phi$$ and $$KE_2 = \frac{hc}{\lambda_2} - \phi = \frac{1230}{500} - \phi$$.

Using the condition that $$KE_2 = 2 \times KE_1$$ leads to the equation $$\frac{1230}{500} - \phi = 2\left(\frac{1230}{800} - \phi\right)$$, or in numeric form $$2.46 - \phi = 2(1.5375 - \phi)$$, which simplifies further to $$2.46 - \phi = 3.075 - 2\phi$$.

Solving for $$\phi$$ in $$2.46 - \phi = 3.075 - 2\phi$$ gives $$\phi = 0.615 \text{ eV}$$.

Hence, the correct answer is Option C.

We need to find the ratio of de-Broglie wavelengths of an alpha particle and a Carbon-12 atom, both having the same kinetic energy $$K$$.

The de-Broglie wavelength is given by:

$$\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}}$$

For the alpha particle (mass number 4):

$$\lambda_\alpha = \frac{h}{\sqrt{2 \cdot 4u \cdot K}}$$

For the Carbon-12 atom (mass number 12):

$$\lambda_{C12} = \frac{h}{\sqrt{2 \cdot 12u \cdot K}}$$

Taking the ratio:

$$\frac{\lambda_\alpha}{\lambda_{C12}} = \frac{\sqrt{2 \cdot 12u \cdot K}}{\sqrt{2 \cdot 4u \cdot K}} = \sqrt{\frac{12}{4}} = \sqrt{3}$$

Therefore, $$\lambda_\alpha : \lambda_{C12} = \sqrt{3} : 1$$.

The correct answer is Option A.

Einstein’s photoelectric equation is stated first:
$$K_{\text{max}} = h\nu - \phi \quad -(1)$$ where $$K_{\text{max}}$$ is the maximum kinetic energy of emitted electrons, $$h$$ is Planck’s constant, $$\nu$$ is the frequency of incident radiation, and $$\phi$$ is the work function of the metal.

For photoelectric emission to occur, we require $$K_{\text{max}} \gt 0$$. From (1):
$$h\nu - \phi \gt 0 \quad\Longrightarrow\quad h\nu \gt \phi$$ If $$h\nu \lt \phi$$ then $$K_{\text{max}} \lt 0$$, which is not physically possible. Hence no electrons are emitted and the photoelectric effect does not take place. This shows Assertion A is correct.

Next consider the case $$h\nu = \phi$$. Substituting into (1) gives:
$$K_{\text{max}} = \phi - \phi = 0$$ This means that at the threshold frequency, electrons are just emitted with zero kinetic energy. Therefore Reason R is also correct.

However, Reason R describes the situation exactly at $$h\nu = \phi$$ and does not explain why emission fails when $$h\nu \lt \phi$$. Thus R is not the correct explanation of A.

Final answer: Both A and R are correct but R is not the correct explanation of A (Option B).

We need to evaluate two statements about the hydrogen atom and Bohr's frequency condition.

Statement I: "In hydrogen atom, the frequency of radiation emitted when an electron jumps from lower energy orbit ($$E_1$$) to higher energy orbit ($$E_2$$), is given as $$hf = E_1 - E_2$$."

This is incorrect. When an electron jumps from a lower energy orbit to a higher energy orbit, it absorbs radiation (not emits). Emission occurs only when an electron transitions from a higher energy orbit to a lower energy orbit. Also, since $$E_2 > E_1$$, the expression $$E_1 - E_2$$ would be negative, which is unphysical for a photon energy.

Statement II: "The jumping of electron from higher energy orbit ($$E_2$$) to lower energy orbit ($$E_1$$) is associated with frequency of radiation given as $$f = \frac{(E_2 - E_1)}{h}$$. This condition is Bohr's frequency condition."

This is correct. When an electron transitions from a higher energy level $$E_2$$ to a lower energy level $$E_1$$, a photon is emitted with energy equal to the difference:

$$hf = E_2 - E_1$$

$$f = \frac{E_2 - E_1}{h}$$

This is indeed Bohr's frequency condition.

Therefore, Statement I is incorrect but Statement II is true.

The correct answer is Option D.

The speed of an electron in the $$n^{th}$$ orbit of a hydrogen-like atom is given by:

$$v_n = \frac{Z e^2}{2 \varepsilon_0 n h}$$

This means the speed is proportional to $$\frac{Z}{n}$$, where $$Z$$ is the atomic number and $$n$$ is the orbit number.

For $$He^+$$ (Z = 2) in the 3rd orbit:

$$v_1 \propto \frac{Z_1}{n_1} = \frac{2}{3}$$

For Hydrogen (Z = 1) in the 3rd orbit:

$$v_2 \propto \frac{Z_2}{n_2} = \frac{1}{3}$$

The ratio of speeds is:

$$\frac{v_1}{v_2} = \frac{2/3}{1/3} = \frac{2}{1}$$

Therefore, the ratio of the speed of the electron in the 3rd orbit of $$He^+$$ to the speed in the 3rd orbit of hydrogen is $$2 : 1$$.

The correct answer is Option D.

We need to find the ratio $$\frac{V_p}{V_d}$$ given that the ratio of de Broglie wavelengths of proton and deuteron is $$1:\sqrt{2}$$.

Write the de Broglie wavelength for a charged particle accelerated through potential V.

$$\lambda = \frac{h}{\sqrt{2mqV}}$$

where $$m$$ is the mass, $$q$$ is the charge, and $$V$$ is the accelerating potential.

Write the wavelengths for proton and deuteron.

For proton: $$\lambda_p = \frac{h}{\sqrt{2m_p e V_p}}$$

For deuteron: $$\lambda_d = \frac{h}{\sqrt{2m_d e V_d}}$$

Note: The deuteron has mass $$m_d = 2m_p$$ and charge $$e$$ (same as proton).

Find the ratio and solve.

$$\frac{\lambda_p}{\lambda_d} = \sqrt{\frac{m_d V_d}{m_p V_p}} = \sqrt{\frac{2m_p V_d}{m_p V_p}} = \sqrt{\frac{2V_d}{V_p}}$$

Given $$\frac{\lambda_p}{\lambda_d} = \frac{1}{\sqrt{2}}$$:

$$\frac{1}{\sqrt{2}} = \sqrt{\frac{2V_d}{V_p}}$$

Squaring both sides:

$$\frac{1}{2} = \frac{2V_d}{V_p}$$

$$\frac{V_p}{V_d} = 4$$

The correct answer is Option D: $$4:1$$.

A hydrogen atom in the ground state (n = 1) absorbs 10.2 eV of energy. We need to find the increase in angular momentum.

The energy levels of hydrogen are given by:

$$E_n = -\frac{13.6}{n^2} \text{ eV}$$

Ground state energy: $$E_1 = -13.6$$ eV

After absorbing 10.2 eV:

$$E_f = -13.6 + 10.2 = -3.4 \text{ eV}$$

Since $$E_2 = -\frac{13.6}{4} = -3.4$$ eV, the electron transitions to n = 2.

The angular momentum in the nth orbit is:

$$L_n = \frac{nh}{2\pi}$$

Change in angular momentum:

$$\Delta L = L_2 - L_1 = \frac{2h}{2\pi} - \frac{h}{2\pi} = \frac{h}{2\pi}$$

Substituting $$h = 6.6 \times 10^{-34}$$ Js:

$$\Delta L = \frac{6.6 \times 10^{-34}}{2\pi} = \frac{6.6 \times 10^{-34}}{6.28} = 1.05 \times 10^{-34} \text{ Js}$$

The correct answer is Option B.

When a hydrogen atom transitions from an excited state with principal quantum number $$n$$ to the ground state ($$n_f = 1$$), the wavelength of the emitted photon is given by the Rydberg formula:

$$\dfrac{1}{\lambda} = R\left(\dfrac{1}{1^2} - \dfrac{1}{n^2}\right)$$

where $$R$$ is the Rydberg constant and $$\lambda$$ is the wavelength of the emitted photon.

Expanding:

$$\dfrac{1}{\lambda} = R - \dfrac{R}{n^2}$$

Isolating the term with $$n$$:

$$\dfrac{R}{n^2} = R - \dfrac{1}{\lambda}$$

Taking $$\lambda$$ as the common denominator on the right side:

$$\dfrac{R}{n^2} = \dfrac{\lambda R - 1}{\lambda}$$

Solving for $$n^2$$:

$$n^2 = \dfrac{R \cdot \lambda}{\lambda R - 1}$$

Taking the square root:

$$n = \sqrt{\dfrac{\lambda R}{\lambda R - 1}}$$

Therefore, the correct answer is Option B.

We are given two photon energies $$E_1 = 3.8$$ eV and $$E_2 = 1.4$$ eV, and the work function $$\phi = 0.6$$ eV. We need to find the ratio of maximum speeds of emitted electrons.

By Einstein's photoelectric equation, the maximum kinetic energy is

$$\frac{1}{2}mv_{max}^2 = E_{photon} - \phi$$

For the first photon, substituting gives

$$KE_1 = E_1 - \phi = 3.8 - 0.6 = 3.2 \text{ eV}$$

Similarly, for the second photon,

$$KE_2 = E_2 - \phi = 1.4 - 0.6 = 0.8 \text{ eV}$$

Since kinetic energy is related to speed by $$KE = \frac{1}{2}mv^2$$, the ratio of squared speeds is

$$\frac{v_1^2}{v_2^2} = \frac{KE_1}{KE_2} = \frac{3.2}{0.8} = 4$$

Taking the square root, we find

$$\frac{v_1}{v_2} = \sqrt{4} = 2$$

From this, the ratio of maximum speeds is $$v_1 : v_2 = 2 : 1$$. Therefore, the correct option is Option A: $$2 : 1$$.

We need to find the momentum of an electron revolving in the $$n^{th}$$ orbit.

Bohr’s quantization condition states that the angular momentum of an electron in the $$n^{th}$$ orbit is quantized:

$$L = mvr = \frac{nh}{2\pi}$$

Since the linear momentum is $$p = mv$$, substituting into the angular momentum relation $$mvr = \frac{nh}{2\pi}$$ yields

$$p \cdot r = \frac{nh}{2\pi}$$

and hence

$$p = \frac{nh}{2\pi r}$$

Hence, the correct answer is Option A.

We begin by recalling the energy expressions in Bohr’s model for the $$n$$-th orbit. The kinetic energy is given by:

$$K_n = \frac{13.6}{n^2} \text{ eV}$$

while the potential energy is

$$P_n = -\frac{27.2}{n^2} \text{ eV}$$

and the total energy is

$$E_n = -\frac{13.6}{n^2} \text{ eV}$$

Note that $$K_n$$ is always positive, $$P_n$$ is always negative, and $$E_n = K_n + P_n$$ remains negative.

Since the electron transitions to a higher energy level, $$n$$ increases and consequently $$n^2$$ increases. Substituting this into the kinetic energy expression,

$$K_n = \frac{13.6}{n^2}$$

we see that as $$n$$ increases, the denominator grows and the positive value of $$K_n$$ decreases.

Similarly, for the potential energy

$$P_n = -\frac{27.2}{n^2}$$

the magnitude $$\frac{27.2}{n^2}$$ becomes smaller, so $$P_n$$ becomes less negative and therefore increases (moves closer to zero).

For the total energy

$$E_n = -\frac{13.6}{n^2}$$

the same reduction in magnitude makes $$E_n$$ less negative, meaning that $$E_n$$ also increases (moves closer to zero).

From the above, when the electron moves to a higher level, the kinetic energy decreases, the potential energy increases, and the total energy increases. Therefore, the correct option is Option B.

A beam of monochromatic light excites the electron in $$Li^{++}$$ from the first orbit (n=1) to the third orbit (n=3). For a hydrogen-like atom with atomic number Z, the energy levels are: $$E_n = -\frac{13.6 Z^2}{n^2} \text{ eV}$$. For $$Li^{++}$$, Z = 3, so $$E_1 = -\frac{13.6 \times 9}{1} = -122.4 \text{ eV}$$ and $$E_3 = -\frac{13.6 \times 9}{9} = -13.6 \text{ eV}$$. Therefore, the energy required for the transition is $$\Delta E = E_3 - E_1 = -13.6 - (-122.4) = 108.8 \text{ eV}$$.

Using $$E = \frac{hc}{\lambda}$$, the wavelength of the photon is $$\lambda = \frac{hc}{\Delta E} = \frac{1242 \text{ eV} \cdot \text{nm}}{108.8 \text{ eV}}$$, which yields $$\lambda = 11.4 \text{ nm} = 114 \times 10^{-10} \text{ m}$$.

The value of x is 114.

The energy of a photon emitted during transition from level $$n_1$$ to level $$n_2$$ in hydrogen is:

$$E = 13.6\left(\dfrac{1}{n_2^2} - \dfrac{1}{n_1^2}\right) \text{ eV}$$

Case (i): Transition from $$n = 3$$ to $$n = 2$$:

$$E_1 = 13.6\left(\dfrac{1}{4} - \dfrac{1}{9}\right) = 13.6 \times \dfrac{9 - 4}{36} = 13.6 \times \dfrac{5}{36}$$

Case (ii): Transition from $$n = \infty$$ to $$n = 2$$ (highest permitted level):

$$E_2 = 13.6\left(\dfrac{1}{4} - 0\right) = 13.6 \times \dfrac{1}{4} = 13.6 \times \dfrac{9}{36}$$

The ratio of the energies:

$$\dfrac{E_1}{E_2} = \dfrac{13.6 \times \frac{5}{36}}{13.6 \times \frac{9}{36}} = \dfrac{5}{9}$$

The given ratio is $$\dfrac{x}{x + 4}$$. Comparing:

$$\dfrac{x}{x + 4} = \dfrac{5}{9}$$

$$9x = 5(x + 4) = 5x + 20$$

$$4x = 20$$

$$x = 5$$

Therefore, the value of $$x$$ is $$\boxed{5}$$.

We need to find the wavelength difference between the 3rd transition line of the Paschen series and the 2nd transition line of the Balmer series in terms of $$\lambda$$, the wavelength of the first transition line of the Lyman series.

We start by finding the wavelength of the first transition line of the Lyman series ($$n = 2 \to n = 1$$):

$$\dfrac{1}{\lambda} = R\left(\dfrac{1}{1^2} - \dfrac{1}{2^2}\right) = R\left(1 - \dfrac{1}{4}\right) = \dfrac{3R}{4}$$

This gives $$\lambda = \dfrac{4}{3R}$$.

Next, for the 3rd transition line of the Paschen series ($$n = 6 \to n = 3$$), we have:

$$\dfrac{1}{\lambda_1} = R\left(\dfrac{1}{3^2} - \dfrac{1}{6^2}\right) = R\left(\dfrac{1}{9} - \dfrac{1}{36}\right) = R\left(\dfrac{4 - 1}{36}\right) = \dfrac{R}{12}$$

Therefore, $$\lambda_1 = \dfrac{12}{R}$$.

Now, considering the 2nd transition line of the Balmer series ($$n = 4 \to n = 2$$):

$$\dfrac{1}{\lambda_2} = R\left(\dfrac{1}{2^2} - \dfrac{1}{4^2}\right) = R\left(\dfrac{1}{4} - \dfrac{1}{16}\right) = R\left(\dfrac{4 - 1}{16}\right) = \dfrac{3R}{16}$$

This gives $$\lambda_2 = \dfrac{16}{3R}$$.

Calculating the difference yields:

$$\lambda_1 - \lambda_2 = \dfrac{12}{R} - \dfrac{16}{3R} = \dfrac{36 - 16}{3R} = \dfrac{20}{3R}$$

Since $$\lambda = \dfrac{4}{3R}$$, it follows that $$\dfrac{1}{R} = \dfrac{3\lambda}{4}$$, and hence:

$$\lambda_1 - \lambda_2 = \dfrac{20}{3R} = \dfrac{20}{3} \times \dfrac{3\lambda}{4} = \dfrac{20\lambda}{4} = 5\lambda$$

Therefore, $$a = \boxed{5}$$.

The value of $$a$$ is 5.

We need to find the threshold frequency for the photoelectric effect.

Wavelength of incident light, $$\lambda = 6630$$ Å $$= 6630 \times 10^{-10}$$ m

Stopping potential, $$V_0 = 0.42$$ V

Speed of light, $$c = 3 \times 10^8$$ m/s

Planck's constant, $$h = 6.63 \times 10^{-34}$$ J s

$$h\nu = h\nu_0 + eV_0$$

$$h\nu_0 = h\nu - eV_0$$

$$\nu = \frac{c}{\lambda} = \frac{3 \times 10^8}{6630 \times 10^{-10}} = \frac{3 \times 10^8}{6.63 \times 10^{-7}}$$

$$= 4.525 \times 10^{14}$$ Hz

$$h\nu = 6.63 \times 10^{-34} \times 4.525 \times 10^{14} = 3.0 \times 10^{-19}$$ J

$$eV_0 = 1.6 \times 10^{-19} \times 0.42 = 0.672 \times 10^{-19}$$ J

$$h\nu_0 = 3.0 \times 10^{-19} - 0.672 \times 10^{-19} = 2.328 \times 10^{-19}$$ J

$$\nu_0 = \frac{2.328 \times 10^{-19}}{6.63 \times 10^{-34}} = 3.51 \times 10^{14}$$ Hz

$$= 35.1 \times 10^{13}$$ s$$^{-1}$$

Rounding to the nearest integer: $$x = 35$$.

The value of $$x$$ is 35.

We use the photoelectric equation: $$h\nu = h\nu_0 + \frac{1}{2}mv^2$$, where $$\nu_0$$ is the threshold frequency.

When the frequency is $$2\nu_0$$, the maximum velocity is $$v_1$$, so

$$ h(2\nu_0) = h\nu_0 + \frac{1}{2}mv_1^2 $$

$$ \frac{1}{2}mv_1^2 = h\nu_0 \quad \cdots (1) $$

When the frequency is $$5\nu_0$$, the maximum velocity is $$v_2$$, so

$$ h(5\nu_0) = h\nu_0 + \frac{1}{2}mv_2^2 $$

$$ \frac{1}{2}mv_2^2 = 4h\nu_0 \quad \cdots (2) $$

Dividing equation (2) by equation (1) gives

$$ \frac{v_2^2}{v_1^2} = \frac{4h\nu_0}{h\nu_0} = 4 $$ and taking the square root yields $$ \frac{v_2}{v_1} = 2 $$.

Since $$v_2 = xv_1$$, we get x = 2.

The de Broglie wavelength of a particle with mass $$m$$ moving with speed $$v$$ is given by $$\lambda = \frac{h}{mv}$$, where $$h$$ is Planck's constant.

For the electron: $$\lambda_{electron} = \frac{h}{m_e v}$$.

For the proton: $$\lambda_{proton} = \frac{h}{m_p v}$$.

Since both particles are moving with the same speed $$v$$, the ratio of their de Broglie wavelengths is:

$$\frac{\lambda_{electron}}{\lambda_{proton}} = \frac{h/(m_e v)}{h/(m_p v)} = \frac{m_p}{m_e}$$

Given that $$m_p = 1836 \, m_e$$, we get:

$$\frac{\lambda_{electron}}{\lambda_{proton}} = \frac{1836 \, m_e}{m_e} = 1836$$

Therefore, the ratio of the de Broglie wavelengths is 1836.

We begin with the de-Broglie relation, which connects the wavelength $$\lambda$$ of a particle with its linear momentum $$p$$:

$$\lambda \;=\; \frac{h}{p},$$

where $$h$$ is Planck’s constant.

All the three particles mentioned—electron, proton and $$\alpha$$-particle—are said to possess the same kinetic energy. For a non-relativistic particle, the kinetic energy $$K$$ and the momentum $$p$$ are related through the familiar formula from elementary mechanics:

$$K \;=\; \frac{p^{2}}{2m},$$

where $$m$$ denotes the mass of the particle. We now solve this expression for the momentum $$p$$ because the de-Broglie wavelength depends directly on $$p$$.

Multiplying both sides by $$2m$$ gives

$$2mK \;=\; p^{2}.$$

Taking the positive square root (momentum magnitude) on both sides, we get

$$p \;=\; \sqrt{\,2mK\,}.$$

We now substitute this explicit form of $$p$$ into the de-Broglie relation:

$$\lambda \;=\; \frac{h}{p} \;=\; \frac{h}{\sqrt{\,2mK\,}}.$$

Because all three particles are given to have the same kinetic energy $$K$$, the symbols $$h$$ and $$K$$ are identical for each. Therefore, the only variable determining the relative size of the wavelength is the mass $$m$$ present under the square root in the denominator.

Let us state this dependence clearly:

$$\lambda \,\propto\, \frac{1}{\sqrt{m}}.$$

That is, the de-Broglie wavelength is inversely proportional to the square root of the particle’s mass. A smaller mass yields a larger wavelength, and a larger mass yields a smaller wavelength.

Now we list the approximate masses of the three particles involved:

• Electron mass: $$m_e \approx 9.1 \times 10^{-31}\ \text{kg}.$$
• Proton mass: $$m_p \approx 1.67 \times 10^{-27}\ \text{kg}.$$
• $$\alpha$$-particle mass: $$m_\alpha \approx 4\,m_p \approx 6.68 \times 10^{-27}\ \text{kg}.$$

Plainly,

$$m_e \;<\; m_p \;<\; m_\alpha.$$

Because the wavelength varies as the inverse square root of the mass, the inequality in wavelengths runs in the opposite sense:

$$\lambda_e \;>\; \lambda_p \;>\; \lambda_\alpha.$$

Translating this into the order requested, we have

$$\lambda_e \;>\; \lambda_p \;>\; \lambda_\alpha.$$

Among the given choices, this ordering corresponds precisely to Option C.

Hence, the correct answer is Option C.

We have a single particle whose initial mass is $$4M$$ and it is stated to be at rest. “At rest’’ means that its initial linear momentum is zero, i.e. $$\vec p_{\text{initial}} = 0$$.

This particle suddenly disintegrates into two separate particles. The first fragment has mass $$M$$ and the second fragment has mass $$3M$$. Let their velocities after disintegration be $$\vec v_1$$ (for the particle of mass $$M$$) and $$\vec v_2$$ (for the particle of mass $$3M$$). Because the original body was isolated and at rest, the law of conservation of linear momentum must hold.

Conservation of momentum (statement). For an isolated system, the vector sum of the momenta before and after any internal interaction remains unchanged.

Applying this law, we write

$$M\,\vec v_1 \;+\; 3M\,\vec v_2 \;=\; \vec 0.$$

We now proceed algebraically, isolating one velocity in terms of the other. Dividing every term by the common factor $$M$$ gives

$$\vec v_1 \;+\; 3\,\vec v_2 \;=\; \vec 0.$$

Rearranging, we get

$$\vec v_1 \;=\; -\,3\,\vec v_2.$$

This result tells us that the two velocities are in opposite directions (the minus sign) and that the magnitude of $$\vec v_1$$ is three times the magnitude of $$\vec v_2$$:

$$|\vec v_1| \;=\; 3\,|\vec v_2|.$$

Next, we write the individual momenta after the disintegration:

First particle (mass $$M$$): $$\vec p_1 \;=\; M\,\vec v_1.$$

Second particle (mass $$3M$$): $$\vec p_2 \;=\; 3M\,\vec v_2.$$

We take magnitudes to compare them. Using $$|\vec v_1| = 3\,|\vec v_2|$$, we have

$$|\vec p_1| \;=\; M\,|\vec v_1| \;=\; M\,(3\,|\vec v_2|) \;=\; 3M\,|\vec v_2|.$$

Similarly,

$$|\vec p_2| \;=\; 3M\,|\vec v_2|.$$

From the last two lines we observe

$$|\vec p_1| = |\vec p_2|.$$

Thus, although the particles have different masses and different speeds, the magnitudes of their linear momenta are identical, fulfilling momentum conservation.

Now we invoke the de-Broglie relation.

de-Broglie wavelength formula. The wavelength associated with a particle of momentum $$p$$ is

$$\lambda = \dfrac{h}{p},$$

where $$h$$ is Planck’s constant.

Let $$\lambda_1$$ be the wavelength of the particle of mass $$M$$ and $$\lambda_2$$ be that of the particle of mass $$3M$$. Then

$$\lambda_1 = \dfrac{h}{|\vec p_1|}, \qquad \lambda_2 = \dfrac{h}{|\vec p_2|}.$$

We now form the required ratio:

$$\dfrac{\lambda_1}{\lambda_2} = \dfrac{h/|\vec p_1|}{\,h/|\vec p_2|\,} = \dfrac{|\vec p_2|}{|\vec p_1|}.$$

But we have already shown that $$|\vec p_1| = |\vec p_2|$$, so

$$\dfrac{\lambda_1}{\lambda_2} = 1.$$

This translates to the numerical ratio

$$\lambda_1 : \lambda_2 = 1 : 1.$$

Hence, the correct answer is Option D.

The de Broglie wavelength of a particle of mass $$m$$ and charge $$q$$ accelerated through a potential difference $$V$$ is given by $$\lambda = \frac{h}{\sqrt{2mqV}}$$.

For the proton: $$\lambda_p = \frac{h}{\sqrt{2m_p \cdot e \cdot V}}$$

For the $$\alpha$$ particle (mass $$= 4m_p$$, charge $$= 2e$$): $$\lambda_\alpha = \frac{h}{\sqrt{2 \cdot 4m_p \cdot 2e \cdot V}} = \frac{h}{\sqrt{16m_p eV}}$$

Taking the ratio:

$$\frac{\lambda_p}{\lambda_\alpha} = \frac{\sqrt{16m_p eV}}{\sqrt{2m_p eV}} = \sqrt{\frac{16}{2}} = \sqrt{8} = 2\sqrt{2} \approx 2.83$$

Rounding to one decimal place, this gives approximately $$2.8$$.

The correct answer is Option (3): 2.8.

The de Broglie wavelength of an electron with energy $$E$$ is $$\lambda_e = \frac{h}{p} = \frac{h}{\sqrt{2mE}}$$, since the kinetic energy $$E = \frac{p^2}{2m}$$ gives momentum $$p = \sqrt{2mE}$$.

For a photon with energy $$E$$, we have $$E = \frac{hc}{\lambda_p}$$, so $$\lambda_p = \frac{hc}{E}$$.

The ratio of wavelengths is $$\frac{\lambda_e}{\lambda_p} = \frac{h/\sqrt{2mE}}{hc/E} = \frac{h}{\sqrt{2mE}} \times \frac{E}{hc} = \frac{E}{c\sqrt{2mE}} = \frac{\sqrt{E}}{c\sqrt{2m}} = \frac{1}{c}\sqrt{\frac{E}{2m}}$$.

This can be written as $$\frac{1}{c}\left(\frac{E}{2m}\right)^{1/2}$$, so the correct answer is option 2.

We can calculate the change in stopping potential by applying Einstein’s photoelectric equation to both wavelengths.

1. Einstein’s Photoelectric Equation

The maximum kinetic energy ($$K_{\text{max}}$$) of an emitted electron is the difference between the incident photon energy ($$h\nu$$) and the work function ($$\phi$$) of the metal:

$$K_{\text{max}} = h\nu - \phi$$

Since the stopping potential ($$V_0$$) is the potential required to stop the most energetic electrons, we have:

$$eV_0 = K_{\text{max}}$$

Combining these, and using $$\nu = \frac{c}{\lambda}$$:

$$eV_0 = \frac{hc}{\lambda} - \phi$$

2. Numerical Constants

To simplify the calculation, we use the value of the product $$hc$$ in electron-volt nanometers ($$\text{eV}\cdot\text{nm}$$):

$$\frac{hc}{e} \approx 1240\ \text{eV}\cdot\text{nm}$$

Thus, the energy of a photon ($$E_p$$) in eV is:

$$E_p = \frac{1240}{\lambda \text{ (nm)}}$$

3. Step-by-Step Calculation

For the first wavelength ($$\lambda_1 = 280$$ nm):

• Photon Energy ($$E_1$$): $$\frac{1240}{280} \approx 4.43\ \text{eV}$$
• Max Kinetic Energy ($$K_{\text{max},1}$$): $$4.43\ \text{eV} - 2.5\ \text{eV} = 1.93\ \text{eV}$$
• Stopping Potential ($$V_{0,1}$$): $$1.93\ \text{V}$$
• Photon Energy ($$E_2$$): $$\frac{1240}{400} = 3.10\ \text{eV}$$
• Max Kinetic Energy ($$K_{\text{max},2}$$): $$3.10\ \text{eV} - 2.5\ \text{eV} = 0.60\ \text{eV}$$
• Stopping Potential ($$V_{0,2}$$): $$0.60\ \text{V}$$

For the second wavelength ($$\lambda_2 = 400$$ nm):

4. Change in Stopping Potential ($$\Delta V$$)

The decrease in stopping potential when switching from $$280$$ nm to $$400$$ nm is:

$$\Delta V = V_{0,1} - V_{0,2}$$

$$\Delta V = 1.93\ \text{V} - 0.60\ \text{V} = 1.33\ \text{V}$$

Conclusion:

Rounding to two significant figures, the change is:

$$\boxed{\Delta V \approx 1.3\ \text{V}}$$

This corresponds to Option C.

The de Broglie wavelength of a particle accelerated through a potential difference $$V$$ is $$\lambda = \frac{h}{\sqrt{2mqV}}$$, where $$m$$ is the mass and $$q$$ is the charge of the particle.

For an electron and a proton accelerated through the same potential $$V = 100$$ V, both having the same charge magnitude $$e$$, the ratio of their wavelengths is: $$\frac{\lambda_e}{\lambda_p} = \frac{\sqrt{2m_p eV}}{\sqrt{2m_e eV}} = \sqrt{\frac{m_p}{m_e}}$$.

Using the given masses: $$m_p = 1.00727$$ u and $$m_e = 0.00055$$ u, we get $$\frac{m_p}{m_e} = \frac{1.00727}{0.00055} = 1831.4$$.

Therefore, $$\frac{\lambda_e}{\lambda_p} = \sqrt{1831.4} \approx 42.8 \approx 43$$.

The ratio of the de Broglie wavelengths is approximately $$43 : 1$$.

The de Broglie wavelength of a particle is given by $$\lambda = \frac{h}{mv}$$, where $$h$$ is Planck's constant, $$m$$ is the mass, and $$v$$ is the velocity of the particle.

For the proton, $$\lambda_p = \frac{h}{m_p v_p}$$, and for the alpha particle, $$\lambda_\alpha = \frac{h}{m_\alpha v_\alpha}$$.

Since the de Broglie wavelengths are equal, $$\lambda_p = \lambda_\alpha$$, we have $$\frac{h}{m_p v_p} = \frac{h}{m_\alpha v_\alpha}$$, which gives $$m_p v_p = m_\alpha v_\alpha$$.

An alpha particle has mass $$m_\alpha = 4m_p$$. Substituting, $$m_p v_p = 4m_p v_\alpha$$, so $$\frac{v_p}{v_\alpha} = 4$$.

Therefore, the ratio of their velocities is $$v_p : v_\alpha = 4 : 1$$.

The correct answer is 4 : 1.

Using Einstein's photoelectric equation, the stopping potential $$V_0$$ is related to the wavelength $$\lambda$$ of incident light by: $$eV_0 = \frac{hc}{\lambda} - \phi$$, where $$\phi$$ is the work function of the metal.

For the first case: $$eV_1 = \frac{hc}{\lambda_1} - \phi$$, with $$\lambda_1 = 491$$ nm and $$V_1 = 0.710$$ V.

For the second case: $$eV_2 = \frac{hc}{\lambda_2} - \phi$$, with $$V_2 = 1.43$$ V and $$\lambda_2$$ unknown.

Subtracting the first equation from the second: $$e(V_2 - V_1) = hc\left(\frac{1}{\lambda_2} - \frac{1}{\lambda_1}\right)$$.

Using $$hc = 1240$$ eV·nm: $$1.43 - 0.710 = 1240\left(\frac{1}{\lambda_2} - \frac{1}{491}\right)$$.

$$0.72 = 1240\left(\frac{1}{\lambda_2} - \frac{1}{491}\right)$$

$$\frac{1}{\lambda_2} = \frac{0.72}{1240} + \frac{1}{491} = 0.000581 + 0.002037 = 0.002618 \; \text{nm}^{-1}$$

$$\lambda_2 = \frac{1}{0.002618} \approx 382$$ nm.

Therefore, the new wavelength is approximately 382 nm.

We begin with the de-Broglie relation, which states that the wavelength $$\lambda$$ of any moving particle is connected to its linear momentum $$P$$ through the formula $$\lambda=\dfrac{h}{P}$$, where $$h$$ is Planck’s constant.

The problem tells us that the proton and the electron have the same de-Broglie wavelength. Using the above relation, we write for the proton (subscript $$p$$) and for the electron (subscript $$e$$)

$$\lambda_p=\dfrac{h}{P_p}\quad\text{and}\quad\lambda_e=\dfrac{h}{P_e}$$.

Since the wavelengths are equal, $$\lambda_p=\lambda_e$$, so

$$\dfrac{h}{P_p}=\dfrac{h}{P_e}.$$

Canceling the common factor $$h$$ on both sides, we get

$$P_p=P_e.$$

Hence the momenta of the proton and the electron are identical.

Next, we compare their kinetic energies. For non-relativistic motion, the kinetic energy $$K$$ of a particle of mass $$m$$ and momentum $$P$$ is given by the formula

$$K=\dfrac{P^{2}}{2m}.$$

Applying this to each particle, we have

$$K_p=\dfrac{P_p^{2}}{2m_p}\quad\text{and}\quad K_e=\dfrac{P_e^{2}}{2m_e}.$$

We have just shown $$P_p=P_e$$. Let us denote this common value simply by $$P$$. Substituting,

$$K_p=\dfrac{P^{2}}{2m_p},\qquad K_e=\dfrac{P^{2}}{2m_e}.$$

Now, the rest mass of a proton $$m_p$$ is much larger than the rest mass of an electron $$m_e$$; numerically, $$m_p\approx1836\,m_e$$. Because the denominator in $$K_p$$ is larger, the value of $$K_p$$ is smaller. Thus,

$$K_p<K_e.$$

Combining our two conclusions,

$$K_p<K_e\quad\text{and}\quad P_p=P_e.$$

This matches Option B in the given list.

Hence, the correct answer is Option B.

We begin with the energy-frequency relation for any photon: $$\Delta E = h\,\nu,$$ where $$\Delta E$$ is the change in energy of the electron and $$\nu$$ is the frequency of the emitted radiation.

For a hydrogen-like ion (single electron, nuclear charge $$Z$$) the energy of the level having principal quantum number $$n$$ is given by the Bohr formula

$$E_n = -\,13.6\ \text{eV}\; \dfrac{Z^{2}}{n^{2}}.$$(The factor $$13.6\ \text{eV}$$ is the ground-state energy of hydrogen.)

When an electron jumps from an initial level $$n_i$$ to a lower level $$n_f$$, the energy released is

$$\Delta E = E_{n_f}-E_{n_i} = -\,13.6\,Z^{2}\!\left(\dfrac{1}{n_f^{2}}-\dfrac{1}{n_i^{2}}\right)\text{eV}.$$

Because the negative sign simply denotes that energy is released, we may write the magnitude of the frequency as

$$\nu = \dfrac{\Delta E}{h} = \dfrac{13.6\,Z^{2}}{h}\!\left(\dfrac{1}{n_f^{2}}-\dfrac{1}{n_i^{2}}\right).$$

Now, for the given ion we know the frequency for the transition $$n=3 \rightarrow n=1$$:

$$\nu_{31} = 2.92 \times 10^{15}\ \text{Hz}.$$

Using the above formula with $$n_i = 3$$ and $$n_f = 1$$ we have

$$\nu_{31}= \dfrac{13.6\,Z^{2}}{h}\!\left(\dfrac{1}{1^{2}}-\dfrac{1}{3^{2}}\right) = \dfrac{13.6\,Z^{2}}{h}\!\left(1-\dfrac{1}{9}\right) = \dfrac{13.6\,Z^{2}}{h}\!\left(\dfrac{8}{9}\right).$$

For the unknown $$Z$$ we retain the factor $$\dfrac{13.6\,Z^{2}}{h}$$ as it will cancel out. Next, the required transition is $$n=2 \rightarrow n=1$$. Its frequency will be

$$\nu_{21}= \dfrac{13.6\,Z^{2}}{h}\!\left(\dfrac{1}{1^{2}}-\dfrac{1}{2^{2}}\right) = \dfrac{13.6\,Z^{2}}{h}\!\left(1-\dfrac{1}{4}\right) = \dfrac{13.6\,Z^{2}}{h}\!\left(\dfrac{3}{4}\right).$$

Because the prefactor $$\dfrac{13.6\,Z^{2}}{h}$$ is common to both expressions, we can write the ratio of the two frequencies directly:

$$\dfrac{\nu_{21}}{\nu_{31}} = \dfrac{\dfrac{3}{4}}{\dfrac{8}{9}} = \dfrac{3}{4}\times\dfrac{9}{8} = \dfrac{27}{32}.$$

Hence,

$$\nu_{21} = \nu_{31}\times\dfrac{27}{32} = \left(2.92 \times 10^{15}\ \text{Hz}\right)\times\dfrac{27}{32}.$$

Now we perform the multiplication step by step. First calculate the numerator:

$$2.92 \times 27 = 78.84.$$

Next divide by 32:

$$\dfrac{78.84}{32} = 2.46375.$$

So,

$$\nu_{21} \approx 2.46 \times 10^{15}\ \text{Hz}.$$

Therefore, the emitted radiation for the $$n=2 \rightarrow n=1$$ transition has a frequency of approximately $$2.46 \times 10^{15}$$ Hz.

Hence, the correct answer is Option B.

According to the Bohr model, when an electron transitions from a higher energy level $$n_i$$ to a lower energy level $$n_f$$, the frequency of the emitted photon is given by $$\nu = \frac{E_i - E_f}{h} = \frac{13.6 \text{ eV}}{h}\left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right)$$.

For the transition $$n = 2 \to n = 1$$: $$\nu \propto \left(\frac{1}{1^2} - \frac{1}{2^2}\right) = 1 - \frac{1}{4} = \frac{3}{4} = 0.75$$.

For the transition $$n = 3 \to n = 2$$: $$\nu \propto \left(\frac{1}{4} - \frac{1}{9}\right) = \frac{5}{36} \approx 0.139$$.

For the transition $$n = 4 \to n = 3$$: $$\nu \propto \left(\frac{1}{9} - \frac{1}{16}\right) = \frac{7}{144} \approx 0.049$$.

For the transition $$n = 5 \to n = 4$$: $$\nu \propto \left(\frac{1}{16} - \frac{1}{25}\right) = \frac{9}{400} = 0.0225$$.

Comparing all values, the transition $$n = 2 \to n = 1$$ gives the maximum value of $$\frac{3}{4}$$, and hence the maximum frequency.

The correct answer is the transition from $$n = 2$$ to $$n = 1$$.

First, we recall the Heisenberg uncertainty principle which relates the uncertainty in position $$\Delta x$$ to the uncertainty in momentum $$\Delta p$$ of a particle:

$$\Delta x \,\Delta p \;\ge\; \dfrac{\hbar}{2}.$$

When we want only the order of magnitude or a proportionality, we may write

$$\Delta x \;\propto\; \dfrac{\hbar}{\Delta p}.$$

So, to compare the positional uncertainties of different particles kept under identical conditions, we must compare the corresponding momentum uncertainties $$\Delta p$$.

Now, each gas is ideal and in thermal equilibrium at a common temperature $$T$$. For one translational degree of freedom, the equipartition theorem states that the mean translational kinetic energy is

$$\dfrac{1}{2} m \langle v^{2}\rangle \;=\;\dfrac{1}{2}k_{B}T.$$

Re-arranging this, we obtain the root-mean-square (rms) speed:

$$\langle v^{2}\rangle^{1/2} \;=\; \sqrt{\dfrac{k_{B}T}{m}}.$$

The momentum of a particle is $$p = mv$$, so its rms value is

$$p_{\text{rms}} \;=\; m \,\langle v^{2}\rangle^{1/2} \;=\; m \sqrt{\dfrac{k_{B}T}{m}} \;=\; \sqrt{m\,k_{B}T}.$$

The spread of momentum values in the gas, that is, the uncertainty $$\Delta p$$, is of the same order as this rms momentum, so we write

$$\Delta p\;\propto\;\sqrt{m\,k_{B}T}.$$

Because both gases are at the same temperature, the factor $$k_{B}T$$ is common and cancels out when we form a ratio. Hence,

$$\Delta p \;\propto\;\sqrt{m}.$$

Substituting this proportionality into the uncertainty-principle expression gives

$$\Delta x \;\propto\;\dfrac{1}{\Delta p} \;\propto\;\dfrac{1}{\sqrt{m}}.$$

Therefore, the positional uncertainty of a particle varies inversely as the square root of its mass.

Let $$\Delta x_e$$ denote the uncertainty for an electron of mass $$m_e$$ and $$\Delta x_p$$ for a proton of mass $$m_p$$. Using the above proportionality, we write

$$ \dfrac{\Delta x_e}{\Delta x_p} \;=\; \dfrac{1/\sqrt{m_e}}{1/\sqrt{m_p}} \;=\; \sqrt{\dfrac{m_p}{m_e}}. $$

This shows that the required ratio is proportional to $$\sqrt{m_p/m_e}$$.

Hence, the correct answer is Option A.

In the Bohr model of the hydrogen atom, the electron moves in a circular orbit around the nucleus. The key condition is that the angular momentum of the electron is quantised: $$m v_n r_n = \frac{nh}{2\pi}$$, where $$m$$ is the electron mass, $$v_n$$ is the velocity in the $$n^{th}$$ orbit, $$r_n$$ is the radius of the $$n^{th}$$ orbit, $$n$$ is the principal quantum number, and $$h$$ is Planck's constant.

For the electron in a circular orbit, the centripetal force is provided by the Coulomb attraction: $$\frac{mv_n^2}{r_n} = \frac{e^2}{4\pi\varepsilon_0 r_n^2}$$. This simplifies to $$mv_n^2 = \frac{e^2}{4\pi\varepsilon_0 r_n}$$ $$-(1)$$.

From the quantisation condition, $$r_n = \frac{nh}{2\pi m v_n}$$ $$-(2)$$. Substituting $$(2)$$ into $$(1)$$: $$mv_n^2 = \frac{e^2}{4\pi\varepsilon_0} \cdot \frac{2\pi m v_n}{nh}$$.

Cancelling $$m v_n$$ from both sides (since $$v_n \neq 0$$): $$v_n = \frac{e^2}{4\pi\varepsilon_0} \cdot \frac{2\pi}{nh} = \frac{e^2}{2\varepsilon_0 h} \cdot \frac{1}{n}$$.

Since $$\frac{e^2}{2\varepsilon_0 h}$$ is a collection of constants, we conclude that $$v_n \propto \frac{1}{n}$$. As the quantum number increases, the electron moves more slowly.

The correct answer is option 2: $$v_n \propto \frac{1}{n}$$.

We start with the well-known de-Broglie relation for any material particle, which states that the wavelength associated with a particle is given by

$$\lambda=\frac{h}{p},$$

where $$h$$ is Planck’s constant and $$p$$ is the linear momentum of the particle.

For a non-relativistic particle, the kinetic energy $$E$$ is related to the momentum by the formula

$$E=\frac{p^{2}}{2m},$$

where $$m$$ is the mass of the particle. Solving this equation for $$p$$ gives

$$p=\sqrt{2mE}.$$

Substituting this expression for $$p$$ back into the de-Broglie formula, we obtain

$$\lambda=\frac{h}{\sqrt{2mE}}.$$

From this last equation we notice an important proportionality:

$$\lambda\propto\frac{1}{\sqrt{E}}.$$

Now, let the initial kinetic energy be $$E$$ and the corresponding wavelength be $$\lambda$$. Suppose we supply extra energy so that the new kinetic energy becomes $$E'$$ and the new wavelength becomes $$\lambda'$$. According to the problem statement, the wavelength is reduced to 75 % of its initial value, so we have

$$\lambda' = 0.75\,\lambda=\frac{3}{4}\,\lambda.$$

Using the proportionality $$\lambda\propto\frac{1}{\sqrt{E}}$$ for both the initial and final states, we can write

$$\frac{\lambda'}{\lambda}=\sqrt{\frac{E}{E'}}.$$

Substituting $$\lambda'/\lambda = 3/4$$ into this equation gives

$$\frac{3}{4}=\sqrt{\frac{E}{E'}}.$$

Now we square both sides to remove the square root:

$$\left(\frac{3}{4}\right)^{2} = \frac{E}{E'}.$$

This simplifies to

$$\frac{9}{16} = \frac{E}{E'}.$$

Rearranging to solve for $$E'$$ gives

$$E' = \frac{16}{9}\,E.$$

The extra energy that must be supplied to the particle is the difference between the final and initial energies:

$$\text{Extra energy}=E' - E=\frac{16E}{9}-E=\frac{16E}{9}-\frac{9E}{9}=\frac{7E}{9}.$$

Hence, the correct answer is Option B.

The energy of the photon emitted in the $$3 \to 2$$ transition of hydrogen is $$E_{photon} = 13.6\left(\frac{1}{2^2} - \frac{1}{3^2}\right) = 13.6 \times \frac{5}{36} \approx 1.89$$ eV.

These photons eject photoelectrons from gold. The maximum kinetic energy of photoelectrons is $$K_{max} = E_{photon} - \phi$$, where $$\phi$$ is the work function.

The electrons with maximum kinetic energy follow the largest circular path in the magnetic field. Using $$r = \frac{mv}{eB}$$, the maximum kinetic energy is $$K_{max} = \frac{p^2}{2m} = \frac{(eBr)^2}{2m}$$.

Substituting $$e = 1.6 \times 10^{-19}$$ C, $$B = 5 \times 10^{-4}$$ T, $$r = 7 \times 10^{-3}$$ m, $$m = 9.1 \times 10^{-31}$$ kg:

$$K_{max} = \frac{(1.6 \times 10^{-19} \times 5 \times 10^{-4} \times 7 \times 10^{-3})^2}{2 \times 9.1 \times 10^{-31}} = \frac{(5.6 \times 10^{-25})^2}{1.82 \times 10^{-30}} = \frac{3.136 \times 10^{-49}}{1.82 \times 10^{-30}} \approx 1.723 \times 10^{-19}$$ J $$\approx 1.077$$ eV.

The work function is $$\phi = E_{photon} - K_{max} = 1.89 - 1.077 \approx 0.81 \approx 0.82$$ eV.

The wavelength of a photon emitted during a transition in the hydrogen atom is given by the Rydberg formula: $$\frac{1}{\lambda} = R_H\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$$, where $$R_H = 1.097 \times 10^7$$ m$$^{-1}$$ is the Rydberg constant, $$n_1$$ is the lower energy level, and $$n_2$$ is the higher energy level.

For the transition from $$n = 2$$ to $$n = 1$$ (Lyman series): $$\frac{1}{\lambda} = R_H\left(\frac{1}{1^2} - \frac{1}{2^2}\right) = R_H\left(1 - \frac{1}{4}\right) = \frac{3R_H}{4}$$.

$$\frac{1}{\lambda} = \frac{3 \times 1.097 \times 10^7}{4} = \frac{3.291 \times 10^7}{4} = 8.228 \times 10^6 \; \text{m}^{-1}$$

$$\lambda = \frac{1}{8.228 \times 10^6} = 1.215 \times 10^{-7}$$ m $$= 121.5$$ nm.

This is approximately 121.8 nm, which falls in the ultraviolet region and is the first line of the Lyman series.

By Einstein's photoelectric equation, the maximum kinetic energy of a photoelectron is $$\frac{1}{2}mv^2 = hf - \phi$$, where $$\phi$$ is the work function of the photocathode.

For the first photocathode: $$\frac{1}{2}mv_1^2 = hf_1 - \phi$$. For the second photocathode (identical, so same $$\phi$$): $$\frac{1}{2}mv_2^2 = hf_2 - \phi$$.

Subtracting the second equation from the first: $$\frac{1}{2}m(v_1^2 - v_2^2) = h(f_1 - f_2)$$, which gives $$v_1^2 - v_2^2 = \frac{2h}{m}(f_1 - f_2)$$.

We start from Einstein’s photo-electric equation, which relates the incident photon energy to the maximum kinetic energy of the emitted electrons:

$$h\nu \;=\; \phi \;+\; K_{\text{max}},$$

where $$\phi$$ is the work-function of the material and $$K_{\text{max}} = eV_s$$ with $$V_s$$ being the stopping (cut-off) potential and $$e$$ the electronic charge.

The photon energy can be written in terms of the wavelength by using $$\nu = \dfrac{c}{\lambda}$$. Hence

$$h\nu = h\frac{c}{\lambda}.$$

For numerical work it is convenient to recall the combined constant

$$hc = 1240\;\text{eV·nm}.$$

Step 1: Calculate the work-function using the first lamp.

For the wavelength $$\lambda_1 = 670.5\;\text{nm}$$ the photon energy is

$$E_1 = \frac{1240\;\text{eV·nm}}{670.5\;\text{nm}} = 1.85\;\text{eV}\;(\text{approximately}).$$

The given stopping potential is $$V_{s1}=0.48\;\text{V}$$, so

$$K_{\text{max}\,1}=eV_{s1}=0.48\;\text{eV}.$$

Substituting into Einstein’s equation,

$$\phi = E_1 - K_{\text{max}\,1} = 1.85\;\text{eV} - 0.48\;\text{eV} = 1.37\;\text{eV}.$$

Step 2: Use the same work-function for the second lamp to find the new stopping potential.

The new wavelength is $$\lambda_2 = 474.6\;\text{nm}$$, giving the photon energy

$$E_2 = \frac{1240\;\text{eV·nm}}{474.6\;\text{nm}} = 2.61\;\text{eV}\;(\text{approximately}).$$

The maximum kinetic energy for this light is therefore

$$K_{\text{max}\,2}=E_2-\phi = 2.61\;\text{eV}-1.37\;\text{eV} = 1.24\;\text{eV}.$$

Finally, the stopping potential is obtained from $$K_{\text{max}\,2}=eV_{s2}$$:

$$V_{s2}= \frac{K_{\text{max}\,2}}{e} = \frac{1.24\;\text{eV}}{1\;\text{e}} = 1.24\;\text{V} \approx 1.25\;\text{V}.$$

Hence, the correct answer is Option C.

We begin with an electron and a proton that are initially so far apart that the electrostatic potential energy between them is zero. The electron is supplied with kinetic energy $$3\ \text{eV}$$ and is allowed to approach the proton.

When the proton captures the electron, a hydrogen atom is formed. The question tells us that the atom is left in the “second excited state.” Counting from the ground state $$n=1$$, we have

Ground state : $$n=1$$    First excited state : $$n=2$$    Second excited state : $$n=3$$.

For a hydrogen atom, the stationary‐state energies are given by the well-known Bohr formula

$$E_n = -\dfrac{13.6\ \text{eV}}{n^2}.$$

Substituting $$n = 3$$ we obtain

$$E_3 = -\dfrac{13.6\ \text{eV}}{3^2} = -\dfrac{13.6\ \text{eV}}{9} = -1.511\ \text{eV}\;(\text{to three significant figures}).$$

The total mechanical energy before capture is purely the kinetic energy of the electron, because the potential energy at infinite separation is zero:

$$E_{\text{initial}} = +3.000\ \text{eV}.$$

The total mechanical energy after capture is the hydrogen-atom energy in level $$n=3$$:

$$E_{\text{final}} = -1.511\ \text{eV}.$$

By energy conservation, the excess energy must be emitted as a photon when the atom is formed. Thus,

$$E_{\text{photon}} = E_{\text{initial}} - E_{\text{final}} = 3.000\ \text{eV} - (-1.511\ \text{eV}) = 4.511\ \text{eV}.$$

This photon is next allowed to strike a photosensitive metal whose threshold wavelength is given as $$\lambda_0 = 4000\ \text{\AA}.$$ (Remember $$1\ \text{\AA}=10^{-10}\ \text{m}$$.)

The work function $$\phi$$ of the metal is obtained from the photoelectric relation $$\phi = h c / \lambda_0$$. Using the convenient constant $$h c = 12400\ \text{eV\;{\AA}}$$, we calculate

$$\phi = \dfrac{12400\ \text{eV\;{\AA}}}{4000\ \text{\AA}} = 3.10\ \text{eV}.$$

Now we apply Einstein’s photoelectric equation

$$K_{\text{max}} = h\nu - \phi = E_{\text{photon}} - \phi.$$

Substituting the numbers just obtained, we get

$$K_{\text{max}} = 4.511\ \text{eV} - 3.10\ \text{eV} = 1.411\ \text{eV}.$$

Carrying the result to three significant figures, the maximum kinetic energy of the emitted photoelectron is therefore

$$K_{\text{max}} \approx 1.41\ \text{eV}.$$

Hence, the correct answer is Option B.

We analyse each statement one by one.

Statement I: The momentum of a photon is given by $$p = \frac{h}{\lambda}$$, where $$h$$ is Planck's constant and $$\lambda$$ is the wavelength. If two photons have equal linear momenta, then $$\frac{h}{\lambda_1} = \frac{h}{\lambda_2}$$, which gives $$\lambda_1 = \lambda_2$$. So two photons with equal linear momenta must have equal wavelengths. Statement I is true.

Statement II: The momentum of a photon is $$p = \frac{h}{\lambda}$$ and its energy is $$E = \frac{hc}{\lambda}$$. If the wavelength $$\lambda$$ decreases, then $$p = \frac{h}{\lambda}$$ increases (since $$\lambda$$ is in the denominator). Similarly, $$E = \frac{hc}{\lambda}$$ also increases. So when wavelength decreases, both momentum and energy increase, not decrease. Statement II is false.

Hence, the correct answer is Option C.

In the photoelectric effect we illuminate a metal surface with light and observe that electrons are emitted. According to Einstein’s explanation, each electron absorbs one whole photon, gains the photon’s energy and then uses part of that energy to escape the metal. The precise energy balance is given by the photoelectric equation

$$h\nu = \phi + K_{\text{max}},$$

where $$h$$ is Planck’s constant, $$\nu$$ (Greek letter “nu”) is the frequency of the incident photon, $$\phi$$ is the work function of the metal (the minimum energy required to pull an electron out of the surface) and $$K_{\text{max}}$$ is the maximum kinetic energy of the emitted electrons.

From this formula we immediately see that $$K_{\text{max}}$$ depends only on $$\nu$$ and not on the intensity of the light. If we keep the frequency fixed and shine brighter light, we are merely sending more photons of exactly the same energy toward the metal. Each individual photon still carries the same energy $$h\nu$$, so every emitted electron still receives the same maximum kinetic energy $$K_{\text{max}}$$.

Intensity, in simple words, tells us how much energy comes per unit area per unit time. For monochromatic (single-frequency) light, intensity is proportional to the number of photons arriving per unit area per unit time. Mathematically,

$$I \propto N_{\text{photons}},$$

where $$I$$ is the intensity and $$N_{\text{photons}}$$ is the photon arrival rate. Thus increasing intensity, with frequency held constant, only increases $$N_{\text{photons}}$$; it does not alter $$\nu$$, and therefore it does not alter $$K_{\text{max}}$$.

Now we examine each option in the light of these facts:

A. “Increases the number of photons incident and also increases the K.E. of the ejected electrons.” This is partly correct (the number of photons does increase) but the statement about kinetic energy is wrong. So Option A is rejected.

B. “Increases the frequency of photons incident and increases the K.E. of the ejected electrons.” Intensity has nothing to do with frequency, so this option is wrong on both counts. Option B is rejected.

C. “Increases the number of photons incident and the K.E. of the ejected electrons remains unchanged.” This matches exactly what we deduced from the photoelectric equation and the definition of intensity. Option C is correct.

D. “Increases the frequency of photons incident and the K.E. of the ejected electrons remains unchanged.” Again, intensity does not affect frequency, so Option D is also wrong.

Hence, the correct answer is Option C.

The hydrogen emission spectrum consists of several series corresponding to transitions ending at different energy levels. The Lyman series (transitions to $$n = 1$$) lies in the ultraviolet region. The Paschen series (transitions to $$n = 3$$), Brackett series (transitions to $$n = 4$$), and higher series lie in the infrared region.

The Balmer series corresponds to transitions from higher energy levels ($$n = 3, 4, 5, \ldots$$) to $$n = 2$$. The wavelengths of the Balmer series range from about 365 nm to 656 nm, which falls in the visible region of the electromagnetic spectrum.

Therefore, the series lying in the visible region is the Balmer series.

The resolving power of a microscope depends on the wavelength of the probe particle — a shorter wavelength allows finer details to be resolved. Specifically, resolving power is inversely proportional to the de Broglie wavelength $$\lambda$$.

The de Broglie wavelength of a particle with mass $$m$$ and speed $$v$$ is given by $$\lambda = \frac{h}{mv}$$, where $$h$$ is Planck's constant. For a given speed $$v$$, a heavier particle will have a shorter wavelength.

For the electron: $$\lambda_e = \frac{h}{m_e v}$$. For the proton: $$\lambda_p = \frac{h}{m_p v}$$. Since both travel at the same speed, the ratio of wavelengths is $$\frac{\lambda_e}{\lambda_p} = \frac{m_p}{m_e} \approx 1837$$.

This means the proton's de Broglie wavelength is $$1837$$ times smaller than the electron's. Since resolving power is inversely proportional to wavelength, the resolving power of the proton microscope is $$1837$$ times greater than that of the electron microscope.

Therefore, the resolving power changes by a factor of $$1837$$.

We are asked to find the de-Broglie wavelength of an electron present in an ideal gas whose temperature is $$T = 300 \text{ K}$$.

The average kinetic energy per molecule of an ideal gas in three dimensions is given by the well-known equipartition theorem:

$$\frac{3}{2}k_B T = \frac{1}{2}m v_{\text{rms}}^{\,2}.$$

Here $$k_B$$ is the Boltzmann constant, $$m$$ is the mass of the particle and $$v_{\text{rms}}$$ is the root-mean-square speed. We first solve this equation for $$v_{\text{rms}}$$.

Multiplying both sides by 2, we get

$$3 k_B T = m v_{\text{rms}}^{\,2}.$$

Dividing by $$m$$ and then taking the square root,

$$v_{\text{rms}} = \sqrt{\frac{3 k_B T}{m}}.$$

Now we substitute the given numerical values: $$k_B = 1.38 \times 10^{-23}\ \text{J K}^{-1},\; T = 300\ \text{K},\; m = 9 \times 10^{-31}\ \text{kg}.$$

First the product $$k_B T$$:

$$k_B T = (1.38 \times 10^{-23})(300) = 1.38 \times 300 \times 10^{-23} = 414 \times 10^{-23} = 4.14 \times 10^{-21}\,\text{J}.$$

Then the factor $$3 k_B T$$ appearing in the numerator:

$$3 k_B T = 3 \times 4.14 \times 10^{-21} = 12.42 \times 10^{-21} = 1.242 \times 10^{-20}\,\text{J}.$$

Now we divide by the mass of the electron:

$$\frac{3 k_B T}{m} = \frac{1.242 \times 10^{-20}}{9 \times 10^{-31}} = \frac{1.242}{9} \times 10^{-20+31} = 0.138 \times 10^{11} = 1.38 \times 10^{10}\ \text{(m}^2\text{/s}^2).$$

Taking the square root gives the rms speed:

$$v_{\text{rms}} = \sqrt{1.38 \times 10^{10}} = \sqrt{1.38}\,\sqrt{10^{10}} \approx 1.175 \times 10^{5}\ \text{m s}^{-1}.$$

With the speed known, we use the de-Broglie relation

$$\lambda = \frac{h}{m v}$$

and for thermal motion we put $$v = v_{\text{rms}}$$. Stating the values, $$h = 6.6 \times 10^{-34}\ \text{J s}$$, $$m = 9 \times 10^{-31}\ \text{kg}$$, and $$v_{\text{rms}} \approx 1.175 \times 10^{5}\ \text{m s}^{-1}$$.

The denominator $$m v_{\text{rms}}$$ is

$$m v_{\text{rms}} = (9 \times 10^{-31})(1.175 \times 10^{5}) = 9 \times 1.175 \times 10^{-31+5} = 10.575 \times 10^{-26} = 1.0575 \times 10^{-25}\ \text{kg m s}^{-1}.$$

Hence the wavelength is

$$\lambda = \frac{6.6 \times 10^{-34}} {1.0575 \times 10^{-25}} = \frac{6.6}{1.0575} \times 10^{-34+25} \approx 6.24 \times 10^{-9}\ \text{m}.$$

Since $$1\ \text{nm} = 10^{-9}\ \text{m}$$, we convert the answer:

$$\lambda \approx 6.24\ \text{nm}.$$

The option that is nearest to this value is 6.26 nm.

Hence, the correct answer is Option D.

The energy of the $$n^{th}$$ level of a hydrogen-like atom with atomic number $$Z$$ is $$E_n = -\frac{13.6\, Z^2}{n^2}$$ eV. The ground state energy of hydrogen ($$Z = 1, n = 1$$) is $$E_1^H = -13.6$$ eV.

For singly ionized carbon, we treat it as the hydrogen-like ion C$$^{5+}$$ with $$Z = 6$$. Its energy levels are $$E_n = -\frac{13.6 \times 36}{n^2}$$ eV. Setting this equal to $$-13.6$$ eV gives $$\frac{13.6 \times 36}{n^2} = 13.6$$, so $$n^2 = 36$$, hence $$n = 6$$.

The correct answer is option 2: $$n = 6$$.

An electron with de-Broglie wavelength $$\lambda$$ has momentum $$p = \frac{h}{\lambda}$$.

The kinetic energy of the electron is $$K = \frac{p^2}{2m} = \frac{h^2}{2m\lambda^2}$$.

When this electron strikes the target and comes to a complete stop, all of its kinetic energy can be converted into a single X-ray photon. This gives the maximum energy photon, i.e., the cut-off (minimum wavelength) X-ray.

Setting the photon energy equal to the kinetic energy of the electron:

$$\frac{hc}{\lambda_0} = \frac{h^2}{2m\lambda^2}$$

Solving for the cut-off wavelength $$\lambda_0$$:

$$\lambda_0 = \frac{hc \cdot 2m\lambda^2}{h^2} = \frac{2mc\lambda^2}{h}$$

We begin with the de-Broglie relation, which connects wavelength $$\lambda$$ with linear momentum $$p$$:

$$\lambda \;=\;\frac{h}{p}$$

Here $$h$$ is Planck’s constant. Because the electron and the photon are said to have the same de-Broglie wavelength, we can write

$$\lambda_{\text{electron}} \;=\;\lambda_{\text{photon}} \;=\;\lambda$$

Substituting the expression for each momentum in the de-Broglie formula, we obtain

For the electron (non-relativistic): $$p_{\text{e}} = m_{\text{e}}\,v$$, so $$\lambda = \dfrac{h}{m_{\text{e}}\,v}$$

For the photon: $$p_{\gamma} = \dfrac{h}{\lambda}$$ by direct inversion of the same formula.

Since the wavelengths are equal, we must have equal momenta:

$$p_{\text{e}} = p_{\gamma}$$

Therefore

$$m_{\text{e}}\,v = \frac{h}{\lambda}$$

but the right-hand side is simply $$p_{\gamma}$$, so we can keep the concise equality

$$p_{\gamma} = m_{\text{e}}\,v$$

Now we write the kinetic energy of each particle.

For the electron (classical, non-relativistic):

$$K_{\text{e}} \;=\;\frac{1}{2}\,m_{\text{e}}\,v^{2}$$

For the photon, the entire energy is kinetic and is given by the relativistic expression $$E = pc$$:

$$K_{\gamma} \;=\;p_{\gamma}\,c$$

We now form the required ratio:

$$\frac{K_{\text{e}}}{K_{\gamma}} \;=\;\frac{\dfrac{1}{2}\,m_{\text{e}}\,v^{2}}{p_{\gamma}\,c}$$

Substituting $$p_{\gamma} = m_{\text{e}}\,v$$ from the equality of momenta, we get

$$\frac{K_{\text{e}}}{K_{\gamma}} \;=\;\frac{\dfrac{1}{2}\,m_{\text{e}}\,v^{2}}{m_{\text{e}}\,v\,c}$$

Now the mass $$m_{\text{e}}$$ cancels out, leaving

$$\frac{K_{\text{e}}}{K_{\gamma}} \;=\;\frac{1}{2}\,\frac{v}{c}$$

Simplifying, we obtain

$$\frac{K_{\text{e}}}{K_{\gamma}} \;=\;\frac{v}{2c}$$

Hence, the correct answer is Option C.

$$\text{Transition A: From the continuum } (n_i = \infty) \text{ to } n_f = 1.$$

$$\text{Transitions ending at } n_f = 1 \text{ belong to the Lyman series. A transition from } \infty$$ represents the shortest wavelength / maximum energy, known as the series limit

$$\therefore \text{Transition A represents the series limit of Lyman series.}$$

$$\text{Transition B: From } n_i = 5 \text{ to } n_f = 2.$$

$$\text{Transitions ending at } n_f = 2 \text{ belong to the Balmer series.}$$

$$\text{The members are ordered by descending energy steps: } n_i=3\,(1^{\text{st}}\text{ member}), \ n_i=4\,(2^{\text{nd}}\text{ member}), \ n_i=5\,(3^{\text{rd}}\text{ member}).$$

$$\therefore \text{Transition B represents the third member of Balmer series.}$$

$$\text{Transition C: From } n_i = 5 \text{ to } n_f = 3.$$

$$\text{Transitions ending at } n_f = 3 \text{ belong to the Paschen series.}$$

$$\text{The members are ordered by descending energy steps: } n_i=4\,(1^{\text{st}}\text{ member}), \ n_i=5\,(2^{\text{nd}}\text{ member}).$$

$$\therefore \text{Transition C represents the second member of Paschen series.}$$

In the photoelectric effect, when light of frequency $$\nu$$ falls on a metal surface with work function $$\phi$$, Einstein's photoelectric equation gives the maximum kinetic energy of the emitted photoelectrons as $$K_{\text{max}} = h\nu - \phi$$, where $$h$$ is Planck's constant.

The stopping potential $$V_0$$ is defined by $$eV_0 = K_{\text{max}} = h\nu - \phi$$, which gives $$V_0 = \frac{h\nu - \phi}{e}$$.

From this relation, the stopping potential depends on the frequency $$\nu$$ of the incident radiation and the work function of the metal. It does not depend on the intensity, amplitude, or phase of the incident light. Increasing the intensity increases the number of photoelectrons (i.e., the photocurrent) but not the maximum kinetic energy or the stopping potential.

Therefore, the stopping potential depends on the frequency of the incident electromagnetic radiation.

We recall the Einstein photoelectric equation. It states that for incident light of wavelength $$\lambda$$ on a metal of threshold wavelength $$\lambda_0$$, the stopping potential $$V_s$$ is related by

$$eV_s \;=\; h\nu \;-\; \phi,$$

where $$\nu=\dfrac{c}{\lambda}$$ is the frequency, $$\phi = h\nu_0 = \dfrac{hc}{\lambda_0}$$ is the work-function, and $$e$$ is the electronic charge. Substituting $$\nu=\dfrac{c}{\lambda}$$ and simplifying gives the working form

$$eV_s \;=\; hc\left(\dfrac{1}{\lambda}-\dfrac{1}{\lambda_0}\right).$$

For convenience we define the constant $$k=\dfrac{hc}{e}$$ so that the equation becomes

$$V_s \;=\; k\left(\dfrac{1}{\lambda}-\dfrac{1}{\lambda_0}\right). \quad -(1)$$

We now apply this relation to the two experimental situations described in the problem.

First illumination: wavelength $$\lambda$$, stopping potential $$V_1 = 4.8\text{ V}$$. Substituting in (1) we get

$$4.8 = k\left(\dfrac{1}{\lambda}-\dfrac{1}{\lambda_0}\right). \quad -(2)$$

Second illumination: wavelength $$2\lambda$$, stopping potential $$V_2 = 1.6\text{ V}$$. Using (1) again, we obtain

$$1.6 = k\left(\dfrac{1}{2\lambda}-\dfrac{1}{\lambda_0}\right). \quad -(3)$$

Our goal is to eliminate $$k$$ and find $$\lambda_0$$ in terms of $$\lambda$$. We start by expressing $$\dfrac{1}{\lambda_0}$$ from equation (2):

$$\dfrac{1}{\lambda_0} = \dfrac{1}{\lambda} - \dfrac{4.8}{k}. \quad -(4)$$

Now we substitute this expression for $$\dfrac{1}{\lambda_0}$$ into equation (3):

$$1.6 = k\left[\dfrac{1}{2\lambda} - \left(\dfrac{1}{\lambda} - \dfrac{4.8}{k}\right)\right].$$

Expanding the bracket gives

$$1.6 = k\left[\dfrac{1}{2\lambda} - \dfrac{1}{\lambda} + \dfrac{4.8}{k}\right].$$

The first two terms inside the bracket combine as

$$\dfrac{1}{2\lambda} - \dfrac{1}{\lambda} = -\dfrac{1}{2\lambda},$$

so we obtain

$$1.6 = k\left[-\dfrac{1}{2\lambda}\right] + 4.8.$$

Rearranging to isolate the term in $$\lambda$$, we move $$4.8$$ to the left side:

$$1.6 - 4.8 = -\,k\,\dfrac{1}{2\lambda}.$$

This simplifies to

$$( -3.2 ) = -\,k\,\dfrac{1}{2\lambda},$$

and therefore

$$\dfrac{1}{2\lambda} = \dfrac{3.2}{k}.$$

Multiplying both sides by 2 gives the reciprocal of $$\lambda$$:

$$\dfrac{1}{\lambda} = \dfrac{6.4}{k}. \quad -(5)$$

With $$\dfrac{1}{\lambda}$$ known, we return to equation (4) to find $$\dfrac{1}{\lambda_0}$$:

$$\dfrac{1}{\lambda_0} = \dfrac{6.4}{k} - \dfrac{4.8}{k} = \dfrac{1.6}{k}.$$

Taking reciprocals to get $$\lambda_0$$ itself, we find

$$\lambda_0 = \dfrac{k}{1.6}.$$

Equation (5) similarly gives

$$\lambda = \dfrac{k}{6.4}.$$

Dividing $$\lambda_0$$ by $$\lambda$$ produces

$$\dfrac{\lambda_0}{\lambda} = \dfrac{k/1.6}{k/6.4} = \dfrac{6.4}{1.6} = 4.$$

Thus

$$\lambda_0 = 4\lambda.$$

The threshold wavelength of the metal is therefore four times the given wavelength. Among the options provided, this corresponds to option B: $$4\lambda$$.

Hence, the correct answer is Option B.

We have a photon of wavelength $$\lambda = 500 \text{ nm}=500 \times 10^{-9}\ \text{m}.$$ The energy of one photon is obtained from the formula $$E=\dfrac{hc}{\lambda}.$$ Given $$hc = 20 \times 10^{-26}\ \text{J m},$$ we substitute:

$$E=\dfrac{20 \times 10^{-26}}{500 \times 10^{-9}} =\dfrac{20}{5}\times 10^{-26+7} =4 \times 10^{-19}\ \text{J}.$$

The work function of the metal is $$\phi =1.25\ \text{eV}.$$ Since $$1\ \text{eV}=1.6 \times 10^{-19}\ \text{J},$$

$$\phi =1.25 \times 1.6 \times 10^{-19} =2.0 \times 10^{-19}\ \text{J}.$$

Einstein’s photo-electric equation states $$\text{K.E.}_{\max}=E-\phi.$$ So,

$$\text{K.E.}_{\max}=4 \times 10^{-19}-2 \times 10^{-19} =2 \times 10^{-19}\ \text{J}.$$

Kinetic energy and speed are related by $$\text{K.E.}=\dfrac{1}{2}mv^{2}.$$ Therefore

$$v=\sqrt{\dfrac{2\,\text{K.E.}_{\max}}{m}} =\sqrt{\dfrac{2 \times 2 \times 10^{-19}} {9 \times 10^{-31}}} =\sqrt{\dfrac{4}{9}\times 10^{12}} =\sqrt{0.444\;4 \times 10^{12}} \approx 6.66 \times 10^{5}\ \text{m s}^{-1}.$$

An electron moving perpendicular to a magnetic field describes a circle of radius $$r$$ given by $$r=\dfrac{mv}{eB},$$ where $$e=1.6 \times 10^{-19}\ \text{C}.$$ Rearranging,

$$B=\dfrac{mv}{er}.$$

The radius is $$r = 30 \text{ cm}=0.30\ \text{m}.$$ Substituting all values,

$$B=\dfrac{9 \times 10^{-31}\; \text{kg}\; \times 6.66 \times 10^{5}\; \text{m s}^{-1}} {1.6 \times 10^{-19}\; \text{C}\; \times 0.30\ \text{m}}.$$

$$B=\dfrac{59.94 \times 10^{-26}} {0.48 \times 10^{-19}} =\dfrac{59.94}{0.48}\times 10^{-7} \approx 125 \times 10^{-7}\ \text{T}.$$

Hence, the correct answer is Option 125.

For a material particle, the de-Broglie relation gives the wavelength as $$\lambda = \frac{h}{p}$$, where $$h$$ is Planck’s constant and $$p$$ is the linear momentum.

Case 1: Particle of mass $$m = 9.1 \times 10^{-31}\,\text{kg}$$ moving with speed $$v = 10^{6}\,\text{m s}^{-1}$$
  Momentum of particle: $$p_{\text{particle}} = m v = (9.1 \times 10^{-31})(10^{6}) = 9.1 \times 10^{-25}\,\text{kg m s}^{-1} \;-(1)$$
  de-Broglie wavelength of particle: $$\lambda_{\text{particle}} = \frac{h}{p_{\text{particle}}}$$

Case 2: Photon with given momentum $$p_{\text{photon}} = 10^{-27}\,\text{kg m s}^{-1}$$
  Wavelength of photon: $$\lambda_{\text{photon}} = \frac{h}{p_{\text{photon}}}$$

Required ratio of wavelengths:
$$\frac{\lambda_{\text{photon}}}{\lambda_{\text{particle}}} = \frac{\frac{h}{p_{\text{photon}}}}{\frac{h}{p_{\text{particle}}}} = \frac{p_{\text{particle}}}{p_{\text{photon}}}$$ Substituting from $$(1)$$:
$$\frac{\lambda_{\text{photon}}}{\lambda_{\text{particle}}} = \frac{9.1 \times 10^{-25}}{10^{-27}} = 9.1 \times 10^{2} = 910$$

Therefore, the wavelength of the photon is 910 times the wavelength of the particle.

Using the photoelectric equation $$eV_s = \frac{hc}{\lambda} - \phi$$, where $$\phi = \frac{hc}{\lambda_0}$$ is the work function and $$\lambda_0$$ is the threshold wavelength.

For radiation of wavelength $$\lambda$$ with stopping potential $$3V_0$$:

$$e(3V_0) = \frac{hc}{\lambda} - \frac{hc}{\lambda_0} \quad \cdots (1)$$

For radiation of wavelength $$2\lambda$$ with stopping potential $$V_0$$:

$$e(V_0) = \frac{hc}{2\lambda} - \frac{hc}{\lambda_0} \quad \cdots (2)$$

Subtracting equation (2) from equation (1):

$$2eV_0 = \frac{hc}{\lambda} - \frac{hc}{2\lambda} = \frac{hc}{2\lambda}$$

So $$eV_0 = \frac{hc}{4\lambda}$$.

Substituting back into equation (2):

$$\frac{hc}{4\lambda} = \frac{hc}{2\lambda} - \frac{hc}{\lambda_0}$$

$$\frac{hc}{\lambda_0} = \frac{hc}{2\lambda} - \frac{hc}{4\lambda} = \frac{hc}{4\lambda}$$

$$\lambda_0 = 4\lambda$$

The threshold wavelength is $$4\lambda$$, so the answer is $$4$$.

In the Balmer series, the spectral lines correspond to transitions from higher energy levels to $$n = 2$$. The wavelength for a transition from level $$n$$ to level 2 is given by $$\frac{1}{\lambda} = R\left(\frac{1}{2^2} - \frac{1}{n^2}\right)$$, where $$R$$ is the Rydberg constant.

For the first spectral line ($$\lambda_1$$), the transition is from $$n = 3$$ to $$n = 2$$: $$\frac{1}{\lambda_1} = R\left(\frac{1}{4} - \frac{1}{9}\right) = R \cdot \frac{5}{36}$$.

For the third spectral line ($$\lambda_3$$), the transition is from $$n = 5$$ to $$n = 2$$: $$\frac{1}{\lambda_3} = R\left(\frac{1}{4} - \frac{1}{25}\right) = R \cdot \frac{21}{100}$$.

Taking the ratio, $$\frac{\lambda_1}{\lambda_3} = \frac{21/100}{5/36} = \frac{21 \times 36}{100 \times 5} = \frac{756}{500} = 1.512$$.

This can be written as approximately $$15.12 \times 10^{-1}$$, so $$x \approx 15$$ to the nearest integer.

Let the work function of the metal be $$\phi$$. The two streams of photons have energies $$E_1 = 2\phi$$ and $$E_2 = 10\phi$$.

By the photoelectric equation, the maximum kinetic energy of the emitted photoelectrons is $$KE_{max} = E - \phi$$.

For the first stream: $$\frac{1}{2}mv_1^2 = 2\phi - \phi = \phi$$.

For the second stream: $$\frac{1}{2}mv_2^2 = 10\phi - \phi = 9\phi$$.

Taking the ratio: $$\frac{v_1^2}{v_2^2} = \frac{\phi}{9\phi} = \frac{1}{9}$$.

Therefore, $$\frac{v_1}{v_2} = \frac{1}{3}$$, which means $$v_1 : v_2 = 1 : 3$$.

Comparing with the given ratio $$x : 3$$, we get $$x = 1$$.

In a hydrogen atom each spectral line arises when the electron falls from an upper level with principal quantum number $$n_i$$ to a lower level $$n_f$$, where $$n_i > n_f$$. Because the energy difference $$\Delta E$$ and, therefore, the wavelength $$\lambda$$ of the emitted photon depend only on these two numbers, every different ordered pair $$(n_i,n_f)$$ gives a unique wavelength.

The atoms in the sample are excited only up to the level $$n_{\text{max}} = 6$$. Hence the possible values of $$n_i$$ are $$6,5,4,3,2$$ (the ground state $$n=1$$ cannot serve as an upper level). For each chosen $$n_i$$, the lower level $$n_f$$ can be any integer from $$1$$ up to $$(n_i-1)$$.

We count the number of different pairs $$(n_i,n_f)$$ by listing them:

For $$n_i = 6: \; n_f = 5,4,3,2,1 \;\Rightarrow\; 5 \text{ lines}$$

For $$n_i = 5: \; n_f = 4,3,2,1 \;\Rightarrow\; 4 \text{ lines}$$

For $$n_i = 4: \; n_f = 3,2,1 \;\Rightarrow\; 3 \text{ lines}$$

For $$n_i = 3: \; n_f = 2,1 \;\Rightarrow\; 2 \text{ lines}$$

For $$n_i = 2: \; n_f = 1 \;\Rightarrow\; 1 \text{ line}$$

Adding all these possibilities, we obtain

$$X = 5 + 4 + 3 + 2 + 1 = 15.$$

This result is often written with the combination formula

$$X = \frac{n_{\text{max}}\,(n_{\text{max}}-1)}{2} = \frac{6 \times 5}{2} = 15,$$

because the total number of distinct pairs among $$n_{\text{max}}$$ levels is the number of ways to choose two different levels, and that is $$\frac{n(n-1)}{2}.$$

So, the answer is $$15$$.

For a circular orbit in the potential $$U(r) = U_0 r^4$$, the force is $$F = -\frac{dU}{dr} = -4U_0 r^3$$. The magnitude of this centripetal force equals $$\frac{mv^2}{r}$$, so $$4U_0 r^3 = \frac{mv^2}{r}$$, giving $$mv^2 = 4U_0 r^4$$ and thus $$v = 2r^2\sqrt{\frac{U_0}{m}}$$.

Applying Bohr's quantization condition $$mvr = n\hbar$$, we get $$m \cdot 2r^2\sqrt{\frac{U_0}{m}} \cdot r = n\hbar$$, which simplifies to $$2r^3\sqrt{mU_0} = n\hbar$$.

Solving for $$r$$: $$r^3 = \frac{n\hbar}{2\sqrt{mU_0}}$$, so $$r \propto n^{1/3}$$. Since $$r_n \propto n^{1/\alpha}$$, we identify $$\alpha = 3$$.

We analyse each statement separately.

Statement I says that in a diatomic molecule, the rotational energy at a given temperature obeys Maxwell's distribution. This is true because, according to the equipartition theorem and statistical mechanics, each quadratic degree of freedom follows the Maxwell-Boltzmann energy distribution. The rotational kinetic energy of a diatomic molecule, being a sum of quadratic terms in angular velocities, obeys this distribution.

Statement II says that the rotational energy at a given temperature equals the translational kinetic energy for each molecule. A diatomic molecule at moderate temperatures has 2 rotational degrees of freedom (rotation about two axes perpendicular to the bond axis) and 3 translational degrees of freedom. By the equipartition theorem, the average energy per degree of freedom is $$\frac{1}{2}k_BT$$. Therefore, the average translational kinetic energy is $$\frac{3}{2}k_BT$$ while the average rotational energy is $$\frac{2}{2}k_BT = k_BT$$. Since $$\frac{3}{2}k_BT \neq k_BT$$, Statement II is false.

Hence, Statement I is true but Statement II is false.

We begin with the photo-electric equation, which states that for each photon

$$h\nu \;=\; \dfrac{1}{2}m_ev_{\text{max}}^{\,2} \;+\; \phi,$$

where $$h$$ is Planck’s constant, $$\nu$$ is the frequency of the incident radiation, $$m_e$$ is the mass of the electron, $$v_{\text{max}}$$ is the maximum speed of the emitted electrons, and $$\phi$$ is the work function of the metal.

First we convert the given wavelength into metres. One angstrom equals $$10^{-10}\,\text{m}$$, so

$$\lambda \;=\; 6561~\text{Å} \;=\; 6561 \times 10^{-10}\,\text{m} \;=\; 6.561 \times 10^{-7}\,\text{m}.$$

Using the relation $$c = \lambda \nu,$$ we have

$$\nu = \dfrac{c}{\lambda} = \dfrac{3.00 \times 10^{8}\,\text{m s}^{-1}}{6.561 \times 10^{-7}\,\text{m}} = 4.573 \times 10^{14}\,\text{s}^{-1}.$$

Hence the photon energy is

$$h\nu = \left(6.626 \times 10^{-34}\,\text{J s}\right)\!\left(4.573 \times 10^{14}\,\text{s}^{-1}\right) = 3.032 \times 10^{-19}\,\text{J}.$$

To express this in electron-volts we divide by $$1\,\text{eV}=1.602\times10^{-19}\,\text{J}$$:

$$h\nu = \dfrac{3.032 \times 10^{-19}\,\text{J}}{1.602 \times 10^{-19}\,\text{J/eV}} = 1.895\,\text{eV}.$$

Next we use the information about the electron’s motion in the magnetic field. An electron entering a uniform field $$B$$ perpendicular to its velocity describes a circle whose radius $$r$$ satisfies the equality of magnetic and centripetal forces:

$$e v_{\text{max}} B = \dfrac{m_e v_{\text{max}}^{\,2}}{r}.$$

Solving for $$v_{\text{max}}$$ gives

$$v_{\text{max}} = \dfrac{e B r}{m_e}.$$

Substituting the numerical values $$e = 1.602 \times 10^{-19}\,\text{C},\; B = 3.0 \times 10^{-4}\,\text{T},\; r = 10\,\text{mm} = 0.010\,\text{m},\; m_e = 9.11 \times 10^{-31}\,\text{kg},$$ we get

$$v_{\text{max}} = \dfrac{(1.602 \times 10^{-19})(3.0 \times 10^{-4})(0.010)}{9.11 \times 10^{-31}} = \dfrac{4.806 \times 10^{-25}}{9.11 \times 10^{-31}} = 5.27 \times 10^{5}\,\text{m s}^{-1}.$$

Now we find the kinetic energy of these fastest electrons:

$$\dfrac{1}{2}m_e v_{\text{max}}^{\,2} = \dfrac{1}{2}(9.11 \times 10^{-31}\,\text{kg})\bigl(5.27 \times 10^{5}\,\text{m s}^{-1}\bigr)^{2}.$$

Calculating the square first, $$\bigl(5.27 \times 10^{5}\bigr)^{2} = 2.777 \times 10^{11},$$ so

$$\dfrac{1}{2}m_e v_{\text{max}}^{\,2} = 0.5 \times 9.11 \times 10^{-31} \times 2.777 \times 10^{11} = 1.265 \times 10^{-19}\,\text{J}.$$

Converting this to electron-volts:

$$\dfrac{1.265 \times 10^{-19}\,\text{J}}{1.602 \times 10^{-19}\,\text{J/eV}} = 0.790\,\text{eV}.$$

Finally, using the photo-electric equation $$h\nu = \dfrac{1}{2}m_ev_{\text{max}}^{\,2} + \phi,$$ we obtain the work function

$$\phi = h\nu - \dfrac{1}{2}m_ev_{\text{max}}^{\,2} = 1.895\,\text{eV} - 0.790\,\text{eV} = 1.105\,\text{eV}.$$

Rounding to the provided precision, the work function is about $$1.1\,\text{eV}.$$

Hence, the correct answer is Option C.

We begin with the de-Broglie relation, which states that for any particle the wavelength associated with its motion is

$$\lambda = \frac{h}{p},$$

where $$h$$ is Planck’s constant and $$p$$ is the linear momentum of the particle.

Momentum and kinetic energy are related by the classical (non-relativistic) formula

$$K = \frac{p^{2}}{2m}\;,$$

where $$K$$ is the kinetic energy and $$m$$ is the mass of the particle. Solving this expression for $$p$$ we have

$$p = \sqrt{2mK}\;.$$

Substituting this expression for $$p$$ back into the de-Broglie formula gives

$$\lambda = \frac{h}{\sqrt{2mK}}\;.$$

In the given problem the electron $$(e^{-})$$, the proton $$(p)$$ and the doubly ionised helium ion $$(He^{++})$$ are all said to possess the same kinetic energy $$K$$. Because $$h$$ and $$K$$ are common to all three particles, the above formula shows that their wavelengths obey the proportionality

$$\lambda \propto \frac{1}{\sqrt{m}}\;.$$

Thus, smaller mass $$\Longrightarrow$$ larger de-Broglie wavelength, while larger mass $$\Longrightarrow$$ smaller wavelength.

Now we list their approximate masses:

$$m_e \approx 9.11 \times 10^{-31}\;{\rm kg},$$

$$m_p \approx 1.67 \times 10^{-27}\;{\rm kg},$$

$$m_{He^{++}} \approx 4m_p \approx 6.68 \times 10^{-27}\;{\rm kg}.$$

Clearly

$$m_e \;<\; m_p \;<\; m_{He^{++}}\;.$$

Using the proportionality $$\lambda \propto 1/\sqrt{m}$$, we invert this mass order to obtain the wavelength order:

$$\lambda_e \;>\; \lambda_p \;>\; \lambda_{He^{++}}\;.$$

This matches the option

$$\lambda_e > \lambda_p > \lambda_{He^{++}}.$$

Hence, the correct answer is Option C.

According to Einstein’s photo-electric equation, the maximum kinetic energy of the emitted photo-electrons is given by

$$K_{\text{max}} = h\nu - \phi,$$

where $$h$$ is Planck’s constant, $$\nu$$ is the frequency of the incident radiation and $$\phi$$ is the work function of the metal.

Since frequency $$\nu$$ and wavelength $$\lambda$$ are related by $$\nu = \dfrac{c}{\lambda}$$ (with $$c$$ the speed of light), we may write

$$K_{\text{max}} = \dfrac{hc}{\lambda} - \phi.$$

Let us call the first wavelength $$\lambda_1 = 500 \text{ nm}$$ and the corresponding maximum kinetic energy $$K_1$$. For the second case $$\lambda_2 = 200 \text{ nm}$$ and the kinetic energy is given to be three times larger, so $$K_2 = 3K_1$$.

Writing Einstein’s equation for each case, we have

$$K_1 = \dfrac{hc}{\lambda_1} - \phi \quad\quad (1)$$ $$K_2 = \dfrac{hc}{\lambda_2} - \phi. \quad\quad (2)$$

Because $$K_2 = 3K_1$$, substituting this into equation (2) gives

$$3K_1 = \dfrac{hc}{\lambda_2} - \phi. \quad\quad (3)$$

Now we subtract equation (1) from equation (3) to eliminate $$\phi$$:

$$3K_1 - K_1 = \left(\dfrac{hc}{\lambda_2} - \phi\right) - \left(\dfrac{hc}{\lambda_1} - \phi\right).$$

This simplifies to

$$2K_1 = hc\left(\dfrac{1}{\lambda_2} - \dfrac{1}{\lambda_1}\right).$$

So the single-energy $$K_1$$ is

$$K_1 = \dfrac{hc}{2}\left(\dfrac{1}{\lambda_2} - \dfrac{1}{\lambda_1}\right).$$

We next substitute numerical values. A convenient constant for photo-electric problems is

$$hc = 1240 \ \text{eV·nm}.$$

The reciprocal wavelengths are

$$\dfrac{1}{\lambda_2} = \dfrac{1}{200 \text{ nm}} = 0.005 \ \text{nm}^{-1},$$ $$\dfrac{1}{\lambda_1} = \dfrac{1}{500 \text{ nm}} = 0.002 \ \text{nm}^{-1}.$$

Thus

$$\dfrac{1}{\lambda_2} - \dfrac{1}{\lambda_1} = 0.005 - 0.002 = 0.003 \ \text{nm}^{-1}.$$

Now we find $$K_1$$:

$$K_1 = \dfrac{1240 \ \text{eV·nm}}{2} \times 0.003 \ \text{nm}^{-1}.$$

$$K_1 = 620 \ \text{eV·nm} \times 0.003 \ \text{nm}^{-1} = 1.86 \ \text{eV}.$$

With $$K_1$$ known, we return to equation (1) to extract the work function:

$$\phi = \dfrac{hc}{\lambda_1} - K_1.$$

The photon energy for $$\lambda_1 = 500 \text{ nm}$$ is

$$\dfrac{hc}{\lambda_1} = \dfrac{1240 \ \text{eV·nm}}{500 \ \text{nm}} = 2.48 \ \text{eV}.$$

Therefore

$$\phi = 2.48 \ \text{eV} - 1.86 \ \text{eV} = 0.62 \ \text{eV}.$$

This value is closest to $$0.61 \ \text{eV}$$ among the given options.

Hence, the correct answer is Option D.

According to de-Broglie, the wavelength associated with any moving particle is given by the formula $$\lambda=\dfrac{h}{mv},$$ where $$h$$ is Planck’s constant, $$m$$ the mass of the particle and $$v$$ its speed.

For the unknown particle we write $$\lambda_p=\dfrac{h}{m_p v_p},$$ and for the electron $$\lambda_e=\dfrac{h}{m_e v_e}.$$

Dividing the two expressions, we obtain

$$\dfrac{\lambda_p}{\lambda_e}=\dfrac{h}{m_p v_p}\,\Big/\,\dfrac{h}{m_e v_e}=\dfrac{m_e v_e}{m_p v_p}.$$

The question states that the particle is moving five times as fast as the electron, so $$v_p = 5\,v_e.$$ Substituting this into the ratio gives

$$\dfrac{\lambda_p}{\lambda_e}=\dfrac{m_e v_e}{m_p\,(5v_e)}=\dfrac{m_e}{5m_p}.$$

We are told that $$\dfrac{\lambda_p}{\lambda_e}=1.878\times10^{-4}.$$ Equating the two expressions and solving for $$m_p$$, we have

$$1.878\times10^{-4}=\dfrac{m_e}{5m_p}.$$

Now cross-multiplying,

$$5m_p\,(1.878\times10^{-4})=m_e.$$

So,

$$m_p=\dfrac{m_e}{5(1.878\times10^{-4})}.$$

The rest mass of the electron is $$m_e = 9.1\times10^{-31}\,\text{kg}.$$ Substituting this numerical value, we get

$$m_p=\dfrac{9.1\times10^{-31}}{5\times1.878\times10^{-4}} =\dfrac{9.1\times10^{-31}}{9.39\times10^{-4}}.$$

Dividing term by term,

$$m_p=\left(\dfrac{9.1}{9.39}\right)\times10^{-31+4} \approx0.969\times10^{-27}\,\text{kg} =9.69\times10^{-28}\,\text{kg}.$$

This numerical value matches most closely with option D, $$9.7\times10^{-28}\,\text{kg}.$$

Hence, the correct answer is Option D.

We begin with an electron that is initially at rest. Because it is placed in a constant, uniform electric field of magnitude $$E$$, it experiences a constant force. According to Newton’s second law,

$$F = m a.$$ But the electric force on a charge $$e$$ in an electric field $$E$$ is

$$F = eE.$$

Equating the two forces gives

$$m a = eE \quad\Longrightarrow\quad a = \frac{eE}{m}.$$

Since the acceleration $$a$$ is constant and the electron starts from rest, the kinematic relation for velocity after time $$t$$ is

$$v = a t.$$

Substituting the value of $$a$$ we just found, we have

$$v = \left(\frac{eE}{m}\right) t.$$

Linear momentum is defined as $$p = m v$$. Substituting for $$v$$, we obtain the momentum of the electron at time $$t$$:

$$p = m \left(\frac{eE}{m}\right) t = eE t.$$

The de-Broglie wavelength formula is stated as

$$\lambda = \frac{h}{p},$$

where $$h$$ is Planck’s constant. Inserting the expression for $$p$$ gives

$$\lambda = \frac{h}{eE t}.$$

We are asked for the rate of change of this wavelength with respect to time. Therefore, we differentiate $$\lambda$$ with respect to $$t$$. Writing $$\lambda(t) = \dfrac{h}{eE}\, t^{-1}$$, we differentiate using the power rule $$\dfrac{d}{dt}\bigl(t^{-1}\bigr) = -t^{-2}$$:

$$\frac{d\lambda}{dt} = \frac{h}{eE}\left(-t^{-2}\right) = -\frac{h}{eE t^{2}}.$$

The negative sign simply indicates that the wavelength decreases as the time increases, which is expected because the momentum is increasing. All algebraic steps are complete and no relativistic corrections were necessary (as instructed).

Hence, the correct answer is Option D.

$$\text{Einstein's photoelectric equation: } eV_s = \frac{hc}{\lambda} - \phi \implies V_s = \left(\frac{hc}{e}\right)\frac{1}{\lambda} - \frac{\phi}{e}$$

$$\text{Comparing with the straight line equation } y = mx + c\text{, the slope is } m = \frac{hc}{e} \text{ and the x-intercept is } \frac{1}{\lambda_0} = \frac{\phi}{hc}.$$

$$\text{Since } h, c, e$$, and the work function $$\phi$$ are constants for a given photosensitive material, the graph is entirely independent of intensity.

Increasing the intensity only increases the number of photons and the saturation current, leaving the stopping potential and the graph unaltered

We have particle A of mass $$m_A=\dfrac{m}{2}$$ moving with initial velocity $$u_1=v_0$$ and particle B of mass $$m_B=\dfrac{m}{3}$$ initially at rest, so $$u_2=0$$.

For a perfectly elastic, head-on collision in one dimension, the standard result for the final velocities (derived from simultaneous conservation of linear momentum and kinetic energy) is first written:

$$v_1=\dfrac{m_1-m_2}{m_1+m_2}\,u_1+\dfrac{2m_2}{m_1+m_2}\,u_2,$$

$$v_2=\dfrac{2m_1}{m_1+m_2}\,u_1+\dfrac{m_2-m_1}{m_1+m_2}\,u_2.$$

Substituting $$u_2=0$$ (because B is at rest) and taking $$m_1=m_A=\dfrac{m}{2},\;m_2=m_B=\dfrac{m}{3}$$ we obtain for particle A:

$$v_1=\dfrac{\dfrac{m}{2}-\dfrac{m}{3}}{\dfrac{m}{2}+\dfrac{m}{3}}\,v_0.$$

Now we simplify the numerator and denominator separately:

Numerator: $$\dfrac{m}{2}-\dfrac{m}{3}=\dfrac{3m-2m}{6}=\dfrac{m}{6}.$$

Denominator: $$\dfrac{m}{2}+\dfrac{m}{3}=\dfrac{3m+2m}{6}=\dfrac{5m}{6}.$$

Hence

$$v_1=\dfrac{\dfrac{m}{6}}{\dfrac{5m}{6}}\,v_0=\dfrac{1}{5}\,v_0.$$

So, after collision, particle A moves along the x-axis with speed $$v_1=\dfrac{v_0}{5}.$$

We now turn to the de-Broglie relation which connects momentum and wavelength:

$$\lambda=\dfrac{h}{p},\quad\text{where }p=mv.$$

Initial momentum of particle A:

$$p_0=m_A\,v_0=\left(\dfrac{m}{2}\right)v_0=\dfrac{m v_0}{2}.$$

Therefore its initial de-Broglie wavelength is

$$\lambda_0=\dfrac{h}{p_0}=\dfrac{h}{\dfrac{m v_0}{2}}=\dfrac{2h}{m v_0}.$$

After the collision the momentum of particle A becomes

$$p_1=m_A\,v_1=\left(\dfrac{m}{2}\right)\left(\dfrac{v_0}{5}\right)=\dfrac{m v_0}{10}.$$

So its final wavelength is

$$\lambda_1=\dfrac{h}{p_1}=\dfrac{h}{\dfrac{m v_0}{10}}=\dfrac{10h}{m v_0}.$$

We now express $$\lambda_1$$ in terms of $$\lambda_0$$. Dividing the two expressions gives

$$\dfrac{\lambda_1}{\lambda_0}=\dfrac{\dfrac{10h}{m v_0}}{\dfrac{2h}{m v_0}}=\dfrac{10}{2}=5.$$

Thus $$\lambda_1=5\lambda_0.$$

The change in wavelength is

$$\Delta\lambda=\lambda_1-\lambda_0=5\lambda_0-\lambda_0=4\lambda_0.$$

Hence, the correct answer is Option D.

We start by recalling the Einstein photo-electric equation. For any metal surface, the incident photon energy equals the sum of the work function of the metal and the maximum kinetic energy of the emitted photo-electrons. Symbolically, for metal A we write

$$E_A = \phi_A + T_A,$$

where $$E_A$$ is the photon energy, $$\phi_A$$ is the work function of metal A, and $$T_A$$ is the maximum kinetic energy of the photoelectrons from A.

Exactly the same relation applies to metal B:

$$E_B = \phi_B + T_B,$$

with analogous meanings for each symbol.

Next, we need the connection between the kinetic energy of an electron and its de-Broglie wavelength. The de-Broglie relation is

$$\lambda = \frac{h}{p},$$

where $$h$$ is Planck’s constant and $$p$$ is the linear momentum of the electron. For a non-relativistic electron, momentum satisfies

$$p = \sqrt{2mK},$$

where $$m$$ is the electron mass and $$K$$ is its kinetic energy. Substituting this into the de-Broglie formula gives the useful working form

$$\lambda = \frac{h}{\sqrt{2mK}}.$$

This shows that $$\lambda$$ is inversely proportional to the square root of the kinetic energy: $$\lambda \propto \dfrac{1}{\sqrt{K}}.$$

Therefore, for two different electrons, the ratio of their wavelengths equals the square root of the inverse ratio of their kinetic energies:

$$\frac{\lambda_B}{\lambda_A} = \sqrt{\frac{T_A}{T_B}}.$$

Now we substitute the numerical condition given in the problem. We are told that the photo-electrons from metal B have wavelength

$$\lambda_B = 2\lambda_A.$$

Putting this into the proportionality relation gives

$$2 = \sqrt{\frac{T_A}{T_B}}.$$

Squaring both sides yields the direct algebraic link between $$T_A$$ and $$T_B$$:

$$4 = \frac{T_A}{T_B}, \qquad\text{so}\qquad T_A = 4T_B.$$

The statement of the problem also connects the two kinetic energies by

$$T_B = T_A - 1.5 \text{ eV}.$$

We now have two equations:

$$T_A = 4T_B,$$

$$T_B = T_A - 1.5.$$

Substituting the first into the second, we replace $$T_A$$ by $$4T_B$$:

$$T_B = 4T_B - 1.5.$$

Bringing like terms together, we find

$$1.5 = 3T_B \quad\Longrightarrow\quad T_B = \frac{1.5}{3} = 0.5 \text{ eV}.$$

Now we immediately compute $$T_A$$ from $$T_A = 4T_B$$:

$$T_A = 4 \times 0.5 = 2.0 \text{ eV}.$$

Armed with these kinetic energies, we return to the Einstein equation to obtain the work function of metal B. The photon energy striking B is $$E_B = 4.50 \text{ eV}$$, so

$$\phi_B = E_B - T_B = 4.50 \text{ eV} - 0.50 \text{ eV} = 4.00 \text{ eV}.$$

Thus the work function of metal B is $$4 \text{ eV}$$.

Hence, the correct answer is Option A.

We start with the de Broglie relation for a non-relativistic particle

$$\lambda=\frac{h}{p},$$

where $$h$$ is Planck’s constant and $$p$$ is the linear momentum of the particle.

The kinetic energy $$E$$ of a non-relativistic particle of mass $$m$$ and momentum $$p$$ is given by the usual formula

$$E=\frac{p^{2}}{2m}.$$

From the de Broglie relation we can express the momentum in terms of the wavelength:

$$p=\frac{h}{\lambda}.$$

Substituting this value of $$p$$ into the kinetic-energy expression, we obtain

$$E=\frac{1}{2m}\left(\frac{h}{\lambda}\right)^{2} =\frac{h^{2}}{2m\lambda^{2}}.$$

Now extra energy $$\Delta E$$ is supplied so that the new de Broglie wavelength becomes

$$\lambda'=\frac{\lambda}{2}.$$

Using the same de Broglie relation, the new momentum $$p'$$ is

$$p'=\frac{h}{\lambda'}=\frac{h}{\lambda/2}=2\,\frac{h}{\lambda}=2p.$$

The new kinetic energy $$E'$$ corresponding to this momentum is

$$E'=\frac{{p'}^{2}}{2m} =\frac{(2p)^{2}}{2m} =\frac{4p^{2}}{2m} =2\left(\frac{p^{2}}{m}\right) =4\left(\frac{p^{2}}{2m}\right) =4E.$$

The additional energy supplied is therefore

$$\Delta E = E' - E = 4E - E = 3E.$$

Hence, the correct answer is Option C.

We are told that an electron of mass $$m$$ starts with velocity $$\vec{v}_0 = v_0\hat{i}+v_0\hat{j}$$ and is placed in a uniform electric field $$\vec{E} = -E_0\hat{k}$$. Remember that the charge of an electron is $$q = -e$$ (negative).

First we write the force on a charged particle in an electric field. The formula is $$\vec{F}=q\vec{E}.$$ Substituting the values of $$q$$ and $$\vec{E}$$ we get

$$\vec{F}=(-e)(-E_0\hat{k}) = +eE_0\hat{k}.$$

This force acts only in the $$\hat{k}$$ (or $$z$$) direction. Therefore:

• In the $$x$$ direction $$F_x = 0 \Rightarrow p_x = \text{constant}.$$
• In the $$y$$ direction $$F_y = 0 \Rightarrow p_y = \text{constant}.$$
• In the $$z$$ direction there is a constant force $$eE_0$$.

The acceleration produced in the $$z$$ direction is

$$a_z = \frac{F_z}{m}= \frac{eE_0}{m}.$$

Initially the electron has no $$z$$-velocity, so $$v_{z0}=0$$. After a time $$t$$ the velocity component in the $$z$$ direction is

$$v_z = v_{z0}+a_zt = 0+\frac{eE_0}{m}t=\frac{eE_0 t}{m}.$$

Now we list the three momentum components at time $$t$$:

$$\begin{aligned} p_x &= m v_0,\\[4pt] p_y &= m v_0,\\[4pt] p_z &= m v_z = m\left(\frac{eE_0 t}{m}\right)=eE_0t. \end{aligned}$$

The magnitude of the total momentum vector $$\vec{p}$$ is therefore

$$\begin{aligned} p &= \sqrt{p_x^2+p_y^2+p_z^2}\\[4pt] &= \sqrt{(mv_0)^2+(mv_0)^2+(eE_0t)^2}\\[4pt] &= \sqrt{2m^2v_0^2+e^2E_0^2t^2}. \end{aligned}$$

We can factor out $$m^2v_0^2$$ from the square root:

$$p = \sqrt{m^2v_0^2}\,\sqrt{2+\frac{e^2E_0^2t^2}{m^2v_0^2}} = mv_0\sqrt{2+\frac{e^2E_0^2t^2}{m^2v_0^2}}.$$

The de-Broglie relation connects wavelength and momentum by

$$\lambda = \frac{h}{p},$$

where $$h$$ is Planck’s constant.

Initially ($$t=0$$) the magnitude of the momentum is

$$p_0 = \sqrt{(mv_0)^2+(mv_0)^2}=mv_0\sqrt{2},$$

so the initial de-Broglie wavelength is

$$\lambda_0 = \frac{h}{p_0} =\frac{h}{mv_0\sqrt{2}}.$$

At a later time $$t$$ the wavelength is

$$\lambda(t)=\frac{h}{p} =\frac{h}{mv_0\sqrt{2+\dfrac{e^2E_0^2t^2}{m^2v_0^2}}}.$$

We now express $$h/(mv_0\sqrt{2})$$ as $$\lambda_0$$, obtained above. Substituting, we have

$$\lambda(t) = \frac{\lambda_0\sqrt{2}} {\sqrt{2+\dfrac{e^2E_0^2t^2}{m^2v_0^2}}}.$$

To simplify further we factor $$\sqrt{2}$$ out of the denominator:

$$\begin{aligned} \sqrt{2+\frac{e^2E_0^2t^2}{m^2v_0^2}} &= \sqrt{2}\sqrt{1+\frac{e^2E_0^2t^2}{2m^2v_0^2}},\\[6pt] \lambda(t) &= \frac{\lambda_0\sqrt{2}} {\sqrt{2}\sqrt{1+\dfrac{e^2E_0^2t^2}{2m^2v_0^2}}} =\frac{\lambda_0} {\sqrt{1+\dfrac{e^2E_0^2t^2}{2m^2v_0^2}}}. \end{aligned}$$

This expression matches option C in the given list:

$$\lambda(t)=\frac{\lambda_0}{\sqrt{1+\dfrac{e^2E_0^2t^2}{2m^2v_0^2}}}.$$

Hence, the correct answer is Option C.

We begin by recalling the two basic relations that connect energy with wavelength for an electron and for a photon.

For a material particle such as an electron, the de-Broglie relation is stated as $$\lambda_e=\frac{h}{p},$$ where $$\lambda_e$$ is the de-Broglie wavelength, $$h$$ is Planck’s constant and $$p$$ is the linear momentum of the electron.

Because the energy given in the problem is only a few electron-volts, it is extremely small when compared with the electron’s rest-mass energy (about $$0.511\ \text{MeV}$$). Hence the electron moves non-relativistically and its kinetic energy $$E$$ is related to its momentum by the classical formula $$E=\frac{p^{2}}{2m},$$ where $$m$$ is the electron’s mass.

Solving this formula for the momentum, we get $$p=\sqrt{2mE}.$$ Substituting this value of $$p$$ into the de-Broglie relation, we obtain $$\lambda_e=\frac{h}{\sqrt{2mE}}.$$

Now, for a photon the connection between its energy and its wavelength is given by the Planck-Einstein relation $$E=\frac{hc}{\lambda_{\gamma}},$$ where $$\lambda_{\gamma}$$ is the photon’s wavelength and $$c$$ is the speed of light in vacuum.

Rearranging this for $$\lambda_{\gamma}$$ gives $$\lambda_{\gamma}=\frac{hc}{E}.$$

We are required to find the ratio of the electron’s de-Broglie wavelength to the photon’s wavelength, namely $$\frac{\lambda_e}{\lambda_{\gamma}} =\frac{\dfrac{h}{\sqrt{2mE}}}{\dfrac{hc}{E}}.$$

Dividing the two fractions, the factor $$h$$ cancels out and we get $$\frac{\lambda_e}{\lambda_{\gamma}} =\frac{E}{c\sqrt{2mE}}.$$

The numerator contains $$E$$ while the denominator contains $$\sqrt{E}$$, so we simplify by writing $$E=\sqrt{E}\times\sqrt{E}$$:

$$ \frac{\lambda_e}{\lambda_{\gamma}} =\frac{\sqrt{E}\,\sqrt{E}}{c\sqrt{2mE}} =\frac{\sqrt{E}}{c\sqrt{2m}}.$$ Combining the square-root terms in the denominator, we finally reach $$ \frac{\lambda_e}{\lambda_{\gamma}} =\frac{1}{c}\sqrt{\frac{E}{2m}}.$$ This matches option C.

Hence, the correct answer is Option C.

Identifying the threshold frequency from the x-intercept:

$$\text{At point } B(5.5, 0) \implies V_{\text{stop}} = 0 \implies f_0 = 5.5 \times 10^{14}\ \text{Hz}$$

Work function equation: $$\phi = h f_0$$

$$\phi = 6.62 \times 10^{-34}\ \text{J}\cdot\text{s} \times 5.5 \times 10^{14}\ \text{Hz} = 3.641 \times 10^{-19}\ \text{J}$$

$$\phi = \frac{3.641 \times 10^{-19}}{1.6 \times 10^{-19}}\ \text{eV} \approx 2.275\ \text{eV}$$

For a hydrogen atom, the energy of the $$n^{\text{th}}$$ level is given by $$E_n = -\frac{13.6}{n^2}$$ eV.

When an electron transitions from the $$(n+1)^{\text{th}}$$ level to the $$n^{\text{th}}$$ level, the energy of the emitted photon is:

$$\Delta E = E_{n+1} - E_n = -\frac{13.6}{(n+1)^2} + \frac{13.6}{n^2} = 13.6\left(\frac{1}{n^2} - \frac{1}{(n+1)^2}\right)$$

Simplifying the expression inside the parentheses:

$$\frac{1}{n^2} - \frac{1}{(n+1)^2} = \frac{(n+1)^2 - n^2}{n^2(n+1)^2} = \frac{2n + 1}{n^2(n+1)^2}$$

So $$\Delta E = 13.6 \times \frac{2n + 1}{n^2(n+1)^2}$$.

Since the frequency of the emitted radiation is $$\nu = \frac{\Delta E}{h}$$, we have $$\nu \propto \frac{2n+1}{n^2(n+1)^2}$$.

Now we apply the approximation for $$n \gg 1$$:

When $$n$$ is very large, $$2n + 1 \approx 2n$$ and $$(n+1)^2 \approx n^2$$.

Therefore: $$\nu \propto \frac{2n}{n^2 \cdot n^2} = \frac{2n}{n^4} = \frac{2}{n^3}$$.

So the frequency of radiation emitted is proportional to $$\frac{1}{n^3}$$, which is Option C.

We have a hydrogen-like ion whose ionisation energy in the ground state is given as $$9$$ Rydbergs. For any hydrogen-like species the ionisation energy from the ground state is given by the formula

$$E_{\text{ion}} = Z^{2}R_{H},$$

where $$Z$$ is the atomic number (nuclear charge) and $$R_{H}$$ is the Rydberg constant expressed in energy units (1 Rydberg = 13.6 eV). Substituting the given value,

$$Z^{2}R_{H}=9\,R_{H}\;\;\Longrightarrow\;\;Z^{2}=9\;\;\Longrightarrow\;\;Z=3.$$

So the ion is $$\text{Li}^{2+}$$ (a 3-proton nucleus with one electron).

The energy of the $$n^{\text{th}}$$ Bohr orbit for any hydrogen-like ion is given by

$$E_{n}=-\dfrac{Z^{2}R_{H}}{n^{2}}.$$

The phrase “second excited state” means $$n=3$$ (ground state $$n=1$$, first excited $$n=2$$, second excited $$n=3$$). When the electron falls from this state to the ground state, the energy released is

$$\Delta E = E_{1}-E_{3}= \left(-\dfrac{Z^{2}R_{H}}{1^{2}}\right)-\left(-\dfrac{Z^{2}R_{H}}{3^{2}}\right)=Z^{2}R_{H}\left(1-\dfrac{1}{9}\right)=Z^{2}R_{H}\left(\dfrac{8}{9}\right).$$

With $$Z^{2}=9$$ we get

$$\Delta E = 9R_{H}\left(\dfrac{8}{9}\right)=8R_{H}.$$

Because $$1\,\text{Rydberg}=13.6\,\text{eV},$$ the photon energy in electron-volts is

$$\Delta E = 8 \times 13.6\,\text{eV}=108.8\,\text{eV}.$$

To convert this energy into wavelength we use the relation

$$E=\dfrac{hc}{\lambda}\quad\Longrightarrow\quad \lambda=\dfrac{hc}{E}.$$

In convenient units, $$hc=1240\,\text{eV nm}.$$ Substituting the numerical values,

$$\lambda=\dfrac{1240\,\text{eV nm}}{108.8\,\text{eV}}\approx11.4\,\text{nm}.$$

Hence, the correct answer is Option B.

We are told that for the ground state (principal quantum number $$n=1$$) the electron in hydrogen takes a time $$T_1 = 1.6 \times 10^{-16}\,\text{s}$$ to complete one revolution.

First we change this into frequency, because frequency is the reciprocal of time period.

By definition, $$\text{frequency} = \dfrac{1}{\text{time period}}.$$

So for the ground state we have

$$f_1 = \dfrac{1}{T_1} = \dfrac{1}{1.6 \times 10^{-16}}.$$

Carrying out the division,

$$f_1 = \dfrac{1}{1.6}\times 10^{16} = 0.625 \times 10^{16} = 6.25 \times 10^{15}\,\text{s}^{-1}.$$

Now we want the frequency for the first excited state, i.e. for $$n = 2.$$

In Bohr’s model the radius of the orbit varies as $$r_n \propto n^2$$ and the speed varies as $$v_n \propto \dfrac{1}{n}.$$

Using the formula for time period $$T_n = \dfrac{2\pi r_n}{v_n},$$ we substitute the proportionalities:

$$T_n \propto \dfrac{n^2}{1/n} = n^3.$$

Thus the time period increases as the cube of the quantum number, or equivalently, the frequency decreases as the cube of the quantum number:

$$f_n \propto \dfrac{1}{n^3}.$$

Therefore, when we move from $$n=1$$ to $$n=2,$$ the new frequency is obtained by dividing the ground-state frequency by $$2^3 = 8.$$

$$f_2 = \dfrac{f_1}{2^3} = \dfrac{6.25 \times 10^{15}}{8}.$$

Performing the division,

$$f_2 = 0.78125 \times 10^{15}\,\text{s}^{-1} = 7.8125 \times 10^{14}\,\text{s}^{-1}.$$

Rounded to the proper significant figures,

$$f_2 \approx 7.8 \times 10^{14}\,\text{s}^{-1}.$$

Hence, the correct answer is Option B.

We begin by noting that the incident beam has an intensity of $$6.4 \times 10^{-5}\;{\rm W/cm^2}$$ and the metal surface has an area of $$1\;{\rm cm^2}$$. Intensity is power per unit area, so the power $$P$$ falling on the plate is simply the product of intensity and area.

Hence, we have

$$P = \bigl(6.4 \times 10^{-5}\;{\rm W/cm^2}\bigr)\times \bigl(1\;{\rm cm^2}\bigr) = 6.4 \times 10^{-5}\;{\rm W}.$$

This power is the energy incident per second. Therefore the energy received by the plate in $$1{\rm \; s}$$ is

$$E_{\text{incident}} = P \times 1{\rm \; s} = 6.4 \times 10^{-5}\;{\rm J}.$$

Next, we calculate the energy of a single photon of the given wavelength. The photon energy is obtained from the relation $$E = \dfrac{hc}{\lambda}.$$

We are provided $$hc = 1240\;{\rm eV\,nm}$$ and $$\lambda = 310\;{\rm nm}.$$ Substituting these values, we get

$$E_{\text{photon}} = \dfrac{1240\;{\rm eV\,nm}}{310\;{\rm nm}} = 4.0\;{\rm eV}.$$

To keep all quantities in the same unit system, we convert this photon energy from electron-volts to joules. The conversion factor is $$1\;{\rm eV} = 1.6 \times 10^{-19}\;{\rm J}.$$ Hence,

$$E_{\text{photon}} = 4.0\;{\rm eV}\times \bigl(1.6 \times 10^{-19}\;{\rm J/eV}\bigr) = 6.4 \times 10^{-19}\;{\rm J}.$$

We now determine the number of photons striking the plate in one second. Because total incident energy in one second is $$E_{\text{incident}}$$ and each photon carries energy $$E_{\text{photon}},$$ the photon count $$N_{\text{incident}}$$ is given by

$$N_{\text{incident}} = \dfrac{E_{\text{incident}}}{E_{\text{photon}}} = \dfrac{6.4 \times 10^{-5}\;{\rm J}}{6.4 \times 10^{-19}\;{\rm J}}.$$

Both the numerator and the denominator contain the same coefficient $$6.4,$$ so they cancel, leaving powers of ten to handle:

$$N_{\text{incident}} = 10^{14}.$$

The problem states that only one photon out of every $$10^3$$ actually ejects an electron. Thus the fraction of photons effective in photo-emission is $$\dfrac{1}{10^3}.$$ Therefore the number of electrons emitted in one second, $$N_e,$$ is

$$N_e = N_{\text{incident}}\times \dfrac{1}{10^3} = 10^{14}\times 10^{-3} = 10^{11}.$$

We are told to write the answer in the form $$10^{x}$$ for the total electrons ejected per second, and we have discovered that $$x = 11.$$

Hence, the correct answer is Option C.

$$\frac{1}{\lambda} = R_{\infty} \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$$

For the Lyman series ($$n_1 = 1$$):

$$\text{Shortest wavelength } (\lambda_{L,\text{min}}) \implies n_2 = \infty \implies \lambda_{L,\text{min}} = \frac{1}{R_{\infty}}$$

$$\text{Largest wavelength } (\lambda_{L,\text{max}}) \implies n_2 = 2 \implies \lambda_{L,\text{max}} = \frac{4}{3R_{\infty}}$$

$$\Delta \lambda_L = \lambda_{L,\text{max}} - \lambda_{L,\text{min}} = \frac{1}{3R_{\infty}} = 304\ \text{Å} \implies \frac{1}{R_{\infty}} = 912\ \text{Å}$$

For the Paschen series ($$n_1 = 3$$):

$$\text{Shortest wavelength } (\lambda_{P,\text{min}}) \implies n_2 = \infty \implies \lambda_{P,\text{min}} = \frac{9}{R_{\infty}}$$

$$\text{Largest wavelength } (\lambda_{P,\text{max}}) \implies n_2 = 4 \implies \lambda_{P,\text{max}} = \frac{144}{7R_{\infty}}$$

$$\Delta \lambda_P = \lambda_{P,\text{max}} - \lambda_{P,\text{min}} = \frac{144}{7R_{\infty}} - \frac{9}{R_{\infty}} = \frac{81}{7R_{\infty}}$$

$$\Delta \lambda_P = \frac{81}{7} \times 912 = \frac{73872}{7} \approx 10553.14\ \text{Å}$$

We have to find the wavelength of the second member of the Balmer series of the hydrogen atom, given that the first member has a wavelength of 6561 Å.

First, we convert the given wavelength from angstrom to nanometre because the final answer is required in nm. We know that

$$1\;\text{\AA}=0.1\;\text{nm}.$$

So,

$$\lambda_1 = 6561\;\text{\AA}=6561\times0.1\;\text{nm}=656.1\;\text{nm}.$$

For hydrogen spectral lines we use the Rydberg formula. We state it first:

$$\frac{1}{\lambda}=R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right),$$

where $$R$$ is the Rydberg constant, $$n_1$$ is the lower energy level and $$n_2$$ is the higher energy level with $$n_2>n_1$$.

In the Balmer series $$n_1=2$$ and $$n_2=3,4,5,\ldots$$ 

The first member (H-α line) corresponds to $$n_2=3$$. Therefore, using the given wavelength $$\lambda_1$$, we write

$$\frac{1}{\lambda_1}=R\left(\frac{1}{2^2}-\frac{1}{3^2}\right) =R\left(\frac{1}{4}-\frac{1}{9}\right).$$

Simplifying the bracket:

$$\frac{1}{4}=\frac{9}{36},\qquad \frac{1}{9}=\frac{4}{36},\qquad \frac{1}{4}-\frac{1}{9}=\frac{9}{36}-\frac{4}{36}=\frac{5}{36}.$$

So,

$$\frac{1}{\lambda_1}=R\left(\frac{5}{36}\right).$$

Re-arranging to obtain $$R$$, we get

$$R=\frac{1}{\lambda_1}\cdot\frac{36}{5}.$$

Now we move to the second member (H-β line) of the Balmer series, for which $$n_2=4$$. Again using the Rydberg formula:

$$\frac{1}{\lambda_2}=R\left(\frac{1}{2^2}-\frac{1}{4^2}\right) =R\left(\frac{1}{4}-\frac{1}{16}\right).$$

Simplifying inside the bracket:

$$\frac{1}{4}=\frac{4}{16},\qquad \frac{1}{16}=\frac{1}{16},\qquad \frac{1}{4}-\frac{1}{16}=\frac{4}{16}-\frac{1}{16}=\frac{3}{16}.$$

Hence,

$$\frac{1}{\lambda_2}=R\left(\frac{3}{16}\right).$$

Substituting the expression of $$R$$ obtained from the first line:

$$\frac{1}{\lambda_2}=\left(\frac{1}{\lambda_1}\cdot\frac{36}{5}\right)\left(\frac{3}{16}\right).$$

Multiplying the numerical factors:

$$\frac{36}{5}\times\frac{3}{16}=\frac{36\times3}{5\times16} =\frac{108}{80}=\frac{27}{20}.$$

Thus,

$$\frac{1}{\lambda_2}=\frac{1}{\lambda_1}\cdot\frac{27}{20}.$$

Taking the reciprocal of both sides to isolate $$\lambda_2$$:

$$\lambda_2=\lambda_1\cdot\frac{20}{27}.$$

Now we substitute $$\lambda_1=656.1\;\text{nm}$$:

$$\lambda_2=656.1\;\text{nm}\times\frac{20}{27}.$$

Carrying out the multiplication-division step by step:

$$\frac{20}{27}\approx0.740740\dots$$

$$\lambda_2=656.1\times0.740740\dots\;\text{nm}\approx486.0\;\text{nm}.$$

So, the wavelength of the second member of the Balmer series is $$486\;\text{nm}$$.

Hence, the correct answer is Option C.

According to Einstein’s photo-electric equation, the maximum kinetic energy of an emitted electron is equal to the energy of the incident photon minus the work function of the metal. Stating the relation in symbols, we write

$$\tfrac12\,m\,v_{\max}^2 \;=\; h\nu \;-\;\phi,$$

where $$m$$ is the electron mass, $$v_{\max}$$ is the maximum speed of the emitted photo-electrons, $$h\nu$$ (or simply $$E_{\text{photon}}$$) is the photon energy and $$\phi$$ is the work function of the metal.

The problem gives two different photon energies:

$$E_1 = 4\;\text{eV}, \qquad E_2 = 2.5\;\text{eV}.$$

Let the corresponding maximum speeds of the photo-electrons be $$v_1$$ and $$v_2$$. Using the formula separately for the two cases, we have

$$\tfrac12\,m\,v_1^{\,2} = E_1 - \phi \qquad\text{and}\qquad \tfrac12\,m\,v_2^{\,2} = E_2 - \phi.$$

The ratio of the maximum speeds is given to be 2, that is,

$$\frac{v_1}{v_2} = 2.$$

Squaring both sides we obtain

$$\left(\frac{v_1}{v_2}\right)^2 = 4.$$

Now we divide the two kinetic-energy equations term by term. The factors $$\tfrac12 m$$ cancel out, yielding

$$\frac{v_1^{\,2}}{v_2^{\,2}} = \frac{E_1 - \phi}{E_2 - \phi}.$$

Substituting the squared ratio just obtained, we write

$$4 = \frac{E_1 - \phi}{E_2 - \phi}.$$

Cross-multiplying gives

$$E_1 - \phi = 4\,(E_2 - \phi).$$

Expanding the right-hand side, we have

$$E_1 - \phi = 4E_2 - 4\phi.$$

Now we collect the $$\phi$$ terms on one side and the numerical terms on the other side:

$$-\,\phi + 4\phi = 4E_2 - E_1.$$

Simplifying the left side,

$$3\phi = 4E_2 - E_1.$$

So, the work function is

$$\phi = \frac{4E_2 - E_1}{3}.$$

Substituting the numerical values $$E_1 = 4\;\text{eV}$$ and $$E_2 = 2.5\;\text{eV},$$ we get

$$\phi = \frac{4 \times 2.5 - 4}{3} = \frac{10 - 4}{3} = \frac{6}{3} = 2\;\text{eV}.$$

So, the answer is $$2\;\text{eV}.$$

We begin with Einstein’s photo-electric equation, stated in terms of stopping potential:

$$eV_s \;=\; h\nu \;-\; \phi$$

Here $$e$$ is the electronic charge, $$V_s$$ the stopping potential, $$h$$ Planck’s constant, $$\nu$$ the frequency of the incident light and $$\phi$$ the work function of the metal. Rewriting the photon energy with wavelength $$\lambda$$ using $$\nu = \dfrac{c}{\lambda}$$ gives

$$eV_s \;=\; \dfrac{hc}{\lambda} \;-\; \dfrac{hc}{\lambda_0}$$

where $$\lambda_0$$ is the threshold wavelength of the surface.

According to the problem, for wavelength $$\lambda = \Lambda$$ the stopping potential is $$V$$. Substituting these values, we have

$$eV \;=\; \dfrac{hc}{\Lambda} \;-\; \dfrac{hc}{\lambda_0} \qquad (1)$$

The same surface is next illuminated with wavelength $$3\Lambda$$ and the stopping potential becomes $$\dfrac{V}{4}$$. Substituting again:

$$e\left(\dfrac{V}{4}\right) \;=\; \dfrac{hc}{3\Lambda} \;-\; \dfrac{hc}{\lambda_0} \qquad (2)$$

For simplicity, let us express the threshold wavelength in the form stated in the question: $$\lambda_0 = n\Lambda$$, where $$n$$ is the unknown we must find. We now rewrite both equations using this relation.

From equation (1):

$$eV = hc\left(\dfrac{1}{\Lambda} - \dfrac{1}{n\Lambda}\right) = hc \left(\dfrac{n-1}{n\Lambda}\right) \qquad (3)$$

From equation (2):

$$e\dfrac{V}{4} = hc\left(\dfrac{1}{3\Lambda} - \dfrac{1}{n\Lambda}\right) = hc \left(\dfrac{n-3}{3n\Lambda}\right) \qquad (4)$$

To eliminate the common factors $$hc$$ and $$\Lambda$$, we divide equation (3) by equation (4). The left-hand side becomes

$$\dfrac{eV}{eV/4} = 4$$

The right-hand side becomes

$$\dfrac{\dfrac{n-1}{n\Lambda}}{\dfrac{n-3}{3n\Lambda}} = \dfrac{n-1}{n\Lambda} \cdot \dfrac{3n\Lambda}{\,n-3\,} = 3\,\dfrac{n-1}{n-3}$$

Equating the two sides gives

$$4 = 3\,\dfrac{n-1}{n-3}$$

Cross-multiplying, we obtain

$$4(n-3) = 3(n-1)$$

Expanding both sides:

$$4n - 12 = 3n - 3$$

Bringing all terms containing $$n$$ to one side:

$$4n - 3n = -3 + 12$$

$$n = 9$$

Thus the threshold wavelength is $$\lambda_0 = 9\Lambda$$.

Hence, the correct answer is Option 9.

We have to determine the de-Broglie wavelength of a nitrogen molecule that is moving with its root-mean-square (r.m.s.) speed at a temperature of $$400\ \text{K}$$.

First we recall the expression for the r.m.s. speed of a gas molecule in an ideal gas:

$$v_{\mathrm{rms}}=\sqrt{\dfrac{3k_{\mathrm{B}}T}{m}}$$

Here $$k_{\mathrm{B}}=1.38\times10^{-23}\,\text{J\,K}^{-1}$$ is the Boltzmann constant, $$T=400\ \text{K}$$ is the absolute temperature, and $$m=4.64\times10^{-26}\,\text{kg}$$ is the mass of one nitrogen molecule. Substituting the given values we get

$$v_{\mathrm{rms}}=\sqrt{\dfrac{3\,(1.38\times10^{-23})\,(400)}{4.64\times10^{-26}}}$$

We first evaluate the numerator inside the square root:

$$3\,(1.38\times10^{-23})\,(400)=3\,(552\times10^{-23})=1656\times10^{-23}=1.656\times10^{-20}$$

Now we form the complete fraction:

$$\dfrac{1.656\times10^{-20}}{4.64\times10^{-26}} =1.656\times10^{-20}\times\dfrac{1}{4.64\times10^{-26}} =\dfrac{1.656}{4.64}\times10^{6} \approx0.3576\times10^{6}=3.576\times10^{5}$$

Taking the square root of the result gives

$$v_{\mathrm{rms}}=\sqrt{3.576\times10^{5}}\ \text{m s}^{-1}\approx5.98\times10^{2}\ \text{m s}^{-1}$$

Next, we recall the de-Broglie relation which links a particle’s momentum to its wavelength:

$$\lambda=\dfrac{h}{p}=\dfrac{h}{mv}$$

where $$h=6.63\times10^{-34}\,\text{J\,s}$$ is Planck’s constant, $$m=4.64\times10^{-26}\,\text{kg}$$ is again the mass and $$v=v_{\mathrm{rms}}$$ is the speed just calculated. Substituting the numbers, we obtain

$$\lambda=\dfrac{6.63\times10^{-34}}{(4.64\times10^{-26})(5.98\times10^{2})}$$

We multiply the denominator term-by-term:

$$(4.64\times10^{-26})(5.98\times10^{2}) =4.64\times5.98\times10^{-26+2} =27.77\times10^{-24} =2.777\times10^{-23}$$

Hence

$$\lambda=\dfrac{6.63\times10^{-34}}{2.777\times10^{-23}} =\dfrac{6.63}{2.777}\times10^{-34+23} \approx2.387\times10^{-11}\ \text{m}$$

Finally, we convert metres to ångströms using $$1\ \text{\AA}=10^{-10}\ \text{m}:$$

$$\lambda=2.387\times10^{-11}\ \text{m} =\dfrac{2.387\times10^{-11}}{10^{-10}}\ \text{\AA} =0.2387\ \text{\AA}\approx0.24\ \text{\AA}$$

Hence, the correct answer is Option A.

We are told that an electron in a hydrogen atom is revolving in its second excited state and that the radius of this orbit is $$4.65\;\text{\AA}$$. In the language of the Bohr model, the ground state corresponds to the principal quantum number $$n = 1$$, the first excited state to $$n = 2$$, and therefore the second excited state corresponds to $$n = 3$$. We shall soon verify that this value of $$n$$ is indeed compatible with the given radius.

First, we recall the Bohr‐radius formula for the radius of the $$n^{\text{th}}$$ orbit of hydrogen:

$$r_n = n^2 a_0,$$

where $$a_0 = 0.529\;\text{\AA}$$ is the Bohr radius. Substituting the given radius $$r_n = 4.65\;\text{\AA}$$ and solving for $$n$$ gives

$$n^2 = \frac{r_n}{a_0} = \frac{4.65\;\text{\AA}}{0.529\;\text{\AA}}.$$

We simplify the numerical fraction:

$$\frac{4.65}{0.529} = 8.79,$$

so

$$n^2 = 8.79 \quad\Longrightarrow\quad n = \sqrt{8.79}\approx 2.97.$$

This value rounds essentially to $$n = 3$$, confirming that the electron is indeed in the second excited state as stated.

Now we need the de Broglie wavelength of the electron in this orbit. The Bohr model imposes the standing‐wave condition on the de Broglie matter wave of the electron. The condition is

$$2\pi r_n = n\lambda,$$

where $$\lambda$$ is the de Broglie wavelength of the electron. This relation says that an integral number $$n$$ of wavelengths must fit exactly along the circumference $$2\pi r_n$$ of the circular orbit. We solve this equation for $$\lambda$$:

$$\lambda = \frac{2\pi r_n}{n}.$$

Substituting $$r_n = 4.65\;\text{\AA}$$ and $$n = 3$$ gives

$$\lambda = \frac{2\pi \times 4.65\;\text{\AA}}{3}.$$

We now carry out the multiplication in the numerator:

$$2\pi \times 4.65 = 2 \times 3.1416 \times 4.65 = 6.2832 \times 4.65 = 29.23088.$$

Hence the wavelength becomes

$$\lambda = \frac{29.23088\;\text{\AA}}{3}.$$

Dividing by $$3$$ yields

$$\lambda = 9.7436\;\text{\AA}.$$

Rounding to two significant figures (matching the precision of the data given), we obtain

$$\lambda \approx 9.7\;\text{\AA}.$$

Looking at the options, $$9.7\;\text{\AA}$$ corresponds to Option C.

Hence, the correct answer is Option C.

First, we recall the Einstein photoelectric equation, which links the stopping potential $$V_s$$ with the frequency of the incident radiation:

$$e\,V_s = h\nu - h\nu_0,$$

where $$e$$ is the magnitude of electronic charge, $$h$$ is Planck’s constant, $$\nu$$ is the frequency of the light used, and $$\nu_0$$ is the threshold frequency of the surface.

The problem gives two experimental situations. In the first, the stopping potential is reported as $$-\dfrac{V_0}{2}$$ when the surface is illuminated with monochromatic light of frequency $$\nu$$. Placing this directly into the formula, we write

$$e\left(-\dfrac{V_0}{2}\right)=h\nu - h\nu_0.$$

Simplifying the left‐hand side gives

$$-\dfrac{eV_0}{2}=h\nu-h\nu_0.\qquad(1)$$

In the second experiment the frequency is halved to $$\dfrac{\nu}{2}$$ and the stopping potential becomes $$-V_0$$. Substituting these values into the photoelectric equation, we get

$$e(-V_0)=h\left(\dfrac{\nu}{2}\right)-h\nu_0,$$

or

$$-eV_0=\dfrac{h\nu}{2}-h\nu_0.\qquad(2)$$

Now we possess two simultaneous equations, (1) and (2). Our next task is to eliminate $$\nu_0$$ by subtracting one equation from the other. Subtracting (2) from (1) yields

$$\left(-\dfrac{eV_0}{2}\right)-(-eV_0)=\bigl(h\nu-h\nu_0\bigr)-\bigl(\dfrac{h\nu}{2}-h\nu_0\bigr).$$

The left side simplifies to

$$-\dfrac{eV_0}{2}+eV_0=\dfrac{eV_0}{2},$$

while the right side simplifies to

$$h\nu-h\nu_0-\dfrac{h\nu}{2}+h\nu_0=h\nu-\dfrac{h\nu}{2}=\dfrac{h\nu}{2}.$$

Equating the simplified results gives us

$$\dfrac{eV_0}{2}=\dfrac{h\nu}{2}\quad\Longrightarrow\quad eV_0=h\nu.\qquad(3)$$

This relation directly connects $$V_0$$ with $$\nu$$. We now substitute (3) back into either of our earlier equations to solve for the threshold frequency. Using equation (1):

$$-\dfrac{eV_0}{2}=h\nu-h\nu_0.$$

Replacing $$h\nu$$ by $$eV_0$$ from (3):

$$-\dfrac{eV_0}{2}=eV_0-h\nu_0.$$

Moving terms involving $$eV_0$$ to the left gives

$$h\nu_0=eV_0+\dfrac{eV_0}{2}=\dfrac{3eV_0}{2}.$$

Once more inserting $$eV_0=h\nu$$ from (3):

$$h\nu_0=\dfrac{3}{2}h\nu,$$

and dividing both sides by $$h$$ finally produces

$$\nu_0=\dfrac{3\nu}{2}.$$

Therefore the threshold frequency is $$\dfrac{3\nu}{2}$$, which matches Option B.

Hence, the correct answer is Option B.

We have a particle of mass $$m$$ moving in a central potential field $$U(r)=\tfrac12 k r^2$$. For a circular orbit of radius $$r$$ the necessary centripetal force must be supplied entirely by the central force obtained from the potential.

First, we write the radial force that comes from the potential. By definition

$$F_r \;=\;-\frac{dU}{dr}.$$

Because $$U(r)=\tfrac12 k r^2,$$ we get

$$F_r \;=\;-\frac{d}{dr}\!\Bigl(\tfrac12 k r^2\Bigr) \;=\; -k r.$$

In a circular orbit the magnitude of this inward force equals the required centripetal force $$\dfrac{m v^2}{r}$$, so

$$\frac{m v^2}{r}\;=\;k r.$$

Multiplying both sides by $$r$$ gives

$$m v^2 \;=\;k r^2,$$

and hence

$$v^2 \;=\;\frac{k}{m}\,r^2 \quad\Longrightarrow\quad v \;=\;\sqrt{\frac{k}{m}}\;r.$$

Now we invoke Bohr’s quantization condition for angular momentum, which states

$$L \;=\;m v r \;=\;n\hbar,$$

where $$n=1,2,3,\dots$$ is the principal quantum number and $$\hbar$$ is the reduced Planck constant.

Substituting the expression for $$v$$ into $$L$$, we obtain

$$L \;=\;m(\sqrt{\tfrac{k}{m}}\,r)\,r \;=\;\sqrt{m k}\;r^2.$$

Setting this equal to $$n\hbar$$ gives

$$\sqrt{m k}\;r^2 \;=\;n\hbar.$$

Solving for $$r^2$$ we find

$$r^2 \;=\;\frac{n\hbar}{\sqrt{m k}}.$$

Taking the square root on both sides,

$$r \;\propto\;\sqrt{n}.$$

Hence the radius of the allowed orbitals increases as the square root of the quantum number $$n$$.

Next, we calculate the total mechanical energy $$E$$ of the particle. The kinetic energy is

$$K \;=\;\tfrac12 m v^2,$$

and the potential energy is

$$U \;=\;\tfrac12 k r^2.$$

Using $$v^2=\dfrac{k}{m}r^2,$$ we get

$$K \;=\;\tfrac12 m\Bigl(\tfrac{k}{m}r^2\Bigr) \;=\;\tfrac12 k r^2.$$

Therefore the total energy becomes

$$E \;=\;K + U \;=\;\tfrac12 k r^2 + \tfrac12 k r^2 \;=\;k r^2.$$

Because we have already shown that $$r^2 \propto n,$$ it follows directly that

$$E \;\propto\;n.$$

Summarizing our results,

$$r_n \propto \sqrt{n}, \qquad E_n \propto n.$$

Comparing with the given options, this corresponds exactly to Option B.

Hence, the correct answer is Option B.

We are told that the de Broglie wavelength of an electron is $$10^{-3}$$ times the wavelength of a photon whose frequency is $$6 \times 10^{14}\,\text{Hz}$$. First, we determine the wavelength of that photon.

The relation between the speed of light $$c$$, frequency $$\nu$$ and wavelength $$\lambda$$ of any photon is stated by the formula

$$c = \lambda_{\text{photon}} \,\nu_{\text{photon}}.$$

So, solving for the photon’s wavelength, we have

$$\lambda_{\text{photon}} = \frac{c}{\nu_{\text{photon}}}.$$

Substituting the given values $$c = 3 \times 10^{8}\,\text{m/s}$$ and $$\nu_{\text{photon}} = 6 \times 10^{14}\,\text{Hz}$$, we get

$$\lambda_{\text{photon}} = \frac{3 \times 10^{8}}{6 \times 10^{14}} = \frac{3}{6} \times 10^{8-14} = 0.5 \times 10^{-6}\,\text{m} = 5 \times 10^{-7}\,\text{m}.$$

Now, the electron’s de Broglie wavelength is stated to be $$10^{-3}$$ times this value, so

$$\lambda_{\text{electron}} = 10^{-3}\,\lambda_{\text{photon}} = 10^{-3}\,(5 \times 10^{-7}) = 5 \times 10^{-10}\,\text{m}.$$

For a material particle such as an electron, the de Broglie relation gives

$$\lambda_{\text{electron}} = \frac{h}{m_{\text{e}}\,v},$$

where $$h$$ is Planck’s constant, $$m_{\text{e}}$$ is the electron mass and $$v$$ is the electron’s speed. We rearrange this formula to solve for $$v$$:

$$v = \frac{h}{m_{\text{e}}\,\lambda_{\text{electron}}}.$$

Substituting the numerical values $$h = 6.63 \times 10^{-34}\,\text{J·s}, \qquad m_{\text{e}} = 9.1 \times 10^{-31}\,\text{kg}, \qquad \lambda_{\text{electron}} = 5 \times 10^{-10}\,\text{m},$$ we obtain

$$v = \frac{6.63 \times 10^{-34}} {(9.1 \times 10^{-31})(5 \times 10^{-10})}.$$

We multiply the two numbers in the denominator first:

$$(9.1 \times 10^{-31})(5 \times 10^{-10}) = 9.1 \times 5 \times 10^{-31-10} = 45.5 \times 10^{-41} = 4.55 \times 10^{-40}.$$

Now we divide:

$$v = \frac{6.63 \times 10^{-34}}{4.55 \times 10^{-40}} = \left(\frac{6.63}{4.55}\right)\times 10^{-34 - (-40)} = 1.457 \times 10^{6}\,\text{m/s}.$$

Rounding to two significant figures, the speed is

$$v \approx 1.45 \times 10^{6}\,\text{m/s}.$$

Hence, the correct answer is Option D.

We have an electron that enters the mercury vapour with a kinetic energy of $$5.6\ \text{eV}$$ and comes out with only $$0.7\ \text{eV}$$.

So the loss in the electron’s energy is

$$\Delta E \;=\; 5.6\ \text{eV} \;-\; 0.7\ \text{eV} \;=\; 4.9\ \text{eV}.$$

In the Frank-Hertz experiment this lost energy is absorbed by a mercury atom, raising the atom from its ground state to its first excited state. When the atom returns to the ground state it emits a photon whose energy is exactly this excitation energy. Hence the photon energy is

$$E_{\text{photon}} \;=\; 4.9\ \text{eV}.$$

For a photon the relation between energy and wavelength is first stated:

$$E \;=\; h \nu \;=\; \frac{h c}{\lambda},$$

where $$h$$ is Planck’s constant and $$c$$ is the speed of light. Solving for the wavelength, we get

$$\lambda \;=\; \frac{h c}{E}.$$

It is convenient to use the product $$h c$$ in the unit “eV·nm”, whose value is well remembered as $$h c \;\approx\; 1240\ \text{eV·nm}$$. Substituting the photon energy,

$$\lambda \;=\; \frac{1240\ \text{eV·nm}}{4.9\ \text{eV}}.$$

Carrying out the division step by step,

$$\lambda \;=\; \frac{1240}{4.9}\ \text{nm} \;=\; 253.06\ \text{nm}\;(\text{approximately}).$$

The closest option to this calculated minimum wavelength is $$250\ \text{nm}$$.

Hence, the correct answer is Option A.

Step 1: Energy formula

$$E=\ \frac{\ 1237}{λ}$$

 Step 2: Work function (threshold energy)

$$ϕ=\ \frac{\ 1237}{380}​\approx3.26eV$$

 Step 3: Incident photon energy

$$E=\ \frac{\ 1237}{260}​\approx4.76eV$$

Step 4: Maximum kinetic energy

$$K_{\max}​=E−ϕ=4.76−3.26=1.50eV$$

The photoelectric equation, first stated by Einstein, relates the stopping potential $$V_0$$ to the frequency $$\nu$$ (or wavelength $$\lambda$$) of the incident light as

$$eV_0 \;=\; h\nu \;-\;\phi_0$$

where $$e$$ is the electronic charge, $$h$$ is Planck’s constant, and $$\phi_0$$ is the work function of the metal. Using the relation between frequency and wavelength $$\nu = \dfrac{c}{\lambda}$$, this can be rewritten in terms of wavelength:

$$eV_0 \;=\; \dfrac{hc}{\lambda} \;-\;\phi_0$$

We perform the experiment at two different wavelengths, $$\lambda_1 = 300 \text{ nm}$$ and $$\lambda_2 = 400 \text{ nm}$$, and obtain two corresponding stopping potentials $$V_1$$ and $$V_2$$. Writing the equation for each wavelength, we have

$$eV_1 \;=\; \dfrac{hc}{\lambda_1} \;-\;\phi_0$$

$$eV_2 \;=\; \dfrac{hc}{\lambda_2} \;-\;\phi_0$$

To eliminate the unknown work function $$\phi_0$$, we subtract the second equation from the first:

$$eV_1 \;-\; eV_2 \;=\;\dfrac{hc}{\lambda_1} \;-\;\dfrac{hc}{\lambda_2}$$

Taking $$e$$ common on the left side, this becomes

$$e\,(V_1 - V_2) \;=\; \dfrac{hc}{\lambda_1} \;-\; \dfrac{hc}{\lambda_2}$$

Dividing by $$e$$ gives the change (decrease) in stopping potential:

$$\Delta V \;=\; V_1 - V_2 \;=\; \dfrac{hc}{e}\left(\dfrac{1}{\lambda_1} - \dfrac{1}{\lambda_2}\right)$$

We are supplied with the convenient constant $$\dfrac{hc}{e} = 1240 \text{ nm}\cdot\text{V}$$. Substituting the given wavelengths, we obtain

$$\Delta V \;=\; 1240 \,\text{nm}\cdot\text{V}\;\Bigl(\dfrac{1}{300 \,\text{nm}} - \dfrac{1}{400 \,\text{nm}}\Bigr)$$

Now we evaluate the term in parentheses carefully, keeping every algebraic step explicit:

$$\dfrac{1}{300 \,\text{nm}} \;=\; 0.003333\;\text{nm}^{-1}$$

$$\dfrac{1}{400 \,\text{nm}} \;=\; 0.002500\;\text{nm}^{-1}$$

Hence,

$$\dfrac{1}{300} - \dfrac{1}{400} \;=\; 0.003333 - 0.002500 \;=\; 0.000833\;\text{nm}^{-1}$$

Multiplying by the constant 1240 nm·V, we get

$$\Delta V \;=\; 1240 \times 0.000833 \;\text{V}$$

Carrying out the multiplication step by step,

$$1240 \times 0.000833 \;=\; 1.033\;\text{V (approximately)}$$

The decrease in the stopping potential is therefore about $$1.0 \text{ V}$$ when expressed to one significant figure, matching the choice closest to our calculated value.

Hence, the correct answer is Option C.

We are told that an electron microscope can resolve details whose size is of the same order as the de-Broglie wavelength of the electrons it uses, that is

$$\lambda \approx 7.5 \times 10^{-12}\ \text{m}.$$

For an electron, the de-Broglie relation gives the momentum in terms of the wavelength:

$$p = \frac{h}{\lambda},$$

where $$h = 6.626 \times 10^{-34}\ \text{J·s}$$ is Planck’s constant.

The kinetic energy of a non-relativistic particle is, by definition,

$$E = \frac{p^{2}}{2m},$$

with $$m = 9.11 \times 10^{-31}\ \text{kg}$$ for the electron mass.

Substituting the momentum expression $$p = \dfrac{h}{\lambda}$$ into the kinetic-energy formula, we obtain

$$E = \frac{1}{2m}\left(\frac{h}{\lambda}\right)^{2}.$$

Now we insert the numerical values step by step:

First compute the square of Planck’s constant:

$$h^{2} = (6.626 \times 10^{-34})^{2} = 4.392 \times 10^{-67}\ \text{J}^{2}\!\cdot\!\text{s}^{2}.$$

Next compute the square of the wavelength:

$$\lambda^{2} = (7.5 \times 10^{-12})^{2} = 5.625 \times 10^{-23}\ \text{m}^{2}.$$

The product in the denominator is

$$2m\lambda^{2} = 2 \times (9.11 \times 10^{-31}) \times (5.625 \times 10^{-23}) = 1.025 \times 10^{-52}\ \text{kg·m}^{2}.$$

We can now evaluate the energy:

$$E = \frac{4.392 \times 10^{-67}}{1.025 \times 10^{-52}} = 4.29 \times 10^{-15}\ \text{J}.$$

To express this energy in electron-volts, we use the conversion

$$1\ \text{eV} = 1.602 \times 10^{-19}\ \text{J}.$$

Hence

$$E = \frac{4.29 \times 10^{-15}\ \text{J}}{1.602 \times 10^{-19}\ \text{J/eV}} \approx 2.68 \times 10^{4}\ \text{eV} = 26.8\ \text{keV}.$$

This value is very close to $$25\ \text{keV}$$ listed among the options.

Hence, the correct answer is Option B.

For a hydrogen atom the stationary‐state energy levels are given by the Bohr formula $$E_n=-\dfrac{13.6\ \text{eV}}{n^{2}},$$ where $$n=1,2,3,\ldots$$ is the principal quantum number.

The electron is stated to start from the third excited state. Counting from the ground state $$n=1$$, we have:

Ground state : $$n=1$$
First excited state : $$n=2$$
Second excited state : $$n=3$$
Third excited state : $$n=4$$

Thus the sequence of transitions is $$n=4 \to n=3$$ followed by $$n=3 \to n=2$$.

The energy released as a photon when the electron falls from a higher level $$n_2$$ to a lower level $$n_1$$ is obtained from the difference of the two level energies:

$$\Delta E = E_{n_1}-E_{n_2}= -\dfrac{13.6}{n_1^{2}} - \Bigl(-\dfrac{13.6}{n_2^{2}}\Bigr)=13.6\Bigl(\dfrac{1}{n_1^{2}}-\dfrac{1}{n_2^{2}}\Bigr)\ \text{eV}.$$

For the first photon of wavelength $$\lambda_1$$ corresponding to $$n=4 \to n=3$$ we have

$$\Delta E_1 = 13.6\Bigl(\dfrac{1}{3^{2}}-\dfrac{1}{4^{2}}\Bigr).$$

We simplify term by term:

$$\dfrac{1}{3^{2}}=\dfrac{1}{9},\quad \dfrac{1}{4^{2}}=\dfrac{1}{16}.$$

So

$$\Delta E_1 = 13.6\Bigl(\dfrac{1}{9}-\dfrac{1}{16}\Bigr)=13.6\Bigl(\dfrac{16-9}{144}\Bigr)=13.6\times\dfrac{7}{144}\ \text{eV}.$$

For the second photon of wavelength $$\lambda_2$$ corresponding to $$n=3 \to n=2$$ we write

$$\Delta E_2 = 13.6\Bigl(\dfrac{1}{2^{2}}-\dfrac{1}{3^{2}}\Bigr).$$

Calculating the fractions:

$$\dfrac{1}{2^{2}}=\dfrac{1}{4},\quad \dfrac{1}{3^{2}}=\dfrac{1}{9}.$$

Hence

$$\Delta E_2 = 13.6\Bigl(\dfrac{1}{4}-\dfrac{1}{9}\Bigr)=13.6\Bigl(\dfrac{9-4}{36}\Bigr)=13.6\times\dfrac{5}{36}\ \text{eV}.$$

The energy of a photon and its wavelength are connected by the Planck relation $$E=\dfrac{hc}{\lambda}.$$ Therefore for two photons, $$\lambda\propto\dfrac{1}{E}$$, and the ratio of wavelengths is the inverse ratio of the corresponding energy differences:

$$\dfrac{\lambda_1}{\lambda_2}=\dfrac{\Delta E_2}{\Delta E_1}.$$

Substituting the expressions we have just obtained:

$$\dfrac{\lambda_1}{\lambda_2}=\dfrac{13.6\times\dfrac{5}{36}}{13.6\times\dfrac{7}{144}}.$$

The common factor $$13.6$$ cancels out, giving

$$\dfrac{\lambda_1}{\lambda_2}=\dfrac{\dfrac{5}{36}}{\dfrac{7}{144}}.$$

Dividing one fraction by another is equivalent to multiplying by its reciprocal, so

$$\dfrac{\lambda_1}{\lambda_2}=\dfrac{5}{36}\times\dfrac{144}{7}.$$

We simplify step by step. First, reduce $$144/36$$:

$$\dfrac{144}{36}=4.$$

Hence

$$\dfrac{\lambda_1}{\lambda_2}=5\times\dfrac{4}{7}=\dfrac{20}{7}.$$

Thus the required ratio of wavelengths is $$\dfrac{20}{7}.$$

Hence, the correct answer is Option B.

We have been told that the laser gives a continuous power of $$P = 2 \text{ mW}$$.

First, let us convert this power into the SI unit watt:

$$P = 2 \text{ mW} = 2 \times 10^{-3}\,\text{W}.$$

A power of one watt means one joule of energy is delivered every second, so a power of $$P$$ watts means $$P$$ joules are emitted each second.

The energy carried by a single photon is found from Planck’s relation. We first state the formula:

$$E_{\text{photon}} = h\nu,$$

where $$h$$ is Planck’s constant and $$\nu$$ (nu) is the frequency of the light. Frequency and wavelength are connected through the speed of light $$c$$ by $$c = \lambda \nu$$, so $$\nu = \dfrac{c}{\lambda}$$. Substituting this into Planck’s relation gives an equivalent formula that uses the wavelength directly:

$$E_{\text{photon}} = \dfrac{h\,c}{\lambda}.$$

We now insert the numerical values. The wavelength is $$\lambda = 500\ \text{nm} = 500 \times 10^{-9}\,\text{m}.$$ The given constants are $$h = 6.6 \times 10^{-34}\,\text{J\,s}$$ and $$c = 3.0 \times 10^{8}\,\text{m/s}.$$ Substituting, we have

$$E_{\text{photon}} = \dfrac{(6.6 \times 10^{-34}\,\text{J\,s})(3.0 \times 10^{8}\,\text{m/s})} {500 \times 10^{-9}\,\text{m}}.$$

First multiply the numerator:

$$h\,c = (6.6 \times 10^{-34})(3.0 \times 10^{8}) = 19.8 \times 10^{-26}\,\text{J\,m}.$$

Now place this over the denominator:

$$E_{\text{photon}} = \dfrac{19.8 \times 10^{-26}} {500 \times 10^{-9}}\,\text{J}.$$

Rewrite $$19.8 = 1.98 \times 10^{1}$$ and $$500 = 5.00 \times 10^{2}$$ so that powers of ten are clearer:

$$E_{\text{photon}} = \dfrac{1.98 \times 10^{1} \times 10^{-26}} {5.00 \times 10^{2} \times 10^{-9}}\,\text{J}.$$

Combine powers of ten in the numerator and denominator: the denominator carries $$10^{2}\times10^{-9}=10^{-7}.$$ Hence,

$$E_{\text{photon}} = \dfrac{1.98 \times 10^{-25}}{5.00 \times 10^{-7}}\,\text{J}.$$

Dividing the coefficients and subtracting exponents gives

$$E_{\text{photon}} = \left(\dfrac{1.98}{5.00}\right)\times 10^{-25 + 7}\,\text{J} = 0.396 \times 10^{-18}\,\text{J}.$$

Shifting the decimal once to the right puts the coefficient between 1 and 10:

$$E_{\text{photon}} = 3.96 \times 10^{-19}\,\text{J}.$$

Now we know the energy released every second (the power) and the energy carried by each photon, so the number of photons emitted per second, $$N$$, is obtained by the simple ratio

$$N = \dfrac{\text{energy per second}}{\text{energy per photon}} = \dfrac{P}{E_{\text{photon}}}.$$

Substituting $$P = 2 \times 10^{-3}\,\text{J/s}$$ and $$E_{\text{photon}} = 3.96 \times 10^{-19}\,\text{J},$$ we get

$$N = \dfrac{2 \times 10^{-3}}{3.96 \times 10^{-19}} = \dfrac{2}{3.96}\times 10^{-3+19} = 0.505 \times 10^{16}.$$

One more shift of the decimal converts this to standard scientific notation:

$$N = 5.05 \times 10^{15}\,\text{photons per second}.$$

Rounding to one significant figure, as the options are, we have

$$N \approx 5 \times 10^{15}\,\text{photons per second}.$$

This value matches Option B in the list provided.

Hence, the correct answer is Option B.

For a hydrogen atom the energy of the level having principal quantum number $$n$$ is given by the Bohr formula

$$E_n=-\dfrac{13.6\ \text{eV}}{n^2}.$$

The atom is initially in its ground state, so its initial energy is

$$E_1=-13.6\ \text{eV}.$$

When it absorbs a photon of wavelength $$\lambda=980\ \text{\AA},$$ the energy of that photon is obtained from the relation $$E=\dfrac{hc}{\lambda}.$$

Using the useful constant $$hc\approx12400\ \text{eV\;\AA},$$ we have

$$E=\dfrac{12400\ \text{eV\;\AA}}{980\ \text{\AA}}=12.653\ \text{eV}\;(\text{approximately}).$$

This entire photon energy raises the electron from $$n=1$$ to some higher level $$n=n_f$$, so the energy conservation condition is

$$E=E_{n_f}-E_1.$$

Substituting the Bohr energies we obtain

$$12.653= \left(-\dfrac{13.6}{n_f^{2}}\right)-(-13.6)=13.6\left(1-\dfrac{1}{n_f^{2}}\right).$$

Dividing both sides by $$13.6$$ gives

$$1-\dfrac{1}{n_f^{2}}=\dfrac{12.653}{13.6}=0.9301.$$

Hence

$$\dfrac{1}{n_f^{2}}=1-0.9301=0.0699,$$

and therefore

$$n_f^{2}=\dfrac{1}{0.0699}\approx14.3.$$

The nearest integral value satisfying this is $$n_f=4$$ (because $$4^{2}=16$$ gives almost the same energy, whereas $$n=3$$ would be far off).

Thus the electron is excited to the fourth orbit.

For the Bohr model the radius of the $$n^{\text{th}}$$ orbit is

$$r_n=n^{2}a_{0},$$

so for $$n=4$$ we have

$$r_4=4^{2}a_{0}=16a_{0}.$$

Hence, the correct answer is Option C.

We begin with the given data. The metal plate has an area $$A = 1 \times 10^{-4}\,{\rm m^2}$$ and the radiation that falls on it has an intensity $$I = 16\;{\rm milli\,W/m^2} = 16 \times 10^{-3}\,{\rm W/m^2} = 0.016\,{\rm W/m^2}$$. 1 watt is 1 joule per second, so intensity tells us how many joules reach one square metre each second.

The power (energy per second) actually falling on the small plate is obtained from the definition of intensity: $$P = I \times A$$. Substituting the values, we get

$$P = 0.016\,{\rm W/m^2} \times 1 \times 10^{-4}\,{\rm m^2} = 1.6 \times 10^{-6}\,{\rm W}.$$

Because a watt equals a joule per second, this also means that $$1.6 \times 10^{-6}\,{\rm J}$$ of radiant energy reaches the plate every second.

Now, each incident photon has an energy of $$10\,{\rm eV}$$. Converting this into joules with the relation $$1\,{\rm eV} = 1.6 \times 10^{-19}\,{\rm J}$$, we write

$$E_{\text{photon}} = 10\,{\rm eV} = 10 \times 1.6 \times 10^{-19}\,{\rm J} = 1.6 \times 10^{-18}\,{\rm J}.$$

The number of photons that hit the plate each second is simply the total incident energy per second divided by the energy of one photon:

$$N_{\text{photon/sec}} = \frac{P}{E_{\text{photon}}} = \frac{1.6 \times 10^{-6}\,{\rm J/s}}{1.6 \times 10^{-18}\,{\rm J}} = 1 \times 10^{12}\,{\rm photons/s}.$$

However, only 10 % of these photons actually eject photoelectrons. Therefore the rate of emission of photoelectrons is

$$N_{e^{-}/\text{sec}} = 0.10 \times 1 \times 10^{12} = 1 \times 10^{11}\,{\rm electrons/s}.$$

Next we calculate the maximum kinetic energy of those emitted electrons. According to Einstein’s photoelectric equation,

$$K_{\text{max}} = h\nu - \phi,$$

where $$h\nu$$ is the photon energy and $$\phi$$ is the work function of the metal. Both quantities are conveniently given in electron-volts:

$$K_{\text{max}} = 10\,{\rm eV} - 5\,{\rm eV} = 5\,{\rm eV}.$$

So, each second $$1 \times 10^{11}$$ electrons are emitted, and the most energetic among them possess $$5\,{\rm eV}$$ of kinetic energy.

Hence, the correct answer is Option C.

We are told that the initial nucleus A has a finite de-Broglie wavelength $$\lambda_A$$. The de-Broglie relation states first of all

$$\lambda \;=\;\frac{h}{p},$$

where $$h$$ is Planck’s constant and $$p$$ is the linear momentum of the particle concerned.

Let the mass of each of the two daughter nuclei B and C be $$m$$. Because A splits into two equal-mass fragments, the mass of A is $$2m$$, but we shall not actually need that value; we only need equality of the masses of B and C.

Choose the positive $$x$$-direction as the direction in which A is moving just before fission. Denote the magnitudes of the velocities of B and C by $$v_B$$ and $$v_C$$ respectively. The statement “B flies in the same direction as that of A, while C flies in the opposite direction with a velocity equal to half of that of B” translates into

$$v_C \;=\;\frac{v_B}{2},\qquad$$ and the velocity of C is opposite in sign to that of B.

Thus the momenta of the three nuclei are

$$p_A \;=\;p_A\;(\text{along }+x),$$

$$p_B \;=\;m\,v_B\;(\text{also along }+x),$$

$$p_C \;=\;-\,m\,v_C \;=\;-\,m\left(\frac{v_B}{2}\right)\;(\text{along }-x).$$

Spontaneous fission occurs in free space, so linear momentum is conserved. Therefore

$$p_A \;=\;p_B + p_C.$$

Substituting the expressions just written, we have

$$p_A \;=\;m\,v_B \;-\;m\left(\frac{v_B}{2}\right).$$

Simplifying the right-hand side step by step,

$$p_A \;=\;m\,v_B \;-\;\frac{m\,v_B}{2} \;=\;\frac{m\,v_B}{2}.$$

Now compare this result with the momentum of B:

$$p_B \;=\;m\,v_B \;=\;2\,p_A.$$

Similarly, the magnitude of the momentum of C is

$$|p_C| \;=\;m\left(\frac{v_B}{2}\right) \;=\;p_A.$$

With all three momenta expressed in terms of $$p_A$$, we can write their de-Broglie wavelengths by applying $$\lambda = \dfrac{h}{p}$$ individually:

For A: $$\lambda_A \;=\;\frac{h}{p_A}.$$

For B: $$\lambda_B \;=\;\frac{h}{p_B} \;=\;\frac{h}{2\,p_A} \;=\;\frac{\lambda_A}{2}.$$

For C: $$\lambda_C \;=\;\frac{h}{|p_C|} \;=\;\frac{h}{p_A} \;=\;\lambda_A.$$

Hence the de-Broglie wavelengths of B and C are respectively

$$\lambda_B \;=\;\frac{\lambda_A}{2},\qquad \lambda_C \;=\;\lambda_A.$$

These correspond exactly to Option B (option number 2 in the list).

Hence, the correct answer is Option 2.