JEE Modern Physics Questions

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 $$11.7\;{\rm eV}\;-\;11.9\;{\rm eV}$$.

Solution :

Momentum carried by light pulse of energy E is :

$$p = \frac{E}{c}$$

Since the mirror reflects the light normally, change in momentum of light is :

$$\Delta p = 2\frac{E}{c}$$

Hence momentum imparted to mirror :

$$mv = \frac{2E}{c}$$

Therefore,

$$v = \frac{2E}{mc}$$

After receiving impulse, mirror starts oscillating.

At maximum deflection, kinetic energy converts into potential energy.

So,

$$\frac{1}{2}mv^2 = mgh$$

For small angle $$\theta\ $$,

$$h \approx \frac{l\theta^2}{2}$$

Therefore,

$$\frac{1}{2}mv^2 = mg\frac{l\theta^2}{2}$$

$$v^2 = gl\theta^2$$

Substituting,

$$\left(\frac{2E}{mc}\right)^2 = gl\theta^2$$

$$\theta = \frac{2E}{mc\sqrt{gl}}$$

Final Answer :

$$\theta = \frac{2E}{mc\sqrt{gl}}$$

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.

We need to find the ratio of de Broglie wavelengths of the electron in the third excited state (n=4) to the ground state (n=1) of hydrogen.

The de Broglie wavelength is related to the orbital parameters by the Bohr quantization condition:

$$n\lambda = 2\pi r_n$$

Therefore, the de Broglie wavelength is:

$$\lambda_n = \frac{2\pi r_n}{n}$$

For the ground state (n=1): $$r_1 = 5.3 \times 10^{-11}$$ m

$$\lambda_1 = \frac{2\pi \times 5.3 \times 10^{-11}}{1} = 2\pi \times 5.3 \times 10^{-11}$$

For the third excited state (n=4): $$r_4 = 8.48 \times 10^{-10}$$ m

$$\lambda_4 = \frac{2\pi \times 8.48 \times 10^{-10}}{4} = 2\pi \times 2.12 \times 10^{-10}$$

The ratio of wavelengths:

$$\frac{\lambda_4}{\lambda_1} = \frac{2.12 \times 10^{-10}}{5.3 \times 10^{-11}} = \frac{2.12}{0.53} = 4$$

The correct answer is Option 4: 4.

Assertion A: The Bohr model is applicable to hydrogen and hydrogen-like atoms only. This is TRUE. The Bohr model works for systems with a single electron orbiting a nucleus (hydrogen, He$$^+$$, Li$$^{2+}$$, etc.).

Reason R: The formulation of the Bohr model does not include repulsive force between electrons. This is TRUE. Bohr's model considers only the attractive Coulomb force between the nucleus and a single electron. It does not account for electron-electron repulsion.

The reason R correctly explains the assertion A. The Bohr model is limited to hydrogen-like atoms precisely because it does not incorporate electron-electron repulsive interactions. For multi-electron atoms, the inter-electron repulsion makes the Bohr model inapplicable.

Hence, the correct answer is Option C.

The photoelectric equation is $$eV_0 = h\nu - \phi$$, where $$V_0$$ is the stopping potential, $$\nu$$ is the frequency of incident light and $$\phi$$ is the work function of the metal.

From the above relation we see that $$V_0$$ depends only on the frequency $$\nu$$ (and the work function $$\phi$$), not on the intensity of light. Therefore, even if the intensity is increased, the stopping potential remains unchanged. Hence Assertion A is false.

Intensity of light represents the number of photons incident per unit time. When the frequency $$\nu$$ is kept greater than the threshold frequency $$\nu_0$$ so that each photon has energy $$h\nu \gt \phi$$, an increase in the number of photons results in a proportional increase in the number of emitted photoelectrons. Thus the emission rate (photoelectric current for a given potential) increases. Reason R is therefore true.

We conclude: Assertion A is false but Reason R is true. This corresponds to Option B.

For a hydrogen-like ion (single electron around nucleus of charge $$+Ze$$), Bohr’s model gives the radius of the $$n^{\text{th}}$$ orbit as

$$r_n = \frac{n^2 a_0}{Z}$$

where
$$n$$ = principal quantum number,
$$Z$$ = atomic number of the nucleus,
$$a_0$$ = Bohr radius for the hydrogen ground state.

For $$Li^{2+}$$ the atomic number is $$Z = 3$$ because lithium has three protons. The ground state corresponds to $$n = 1$$.

Substituting $$n = 1$$ and $$Z = 3$$ into the formula,

$$r_1 = \frac{(1)^2 a_0}{3} = \frac{a_0}{3}$$

This can be written as $$\frac{1}{X} a_0$$ with $$X = 3$$.

Hence, $$X = 3$$, which matches Option C.

The energy of an electron in a Bohr orbit of any hydrogen-like species (single-electron ion) is given by
$$E_n = -\dfrac{13.6\,\text{eV}\,Z^{2}}{n^{2}}$$
where $$Z$$ is the atomic number and $$n = 1,2,3,\ldots$$ is the principal quantum number.

For the hydrogen atom $$\left(Z = 1\right)$$ in its ground state $$\left(n = 1\right)$$:
$$E_{\text{H, ground}} = -\dfrac{13.6 \cdot 1^{2}}{1^{2}} = -13.6\ \text{eV}$$

He$$^{+}$$ ion has $$Z = 2$$.
First excited state means $$n = 2$$.
$$E_{\text{He}^{+},\,n=2} = -\dfrac{13.6 \cdot 2^{2}}{2^{2}} = -13.6\ \text{eV}$$
This equals the hydrogen ground-state energy, so statement (A) is correct.

Li$$^{++}$$ ion has $$Z = 3$$.
Second excited state means $$n = 3$$.
$$E_{\text{Li}^{++},\,n=3} = -\dfrac{13.6 \cdot 3^{2}}{3^{2}} = -13.6\ \text{eV}$$
Again equal to the hydrogen ground-state energy, so statement (B) is correct.

He$$^{+}$$ ground state: $$n = 1$$.
$$E_{\text{He}^{+},\,n=1} = -\dfrac{13.6 \cdot 2^{2}}{1^{2}} = -54.4\ \text{eV}$$
This is not equal to $$-13.6\ \text{eV}$$, so statement (C) is incorrect.

He$$^{+}$$ first excited state energy has already been found as $$-13.6\ \text{eV}$$.
Li$$^{++}$$ ground state: $$n = 1$$.
$$E_{\text{Li}^{++},\,n=1} = -\dfrac{13.6 \cdot 3^{2}}{1^{2}} = -122.4\ \text{eV}$$
These are not equal, so statement (D) is incorrect.

Only statements (A) and (B) are correct. Hence the correct option is Option B: (A), (B) only.

For a hydrogen-like ion, the energy of the electron in the $$n^{\text{th}}$$ Bohr orbit is

$$E_n = -\,\frac{13.6\,Z^{2}}{n^{2}}\ \text{eV} \qquad -(1)$$

where $$Z$$ is the atomic number of the ion.

The ground state corresponds to $$n = 1$$.

“2nd excitation energy state’’ means the electron is in the second excited level, i.e. $$n = 3$$ (first excited is $$n = 2$$, second excited is $$n = 3$$).

Using (1):

Energy of ground state ($$n = 1$$): $$E_1 = -\,13.6\,Z^{2}\ \text{eV} \qquad -(2)$$

Energy of 2nd excited state ($$n = 3$$): $$E_3 = -\,\frac{13.6\,Z^{2}}{9}\ \text{eV} \qquad -(3)$$

The energy difference between these two states is

$$\Delta E = E_3 - E_1$$

Substituting (2) and (3):

$$\Delta E = -\,\frac{13.6\,Z^{2}}{9} - \bigl(-\,13.6\,Z^{2}\bigr)$$

$$\Delta E = 13.6\,Z^{2}\left(1 - \frac{1}{9}\right)$$

$$\Delta E = 13.6\,Z^{2}\left(\frac{8}{9}\right)$$

$$\Delta E = \frac{108.8\,Z^{2}}{9}\ \text{eV} \qquad -(4)$$

According to the question, this difference equals $$108.8\ \text{eV}$$, so from (4)

$$\frac{108.8\,Z^{2}}{9} = 108.8$$

$$\Rightarrow \frac{Z^{2}}{9} = 1$$

$$\Rightarrow Z^{2} = 9$$

$$\Rightarrow Z = 3$$

Thus the atomic number of the ion is $$Z = 3$$, which corresponds to Option D.

We are given that the work function of a metal is $$\phi = 3 \, eV$$. We need to find which color of visible light can cause emission of photoelectrons.

For photoelectron emission to occur, the energy of the incident photon must be at least equal to the work function of the metal:
$$E \geq \phi$$

The energy of a photon is related to its wavelength by:
$$E = \frac{hc}{\lambda}$$

where $$hc = 1240 \, eV \cdot nm$$.

For the minimum photon energy of $$3 \, eV$$, the maximum wavelength is:
$$\lambda_{max} = \frac{hc}{\phi} = \frac{1240}{3} \approx 413 \, nm$$

So, only light with wavelength $$\lambda \leq 413 \, nm$$ can cause photoemission.

Now let us check which color of visible light satisfies this condition:

- Red light: $$\lambda \approx 620 - 750 \, nm$$ (energy $$\approx 1.65 - 2.0 \, eV$$) — Not sufficient
- Yellow light: $$\lambda \approx 570 - 590 \, nm$$ (energy $$\approx 2.1 - 2.2 \, eV$$) — Not sufficient
- Green light: $$\lambda \approx 495 - 570 \, nm$$ (energy $$\approx 2.2 - 2.5 \, eV$$) — Not sufficient
- Blue light: $$\lambda \approx 400 - 495 \, nm$$ (energy $$\approx 2.5 - 3.1 \, eV$$) — Sufficient at shorter wavelengths

Blue light at its shorter wavelength end (around $$400 - 413 \, nm$$) has energy of about $$3.0 - 3.1 \, eV$$, which is equal to or greater than the work function of $$3 \, eV$$. Therefore, blue light can cause photoelectron emission.

Red, yellow, and green light all have energies less than $$3 \, eV$$ and cannot cause photoemission from this metal.

Hence, the correct answer is Option B.

The wavelengths of hydrogen spectral lines are obtained from the Rydberg‐formula,

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

where $$R$$ is the Rydberg constant, $$n_f$$ is the lower (final) level of the transition and $$n_i$$ is the upper (initial) level with $$n_i \gt n_f$$.

For any given series, the largest wavelength (smallest energy difference) comes from the transition having the smallest possible gap, that is $$n_i = n_f+1$$.

For the Lyman series, $$n_f = 1$$. Therefore take $$n_i = 2$$.

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

Hence $$\lambda_L=\frac{4}{3R}$$ $$-(1)$$

For the Balmer series, $$n_f = 2$$. Therefore take $$n_i = 3$$.

$$\frac{1}{\lambda_B}=R\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)=R\left(\frac{1}{4}-\frac{1}{9}\right)=R\left(\frac{5}{36}\right)$$

Hence $$\lambda_B=\frac{36}{5R}$$ $$-(2)$$

Taking the ratio of $$(1)$$ to $$(2)$$:

$$\frac{\lambda_L}{\lambda_B}=\frac{\dfrac{4}{3R}}{\dfrac{36}{5R}}=\frac{4}{3R}\times\frac{5R}{36}=\frac{20}{108}=\frac{5}{27}$$

Therefore, the required ratio is $$5 : 27$$.

Correct option: Option B (5 : 27)

The photoelectric effect is described by the equation: $$K_{\text{max}} = h\nu - \phi$$ where $$K_{\text{max}}$$ is the maximum kinetic energy of ejected electrons, $$h$$ is Planck’s constant, $$\nu$$ is the frequency of incident light, and $$\phi$$ is the work function of the metal.

Since frequency $$\nu$$ is related to wavelength $$\lambda$$ by $$\nu = \frac{c}{\lambda}$$ (where $$c$$ is the speed of light), the equation can be rewritten as: $$K_{\text{max}} = \frac{hc}{\lambda} - \phi$$.

In the first case, when the incident light has wavelength $$\lambda$$, the work function is $$\phi = 1 \text{ eV}$$ and the observed maximum kinetic energy is $$K_1 = 2 \text{ eV}$$. Substituting these values into the equation gives $$2 = \frac{hc}{\lambda} - 1$$, which rearranges to $$2 + 1 = \frac{hc}{\lambda}$$ and hence $$3 = \frac{hc}{\lambda}$$. Thus, $$\frac{hc}{\lambda} = 3 \text{ eV}$$ (1).

Next, when the wavelength of the incident light is halved to $$\frac{\lambda}{2}$$ while the work function remains $$\phi = 1 \text{ eV}$$, let the new maximum kinetic energy be $$K_2$$. The photoelectric equation then yields $$K_2 = \frac{hc}{\frac{\lambda}{2}} - \phi$$, which simplifies to $$K_2 = \frac{2hc}{\lambda} - \phi$$. Substituting $$\frac{hc}{\lambda} = 3 \text{ eV}$$ from (1) and $$\phi = 1 \text{ eV}$$ leads to $$K_2 = 2 \times 3 - 1$$, giving $$K_2 = 6 - 1$$ and therefore $$K_2 = 5 \text{ eV}$$.

Therefore, the maximum kinetic energy of ejected electrons when the surface is illuminated by light of wavelength $$\frac{\lambda}{2}$$ is 5 eV.

The options are: A. 3 eV B. 2 eV C. 6 eV D. 5 eV. The correct answer is D. 5 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

For a hydrogen-like ion, Bohr’s radius of the electron in the $$n^{\text{th}}$$ orbit is given by

$$r_n = \frac{n^{2}a_0}{Z}$$

where $$a_0$$ is the Bohr radius of the first orbit of the hydrogen atom and $$Z$$ is the atomic number of the nucleus.

Both $$Li^{2+}$$ and $$He^{+}$$ have a single electron, so the same formula applies.

Case 1: $$Li^{2+}$$
For lithium, $$Z_{Li}=3$$ and for the $$5^{\text{th}}$$ orbit, $$n=5$$.

$$r_{Li} = \frac{(5)^2\,a_0}{3}= \frac{25a_0}{3}$$

Case 2: $$He^{+}$$
For helium, $$Z_{He}=2$$ and again $$n=5$$.

$$r_{He} = \frac{(5)^2\,a_0}{2}= \frac{25a_0}{2}$$

Now compute the required ratio:

$$\frac{r_{Li}}{r_{He}} = \frac{\dfrac{25a_0}{3}}{\dfrac{25a_0}{2}} = \frac{2}{3}$$

Hence, the ratio of the radii is $$\dfrac{2}{3}$$.

Option D is correct.

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: $$n = 2$$.

The energy of an electron in a hydrogen atom is given by
$$E_n = -\frac{13.6\text{ eV}}{n^{2}}$$

“Fourth excited state’’ means the electron starts from the level
$$n_i = 5$$  (because $$n = 1$$ is ground, $$n = 2$$ first excited, …).

If the electron finally reaches an unknown level $$n_f$$ and emits a photon of energy $$E_{\gamma}=2.86\text{ eV}$$, then by energy conservation

$$E_{\gamma}=E_{n_i}-E_{n_f}$$

Substituting the formula for $$E_n$$ gives

$$2.86 = \left(-\frac{13.6}{5^{2}}\right) - \left(-\frac{13.6}{n_f^{2}}\right)$$

Simplify the brackets:

$$2.86 = 13.6\left(\frac{1}{n_f^{2}} - \frac{1}{25}\right)$$

Divide both sides by $$13.6$$ to isolate the bracketed term:

$$\frac{2.86}{13.6} = \frac{1}{n_f^{2}} - \frac{1}{25}$$

Calculate the numerical value:

$$\frac{2.86}{13.6} \approx 0.21$$

Add $$\frac{1}{25}=0.04$$ to both sides:

$$0.21 + 0.04 = \frac{1}{n_f^{2}}$$

$$\frac{1}{n_f^{2}} \approx 0.25$$

Taking reciprocal and square-root:

$$n_f^{2} \approx 4 \quad\Longrightarrow\quad n_f \approx 2$$

Since $$n_f$$ must be an integer quantum number, we take
$$n_f = 2$$.

Therefore, the electron drops to the second energy level. The required integer value of $$n$$ is 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.

UV light of energy 4.13 eV is incident on a metal surface with work function 3.13 eV. We need to find the maximum kinetic energy of ejected photoelectrons.

Recall Einstein's photoelectric equation.

$$ KE_{max} = h\nu - \phi $$

where $$h\nu$$ is the energy of the incident photon and $$\phi$$ is the work function of the metal.

Substitute the given values.

$$ KE_{max} = 4.13 - 3.13 = 1.00 \text{ eV} $$

Explanation: The work function $$\phi = 3.13$$ eV is the minimum energy required to eject an electron from the metal surface. Any photon energy in excess of the work function is converted into kinetic energy of the ejected electron. Since the incident photon has 4.13 eV and 3.13 eV is used to overcome the binding energy, the remaining 1.00 eV becomes the maximum kinetic energy of the photoelectron.

The correct answer is Option (3): 1 eV.

Ground state energy of hydrogen: $$E_1 = -13.6$$ eV. Energy given = 10.2 eV.

Total energy = $$-13.6 + 10.2 = -3.4$$ eV = $$-\frac{13.6}{n^2}$$. So $$n^2 = 4$$, $$n = 2$$.

The electron goes to $$n = 2$$. Spectral lines from transitions: $$n=2 \to n=1$$.

Number of spectral lines = $$\frac{n(n-1)}{2} = \frac{2 \times 1}{2} = 1$$.

The correct answer is Option 4: 1.

A proton and electron have the same de Broglie wavelength. Find the ratio of their kinetic energies.

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

Since both have the same $$\lambda$$, they have the same momentum $$p$$.

$$KE = \frac{p^2}{2m}$$

Since $$p$$ is the same for both:

$$\frac{KE_p}{KE_e} = \frac{p^2/(2m_p)}{p^2/(2m_e)} = \frac{m_e}{m_p} = \frac{m_e}{1836 \, m_e} = \frac{1}{1836}$$

$$KE_p : KE_e = 1 : 1836$$.

The correct answer is Option (4): $$1 : 1836$$.

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

Since both have the same de Broglie wavelength, they both will have the same momentum.

$$ KE = \frac{p^2}{2m} $$

$$KE$$ is inversely proportional to $$m$$.

$$ m_p > m_e $$

$$ K_p < K_e $$

Einstein's photoelectric equation: $$E_k = h\nu - \phi$$

Comparing with the slope-intercept form of a straight line ($$y = mx + c$$):

$$y = E_k, \quad x = \nu$$

$$\implies \text{Slope } (m) = h \quad (\text{Planck's constant})$$

$$\implies y\text{-intercept } (c) = -\phi \quad (-\text{Work function})$$

In a photoelectric experiment, light of energy 2.48 eV is incident on a photosensitive material, and the stopping potential is 0.5 V. We need to find the work function.

Recall Einstein's photoelectric equation.

$$ E_{photon} = \phi + KE_{max} $$

where $$\phi$$ is the work function and $$KE_{max} = eV_0$$ (stopping potential times electron charge).

Express in electron volts.

Since the stopping potential is 0.5 V, the maximum kinetic energy is $$eV_0 = 0.5$$ eV.

Solve for the work function.

$$ \phi = E_{photon} - KE_{max} = 2.48 - 0.5 = 1.98 \text{ eV} $$

The correct answer is Option (3): 1.98 eV.

For a particle of mass $$m$$ and kinetic energy $$E$$, the de-Broglie wavelength is given by

$$\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mE}} \quad -(1)$$
since the linear momentum $$p = \sqrt{2mE}$$ for non-relativistic speeds.

From $$(1)$$ we get the proportionality

$$\lambda \propto \frac{1}{\sqrt{m}} \quad -(2)$$
for particles that possess the same energy $$E$$.

Therefore, for equal energies, the particle with the smaller mass has the larger wavelength.

Masses of the given particles:

Electron mass $$m_e \approx 9.11 \times 10^{-31}\,\text{kg}$$
Proton mass $$m_p \approx 1.67 \times 10^{-27}\,\text{kg}$$
Alpha-particle mass $$m_\alpha \approx 4m_p \approx 6.68 \times 10^{-27}\,\text{kg}$$

Hence

$$m_e \lt m_p \lt m_\alpha$$

Using $$(2)$$, the corresponding de-Broglie wavelengths satisfy

$$\lambda_e \gt \lambda_p \gt \lambda_\alpha$$

Re-ordering to match the option format:

$$\lambda_\alpha \lt \lambda_p \lt \lambda_e$$

This matches Option A.

Therefore, the correct comparison is
Case : $$\lambda_\alpha \lt \lambda_p \lt \lambda_e$$, which is Option A.

Magnetic moment of electron in $$n$$th Bohr orbit: $$\mu_n = \frac{eL}{2m_e}$$ where $$L = n\hbar$$ is the angular momentum.

$$\mu_n = \frac{en\hbar}{2m_e} \propto n$$

So $$\mu_n \propto n^1$$, meaning $$x = 1$$.

The correct answer is Option 2: 1.

We need to evaluate the Assertion and Reason about photons and the photoelectric effect.

Assertion A: "Number of photons increases with increase in frequency of light."

The energy of each photon is $$E = h\nu$$. For a light source of constant power $$P$$, the number of photons per second is $$N = \frac{P}{h\nu}$$. As $$\nu$$ increases, each photon carries more energy, so fewer photons are emitted for the same power. Therefore, A is incorrect.

Reason R: "Maximum kinetic energy of emitted electrons increases with the frequency of incident radiation."

By Einstein's photoelectric equation: $$KE_{\max} = h\nu - \phi$$. This shows $$KE_{\max}$$ increases linearly with $$\nu$$. Therefore, R is correct.

The correct answer is Option (3): A is not correct but R is correct.

We need to determine what happens to photoelectron emission when frequency is halved and intensity is doubled.

Key Concept: Photoelectric Effect

Photoelectrons are emitted only when the frequency of incident light is greater than or equal to the threshold frequency ($$\nu_0$$). This is Einstein's photoelectric equation:

$$KE_{max} = h\nu - h\nu_0$$

Initial condition: Frequency = $$1.5\nu_0 > \nu_0$$ (above threshold, photoelectrons are emitted).

New condition: Frequency halved = $$\frac{1.5\nu_0}{2} = 0.75\nu_0$$.

Since $$0.75\nu_0 < \nu_0$$ (below threshold frequency), no photoelectrons will be emitted, regardless of the intensity. Increasing intensity only increases the number of photons, but each photon still has energy $$h \times 0.75\nu_0 < h\nu_0$$, which is insufficient to eject an electron.

The correct answer is Option 3: Zero.

We are asked to find the ratio of velocities of a proton and an alpha particle given their de Broglie wavelengths as $$\lambda$$ and $$2\lambda$$, respectively. The de Broglie relation is $$\lambda = \frac{h}{mv}$$, which implies $$v = \frac{h}{m\lambda}$$.

Thus, for a proton the velocity is $$v_p = \frac{h}{m_p\lambda}$$, while for an alpha particle with mass $$m_\alpha = 4m_p$$ and wavelength $$2\lambda$$, the velocity becomes $$v_\alpha = \frac{h}{m_\alpha \times 2\lambda} = \frac{h}{4m_p \times 2\lambda} = \frac{h}{8m_p\lambda}$$.

Taking the ratio gives $$\frac{v_p}{v_\alpha} = \frac{h/(m_p\lambda)}{h/(8m_p\lambda)} = \frac{8m_p\lambda}{m_p\lambda} = 8$$ so that $$v_p : v_\alpha = 8 : 1$$. The correct answer is Option D: 8 : 1.

We need to find the stopping potential when UV light of wavelength 300 nm falls on a metal with work function 2.13 eV.

Key Formula: The photoelectric equation (Einstein’s equation) is:

$$ E = W + eV_0 $$

Here $$E = \frac{hc}{\lambda}$$ is the energy of the incident photon, $$W$$ is the work function of the metal, and $$V_0$$ is the stopping potential. Rearranging this equation yields:

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

The energy of the incident photon is calculated as:

$$ E = \frac{hc}{\lambda} = \frac{1240 \, \text{eV}\cdot\text{nm}}{300 \, \text{nm}} = 4.133 \, \text{eV} $$

Substituting into the photoelectric equation to determine the stopping potential gives:

$$ eV_0 = E - W = 4.133 - 2.13 = 2.003 \, \text{eV} $$

Since $$eV_0$$ is expressed in electronvolts, the stopping potential $$V_0$$ is approximately 2 V.

The correct answer is Option C: 2 V.

We need to identify which phenomenon cannot be explained by the wave nature of light.

A. Reflection: Explained by wave theory (Huygens' principle). Each point on a wavefront acts as a source of secondary wavelets, and the reflected wavefront follows the law of reflection.

B. Diffraction: Explained by wave theory. Diffraction is the bending of waves around obstacles, a characteristic wave phenomenon described by Huygens-Fresnel principle.

C. Photoelectric effect: This cannot be explained by the wave nature of light. The wave theory predicts that any frequency of light should eject electrons if given enough time and intensity, but experimentally, there is a threshold frequency below which no electrons are emitted regardless of intensity. Einstein explained this using the particle (photon) nature of light: each photon carries energy $$h\nu$$, and only photons with energy greater than the work function can eject electrons.

D. Interference: Explained by wave theory. Constructive and destructive interference patterns arise from the superposition of waves.

E. Polarization: Explained by wave theory. Polarization demonstrates that light is a transverse wave.

The correct answer is Option (3): C only.

Question: Which statement about stopping potential ($$V_0$$) is NOT true?

In the photoelectric effect, the stopping potential $$V_0$$ is the minimum reverse potential needed to stop the most energetic photoelectrons. From Einstein's photoelectric equation: $$eV_0 = h\nu - \phi$$ where $$h$$ is Planck's constant, $$\nu$$ is the frequency of incident light, and $$\phi$$ is the work function of the metal.

Option 1: "It is $$\frac{1}{e}$$ times the maximum kinetic energy of electrons emitted." Since $$V_0 = \frac{KE_{max}}{e}$$, this is true by definition.

Option 2: "It increases with increase in intensity of the incident light." The stopping potential depends only on the frequency of light, not its intensity. Increasing intensity increases the number of photoelectrons (photocurrent) but does not change the maximum kinetic energy or stopping potential. This statement is NOT true.

Option 3: "It depends on the nature of emitter material." From $$V_0 = \frac{h\nu - \phi}{e}$$, and since the work function $$\phi$$ depends on the material, $$V_0$$ does depend on the emitter material. This is true.

Option 4: "It depends upon frequency of the incident light." From the equation, $$V_0$$ clearly depends on $$\nu$$. This is true.

The statement that is NOT true is Option 2.

According to Bohr's atomic theory, the moment of momentum (which is the angular momentum) of an electron revolving in a stationary orbit is quantized. It must be an integral multiple of $$ \frac{h}{2\pi} $$.

The formula for the moment of momentum $$ L $$ is:

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

Given that the electron is revolving in the 4th orbit:

$$ n = 4 $$

$$ L = \frac{4h}{2\pi} $$

$$ L = \frac{2h}{\pi} $$

Therefore, the moment of momentum of the electron in the 4th orbit of a hydrogen atom is $$ \frac{2h}{\pi} $$.

$$eV_0 = h\nu - \phi \implies V_0 = \left(\frac{h}{e}\right)\nu - \frac{\phi}{e}$$

$$\text{Slope} = \frac{h}{e} \implies \text{Statement I is correct}$$

$$\text{From graph, threshold frequency: } \nu_{02} > \nu_{01} \implies \phi_2 > \phi_1$$

$$K_{\text{max}} = h\nu - \phi \implies K_{\text{max1}} > K_{\text{max2}}\ \text{for same } \nu \implies \text{Statement II is incorrect}$$

We need to find how the angular momentum of an electron in a hydrogen atom depends on the radius of its orbit, using the Bohr model.

In Bohr's model, the angular momentum of the electron in the $$n$$-th orbit is quantised: $$L = mvr = n\hbar = \frac{nh}{2\pi}$$, where $$n$$ is the principal quantum number ($$n = 1, 2, 3, \ldots$$), $$m$$ is the electron mass, $$v$$ is the orbital speed, and $$r$$ is the orbital radius. This tells us that $$L \propto n$$.

In the Bohr model, the radius of the $$n$$-th orbit for hydrogen is given by $$r_n = a_0 n^2$$, where $$a_0 = 0.529$$ Angstrom is the Bohr radius, which shows that $$r \propto n^2$$.

From $$r \propto n^2$$, we can write $$n \propto \sqrt{r}$$.

Since $$L \propto n$$ and $$n \propto \sqrt{r}$$, it follows that $$L \propto \sqrt{r}$$.

The correct answer is Option (2): $$\sqrt{r}$$.

We need to find the longest wavelength associated with the Paschen series of hydrogen. The wavelengths of spectral lines in the hydrogen spectrum are given by the Rydberg formula: $$\frac{1}{\lambda} = R_H\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$$ where $$R_H = 1.097 \times 10^7 \,\text{m}^{-1}$$ is the Rydberg constant, $$n_1$$ is the lower energy level, and $$n_2$$ is the upper energy level.

For the Paschen series, transitions end at $$n_1 = 3$$, and the lines correspond to $$n_2 = 4, 5, 6, \ldots$$ The longest wavelength corresponds to the smallest energy transition, which is the transition from $$n_2 = 4$$ to $$n_1 = 3$$. Thus,

$$\frac{1}{\lambda} = R_H\left(\frac{1}{3^2} - \frac{1}{4^2}\right) = R_H\left(\frac{1}{9} - \frac{1}{16}\right)$$

Computing the difference of fractions gives

$$\frac{1}{9} - \frac{1}{16} = \frac{16 - 9}{144} = \frac{7}{144}$$

Substituting back into the Rydberg formula,

$$\frac{1}{\lambda} = 1.097 \times 10^7 \times \frac{7}{144} = \frac{7.679 \times 10^7}{144} = \frac{7.679}{144} \times 10^7$$

$$\frac{1}{\lambda} = 0.05333 \times 10^7 = 5.333 \times 10^5 \,\text{m}^{-1}$$

Solving for $$\lambda$$ yields

$$\lambda = \frac{144}{7 \times 1.097 \times 10^7} = \frac{144}{7.679 \times 10^7} = 18.752 \times 10^{-7} \,\text{m} = 1.875 \times 10^{-6} \,\text{m}$$

Rounding gives $$1.876 \times 10^{-6}$$ m, so the correct answer is Option B: $$1.876 \times 10^{-6}$$ m.

For Balmer series, the minimum energy transition is n=3 to n=2: E = 13.6(1/4-1/9) = 1.89 eV.

But to emit in Balmer series, atom must first reach n≥3 from ground state n=1: E = 13.6(1-1/9) = 12.09 ≈ 12.1 eV.

The correct answer is Option 4: 12.1 eV.

We need to find the ratio of the shortest wavelength of the Balmer series to the shortest wavelength of the Lyman series for hydrogen.

Recall the Rydberg formula:

$$ \frac{1}{\lambda} = R\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right) $$

For the Balmer series with $$n_1 = 2$$, the shortest wavelength (highest energy) occurs as $$n_2 \to \infty$$. Thus:

$$ \frac{1}{\lambda_B} = R\left(\frac{1}{4} - 0\right) = \frac{R}{4} \implies \lambda_B = \frac{4}{R} $$

For the Lyman series with $$n_1 = 1$$, the shortest wavelength occurs as $$n_2 \to \infty$$. Hence:

$$ \frac{1}{\lambda_L} = R\left(1 - 0\right) = R \implies \lambda_L = \frac{1}{R} $$

Therefore the ratio is given by:

$$ \frac{\lambda_B}{\lambda_L} = \frac{4/R}{1/R} = \frac{4}{1} = 4:1 $$

The correct answer is Option (1): 4:1.

The photoelectric effect equation at threshold (minimum frequency / maximum wavelength) is:

$$E = \phi = \frac{hc}{\lambda_{max}}$$

where $$\phi$$ is the work function, $$h$$ is Planck's constant, $$c$$ is the speed of light, and $$\lambda_{max}$$ is the longest wavelength that can cause photoemission.

Solving for $$\lambda_{max}$$:

$$\lambda_{max} = \frac{hc}{\phi}$$

Using the standard value $$hc = 1240 \text{ eV} \cdot \text{nm}$$:

$$\lambda_{max} = \frac{1240}{3.0} = 413.3 \text{ nm} \approx 414 \text{ nm}$$

The correct answer is $$414 \text{ nm}$$.

Photoelectric effect: $$eV_s = \frac{hc}{\lambda} - \phi$$.

$$8e = \frac{hc}{\lambda} - \phi$$ ... (1). $$2e = \frac{hc}{3\lambda} - \phi$$ ... (2).

From (1)-(2): $$6e = \frac{hc}{\lambda}(1-1/3) = \frac{2hc}{3\lambda} \Rightarrow \frac{hc}{\lambda} = 9e$$.

From (1): $$\phi = 9e - 8e = e$$. Threshold: $$\frac{hc}{\lambda_0} = \phi = e$$.

$$\lambda_0 = \frac{hc}{e} = 9\lambda \cdot \frac{e}{e} = 9\lambda$$.

The answer is Option (3): $$9\lambda$$.

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): $$\frac{27}{32}\lambda$$.

For a hydrogen atom, the energies of an electron in the $$n$$-th state are related by the virial theorem:

$$\text{Kinetic Energy (KE)} = -E_n$$

$$\text{Potential Energy (PE)} = 2E_n$$

$$\text{Total Energy} = E_n$$

where $$E_n$$ is the total energy of the electron in the $$n$$-th state (which is negative).

The ratio of the magnitude of kinetic energy to the magnitude of potential energy is:

$$\frac{|KE|}{|PE|} = \frac{|-E_n|}{|2E_n|} = \frac{|E_n|}{2|E_n|} = \frac{1}{2}$$

This ratio is independent of the quantum number $$n$$ and holds for any state, including the 5th excited state ($$n = 6$$).

The correct answer is $$\frac{1}{2}$$.

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

The energy of an electron in the $$n$$-th level of a hydrogen atom is given by $$E_n = -\frac{13.6}{n^2} \text{ eV}$$. When $$E_n = -0.85 \text{ eV}$$, we have $$-0.85 = -\frac{13.6}{n^2}$$ which leads to $$n^2 = \frac{13.6}{0.85} = 16$$ and hence $$n = 4$$, so the electron occupies the $$n = 4$$ state.

From $$n = 4$$, the electron can transition to any lower energy level $$n = 3, 2, 1$$. The maximum number of allowed transitions (spectral lines) is $$\binom{n}{2} = \binom{4}{2} = \frac{4!}{2! \times 2!} = 6$$, corresponding to the transitions $$4 \to 3$$, $$4 \to 2$$, $$4 \to 1$$, $$3 \to 2$$, $$3 \to 1$$, and $$2 \to 1$$. Therefore, the maximum number of allowed transitions is $$6$$.

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, $$\alpha = 121$$.

We need to find the wavelength of light emitted when hydrogen is excited to the first excitation level (10.2 eV).

In the Franck-Hertz experiment, the first dip at 10.2 V corresponds to the energy required to excite hydrogen from the ground state $$n = 1$$ to the first excited state $$n = 2$$, and when the atom de-excites back to $$n = 1$$, it emits a photon with energy equal to 10.2 eV.

The energy-wavelength relationship for a photon is given by

$$E = \frac{hc}{\lambda} \implies \lambda = \frac{hc}{E}$$

First, the energy of the emitted photon is $$E = 10.2 \, \text{eV}$$.

Next, using the value $$hc = 1245 \, \text{eV·nm}$$, we find

$$\lambda = \frac{1245}{10.2} = 122.06 \approx 122 \, \text{nm}$$

This corresponds to the Lyman-alpha line of hydrogen (the $$n = 2 \to n = 1$$ transition), which lies in the ultraviolet region.

The correct answer is 122 nm.

$$r_n = n^2 a_0$$. Given $$r = 8.48$$ Å: $$n^2 = 8.48/0.529 = 16 \Rightarrow n = 4$$.

$$E_n = E/n^2 = E/16$$. So $$x = 16$$.

The answer is 16.

We are given that the shortest wavelength of the Lyman series is 915 Angstrom, and we need to find the longest wavelength of the Balmer series.

Recall the Rydberg formula for hydrogen spectral lines

The wavelength of spectral lines in hydrogen is given by:

$$\frac{1}{\lambda} = R\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$$

where $$R$$ is the Rydberg constant, $$n_1$$ is the lower energy level, and $$n_2$$ is the upper energy level.

Find the Rydberg constant from the Lyman series data

The Lyman series corresponds to transitions where $$n_1 = 1$$. The shortest wavelength (series limit) occurs when $$n_2 \to \infty$$:

$$\frac{1}{\lambda_{\infty}} = R\left(\frac{1}{1^2} - \frac{1}{\infty}\right) = R$$

Given $$\lambda_{\infty} = 915$$ Angstrom:

$$R = \frac{1}{915} \text{ per Angstrom}$$

Calculate the longest wavelength of the Balmer series

The Balmer series corresponds to transitions where $$n_1 = 2$$. The longest wavelength (smallest energy transition) occurs when $$n_2 = 3$$ (the closest upper level):

$$\frac{1}{\lambda} = R\left(\frac{1}{2^2} - \frac{1}{3^2}\right) = R\left(\frac{1}{4} - \frac{1}{9}\right)$$

Finding the common denominator:

$$\frac{1}{4} - \frac{1}{9} = \frac{9 - 4}{36} = \frac{5}{36}$$

Therefore:

$$\frac{1}{\lambda} = R \times \frac{5}{36} = \frac{1}{915} \times \frac{5}{36}$$

$$\lambda = \frac{36}{5R} = \frac{36 \times 915}{5} = \frac{32940}{5} = 6588 \text{ Angstrom}$$

The answer is 6588 Angstrom.

For any one-electron (hydrogen-like) species, Bohr’s model gives the total energy in the $$n^{\text{th}}$$ orbit as
$$E_n = -\frac{13.6\,\text{eV}\;Z^{2}}{n^{2}}$$
where $$Z$$ is the atomic (nuclear) charge.

The kinetic energy in that orbit is related to the total energy by
$$KE_n = -E_n$$
(because in Bohr’s model $$E_n = KE_n + PE_n$$ with $$PE_n = -2\,KE_n$$).

Hence
$$KE_n = \frac{13.6\,\text{eV}\;Z^{2}}{n^{2}} \quad -(1)$$
so $$KE_n$$ is directly proportional to $$\dfrac{Z^{2}}{n^{2}}$$.

Now evaluate $$\dfrac{Z^{2}}{n^{2}}$$ for each option:

Option A: $$H$$ atom  ($$Z = 1,\; n = 1$$)
$$\dfrac{Z^{2}}{n^{2}} = \dfrac{1^{2}}{1^{2}} = 1$$

Option B: $$He^{+}$$  ($$Z = 2,\; n = 1$$)
$$\dfrac{Z^{2}}{n^{2}} = \dfrac{2^{2}}{1^{2}} = 4$$

Option C: $$He^{+}$$  ($$Z = 2,\; n = 2$$)
$$\dfrac{Z^{2}}{n^{2}} = \dfrac{2^{2}}{2^{2}} = 1$$

Option D: $$Li^{2+}$$  ($$Z = 3,\; n = 2$$)
$$\dfrac{Z^{2}}{n^{2}} = \dfrac{3^{2}}{2^{2}} = \dfrac{9}{4} = 2.25$$

The largest value is $$4$$ for Option B, meaning the electron in the first orbit of $$He^{+}$$ possesses the highest kinetic energy.

Therefore, the correct choice is:
Option B which is: First orbit of $$He^{+}$$

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

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

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

Statement I: "Out of microwaves, infrared rays and ultraviolet rays, ultraviolet rays are the most effective for the emission of electrons from a metallic surface."

This is true. UV rays have the highest frequency (and hence highest photon energy) among these three. Higher energy photons are more effective at causing electron emission (photoelectric effect).

Statement II: "Above the threshold frequency, the maximum kinetic energy of photoelectrons is inversely proportional to the frequency of the incident light."

This is false. From Einstein's photoelectric equation: $$KE_{max} = h\nu - \phi$$. The maximum KE is directly (linearly) proportional to frequency, not inversely proportional.

Statement I is true, Statement II is false.

This matches option 2.

Assertion A: Electrons exhibit wave nature, showing interference and diffraction patterns. This is TRUE, as demonstrated experimentally.

Reason R: The Davisson-Germer experiment (1927) confirmed the wave nature of electrons by observing diffraction of an electron beam from a nickel crystal. This is TRUE and directly verifies the assertion.

Both A and R are correct, and R is the correct explanation of A, since the Davisson-Germer experiment provided the experimental verification of the wave nature of electrons stated in A.

$$\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mE_k}} \implies \lambda \propto \frac{1}{\sqrt{mE_k}}$$

Let the mass of an electron be $$m_e$$, a proton be $$m_p$$, and an $$\alpha$$-particle be $$m_\alpha$$. The standard mass relationships are:  $$m_e \ll m_p \quad \text{and} \quad m_\alpha \approx 4m_p$$

Given kinetic energies: $$E_e = 4K, \quad E_\alpha = 2K, \quad E_p = K$$

For electron: $$m_e \cdot E_e = m_e \cdot 4K = 4m_e K$$

For proton: $$m_p \cdot E_p = m_p \cdot K = m_p K$$

For $$\alpha$$-particle: $$m_\alpha \cdot E_\alpha = (4m_p) \cdot 2K = 8m_p K$$

Since $$m_e$$ is extremely small compared to $$m_p$$ ($$m_p \approx 1836 m_e$$): 

$$4m_e K \ll m_p K < 8m_p K$$ $$\implies (mE_k)_e < (mE_k)_p < (mE_k)_\alpha$$

Taking the inverse square root reverses the inequalities: $$\lambda_e > \lambda_p > \lambda_a \implies \lambda_\alpha < \lambda_p < \lambda_e$$

Threshold wavelength = 5500 Å. We need to identify which sources cause photoelectric emission.

Key Concept: Photoelectric emission occurs only when the incident light has wavelength less than or equal to the threshold wavelength. The power (wattage) doesn't matter — only the wavelength determines whether emission occurs.

Analysis:

- Infrared radiation has wavelength > 5500 Å (typically > 7000 Å). Since this exceeds the threshold, no emission occurs regardless of power.

- Ultraviolet radiation has wavelength < 4000 Å, which is less than 5500 Å. So emission occurs regardless of power.

Therefore:

A. 75 W infrared lamp — NO emission

B. 10 W infrared lamp — NO emission

C. 75 W ultraviolet lamp — YES emission

D. 10 W ultraviolet lamp — YES emission

The correct answer is Option 4: C and D only.

The answer is $$\boxed{\text{C and D only}}$$.

We need to find the work function of a metal surface from the variation of stopping potential $$V_0$$ as a function of the frequency $$\nu$$ of incident light.

Einstein’s photoelectric equation relates the maximum kinetic energy of emitted electrons to the frequency of incident light:

$$eV_0 = h\nu - \phi$$

Here, $$e$$ is the electron charge, $$V_0$$ is the stopping potential, $$h$$ is Planck’s constant, $$\nu$$ is the frequency of incident light, and $$\phi$$ is the work function of the metal.

Rearranging this equation gives $$V_0 = \frac{h}{e}\nu - \frac{\phi}{e}$$, which is a linear relation in $$\nu$$ with slope $$\frac{h}{e}$$ and y-intercept $$-\frac{\phi}{e}$$.

By examining the graph of $$V_0$$ versus $$\nu$$, the threshold frequency $$\nu_0$$, at which $$V_0 = 0$$, corresponds to the x-intercept. At this point, the work function satisfies:

$$\phi = h\nu_0$$

Using Planck’s constant $$h = 6.63 \times 10^{-34}\,\text{J}\,\text{s}$$ and converting joules to electronvolts via $$1\,\text{eV} = 1.6 \times 10^{-19}\,\text{J}$$, and noting from the graph that $$\nu_0 = 5 \times 10^{14}\,\text{Hz}$$, we calculate:

$$\phi = h\nu_0 = 6.63 \times 10^{-34} \times 5 \times 10^{14} = 33.15 \times 10^{-20}\,\text{J}$$

Converting this energy into electronvolts gives:

$$\phi = \frac{33.15 \times 10^{-20}}{1.6 \times 10^{-19}} = \frac{33.15}{16} \approx 2.07\,\text{eV}$$

Therefore, the work function of the metal surface is 2.07 $$\text{ eV}$$.

Using Einstein's photoelectric equation for the two given wavelengths:

$$\frac{hc}{\lambda} = \phi + eV_0 \quad \text{...(i)}$$

$$\frac{hc}{2\lambda} = \phi + \frac{eV_0}{4} \quad \text{...(ii)}$$

From equation (i), we get $$eV_0 = \frac{hc}{\lambda} - \phi$$. Substituting into equation (ii):

$$\frac{hc}{2\lambda} = \phi + \frac{1}{4}\left(\frac{hc}{\lambda} - \phi\right)$$

$$\frac{hc}{2\lambda} = \phi + \frac{hc}{4\lambda} - \frac{\phi}{4}$$

$$\frac{hc}{2\lambda} = \frac{3\phi}{4} + \frac{hc}{4\lambda}$$

Now, solving for $$\phi$$:

$$\frac{hc}{2\lambda} - \frac{hc}{4\lambda} = \frac{3\phi}{4}$$

$$\frac{hc}{4\lambda} = \frac{3\phi}{4}$$

$$\phi = \frac{hc}{3\lambda}$$

Since the work function is also $$\phi = \frac{hc}{\lambda_0}$$ (where $$\lambda_0$$ is the threshold wavelength), we get:

$$\frac{hc}{3\lambda} = \frac{hc}{\lambda_0}$$

$$\lambda_0 = 3\lambda$$

So, the answer is $$3\lambda$$.

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

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

$$\frac{\lambda_p}{\lambda_{\alpha}} = \sqrt{\frac{m_{\alpha} q_{\alpha} V_{\alpha}}{m_p q_p V_p}}$$

$$\frac{\lambda_p}{\lambda_{\alpha}} = \sqrt{\frac{(4m_p)(2q_p)(4)}{(m_p)(q_p)(2)}} = \sqrt{\frac{32}{2}} = \sqrt{16} = 4$$

We have a proton and an alpha particle with the same de Broglie wavelength $$\lambda$$. The de Broglie wavelength is given by:

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

Since both particles have the same wavelength $$\lambda$$, they have the same momentum: $$p_{proton} = p_{\alpha} = \frac{h}{\lambda}$$.

Now, the kinetic energy in terms of momentum is:

$$KE = \frac{p^2}{2m}$$

Since the momentum is the same for both particles:

$$\frac{KE_p}{KE_{\alpha}} = \frac{\frac{p^2}{2m_p}}{\frac{p^2}{2m_{\alpha}}} = \frac{m_{\alpha}}{m_p}$$

The mass of an alpha particle is approximately 4 times the mass of a proton ($$m_{\alpha} = 4m_p$$), so:

$$\frac{KE_p}{KE_{\alpha}} = \frac{4m_p}{m_p} = \frac{4}{1}$$

Hence, the ratio of kinetic energy of the proton to that of the alpha particle is $$4 : 1$$.

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

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

$$m_e < m_p < m_\alpha$$

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

The de Broglie wavelength of a particle accelerated through a voltage $$V$$ is given by the relation $$\lambda = \frac{h}{\sqrt{2meV}}$$, so that $$\lambda \propto \frac{1}{\sqrt{V}}$$.

When the accelerating voltage changes from 20 kV to 40 kV, the ratio of the new wavelength $$\lambda'$$ to the original $$\lambda_0$$ becomes $$\frac{\lambda'}{\lambda_0} = \sqrt{\frac{20}{40}} = \frac{1}{\sqrt{2}}$$.

It follows that $$\lambda' = \frac{\lambda_0}{\sqrt{2}}$$.

The correct answer is Option D: $$\frac{\lambda_0}{\sqrt{2}}$$.

A: Stopping potential depends on frequency AND work function, not just work function. FALSE.

B: Saturation current increases with intensity of light. TRUE.

C: Maximum KE depends on frequency, not intensity. FALSE.

D: Photoelectric effect cannot be explained by wave theory. FALSE.

Only B is correct.

Evaluate each statement about the photoelectric effect:

Statement A: The stopping potential depends only on the work function of the metal.

From Einstein's equation: $$eV_0 = h\nu - \phi$$. The stopping potential depends on both the frequency of incident light ($$\nu$$) and the work function ($$\phi$$). It does NOT depend only on the work function. Statement A is incorrect.

Statement B: The saturation current increases as the intensity of incident light increases.

Higher intensity means more photons per second, which means more photoelectrons are emitted, leading to a higher saturation current. Statement B is correct.

Statement C: The maximum kinetic energy of a photoelectron depends on the intensity of the incident light.

The maximum kinetic energy $$KE_{max} = h\nu - \phi$$ depends on frequency, not intensity. Increasing intensity increases the number of electrons but not their maximum energy. Statement C is incorrect.

Statement D: Photoelectric effect can be explained using the wave theory of light.

The wave theory fails to explain key features of the photoelectric effect (instantaneous emission, frequency threshold, independence of $$KE_{max}$$ from intensity). It requires the particle (quantum) theory. Statement D is incorrect.

Only Statement B is correct, which corresponds to Option 3.

Two metals A and B are exposed to radiation of wavelength $$\lambda = 350$$ nm. Work functions are $$\phi_A = 4.8$$ eV and $$\phi_B = 2.2$$ eV.

$$ E = \frac{hc}{\lambda} = \frac{1240 \text{ eV·nm}}{350 \text{ nm}} = 3.54 \text{ eV} $$

For photoelectron emission, the photon energy must exceed the work function: $$E \geq \phi$$.

For metal A: $$E = 3.54$$ eV $$< \phi_A = 4.8$$ eV → Metal A will NOT emit photoelectrons.

For metal B: $$E = 3.54$$ eV $$> \phi_B = 2.2$$ eV → Metal B WILL emit photoelectrons.

Only metal A fails to emit photoelectrons, while metal B successfully emits them.

The correct answer is Option D: Metal A will not emit photo-electrons.

Let us analyze each statement about the photoelectric effect:

A. The photocurrent is proportional to the intensity of the incident radiation.

This is correct. Higher intensity means more photons per second, which means more photoelectrons are emitted, leading to higher photocurrent.

B. Maximum kinetic energy with which photoelectrons are emitted depends on the intensity of incident light.

This is incorrect. The maximum KE depends on the frequency, not the intensity.

C. Max K.E. with which photoelectrons are emitted depends on the frequency of incident light.

This is correct. From Einstein's equation: $$KE_{max} = h\nu - \phi$$, where $$\nu$$ is the frequency.

D. The emission of photoelectrons require a minimum threshold intensity of incident radiation.

This is incorrect. Photoelectron emission requires a minimum threshold frequency, not intensity.

E. Max K.E. of the photoelectrons is independent of the frequency of the incident light.

This is incorrect. As seen from Einstein's equation, max KE depends on frequency.

The correct statements are A and C only.

The correct answer is Option 3.

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

For the proton and electron to have the same de Broglie wavelength, we set $$ \lambda_p = \lambda_e $$ which implies $$ \frac{h}{p_p} = \frac{h}{p_e} $$ and hence $$ p_p = p_e $$. Therefore, the ratio of their momenta is $$ \frac{p_p}{p_e} = \frac{1}{1} $$. The correct answer is 1 : 1.

The de Broglie wavelength is: $$\lambda = \frac{h}{mv} = \frac{h}{\sqrt{2mKE}}$$

For gas molecules, the average KE is: $$KE = \frac{3}{2}k_BT$$

So: $$\lambda = \frac{h}{\sqrt{3mk_BT}} \propto \frac{1}{\sqrt{T}}$$

$$\frac{\lambda_2}{\lambda_1} = \sqrt{\frac{T_1}{T_2}} = \sqrt{\frac{300}{600}} = \frac{1}{\sqrt{2}}$$

$$\lambda_2 = \frac{\lambda_1}{\sqrt{2}}$$

The correct answer is Option 3.

Given $$\phi_A = 9$$ eV, $$\phi_B = 4.5$$ eV and $$hc = 1242$$ eV·nm, the threshold wavelength $$\lambda_0$$ is given by $$\lambda_0 = \frac{hc}{\phi}\,.$$

Substituting these values yields $$\lambda_A = \frac{1242}{9} = 138$$ nm and $$\lambda_B = \frac{1242}{4.5} = 276$$ nm.

Therefore, $$\lambda_B - \lambda_A = 276 - 138 = 138$$ nm, so the correct answer is Option D: $$138$$ nm.

The energy of a photon emitted during a transition is given by Planck's relation: $$\Delta E = \frac{hc}{\lambda} \implies \lambda \propto \frac{1}{\Delta E}$$

To find the shortest wavelength ($$\lambda_{\text{min}}$$), we look for the transition with the largest energy difference ($$\Delta E_{\text{max}}$$).

Energy levels of a hydrogen atom ($$E_n = -\frac{13.6}{n^2}\ \text{eV}$$):

$$E_1 = -13.6\ \text{eV}$$,  $$E_2 = -3.4\ \text{eV}$$,  $$E_3 = -1.51\ \text{eV}$$,  $$E_4 = -0.85\ \text{eV}$$

Transition A ($$n=4 \rightarrow n=3$$): $$\Delta E = -0.85 - (-1.51) = 0.66\ \text{eV}$$

Transition B ($$n=4 \rightarrow n=2$$): $$\Delta E = -0.85 - (-3.4) = 2.55\ \text{eV}$$

Transition C ($$n=3 \rightarrow n=2$$): $$\Delta E = -1.51 - (-3.4) = 1.89\ \text{eV}$$

Transition D ($$n=3 \rightarrow n=1$$): $$\Delta E = -1.51 - (-13.6) = 12.09\ \text{eV}$$

Since transition D involves the ground state ($$n=1$$), it has the largest energy gap, which results in the shortest emitted wavelength.

The de-Broglie wavelength is given by:

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

where $$K$$ is the kinetic energy.

For same kinetic energy:

$$\frac{\lambda_p}{\lambda_e} = \sqrt{\frac{m_e}{m_p}}$$

Given $$m_p = 1849 \times m_e$$:

$$\frac{\lambda_p}{\lambda_e} = \sqrt{\frac{m_e}{1849 \times m_e}} = \frac{1}{\sqrt{1849}} = \frac{1}{43}$$

The ratio $$\lambda_p : \lambda_e = 1 : 43$$.

Using Einstein's photoelectric equation: $$KE_{max} = h\nu - h\nu_0$$

For the first case, $$\nu = 2f_0$$

For the second case, $$\nu = 5f_0$$

Taking the ratio:

Therefore, $$v_1 : v_2 = 1 : 2$$.

$$\text{Energy levels of a hydrogen atom: } E_n = -\frac{13.6}{n^2}\text{ eV}$$

$$E_1 = -13.6\text{ eV}, \quad E_2 = -\frac{13.6}{4} = -3.4\text{ eV}, \quad E_3 = -\frac{13.6}{9} \approx -1.51\text{ eV}, \quad E_4 = -\frac{13.6}{16} \approx -0.85\text{ eV}$$

$$\text{Maximum energy of the excited state from the ground state } (n=1)\text{: } E_{\text{max}} = E_1 + \Delta E = -13.6 + 12.5 = -1.1\text{ eV}$$

$$\text{Since } E_3 = -1.51\text{ eV} > -1.1\text{ eV} \text{ and } E_4 = -0.85\text{ eV} < -1.1\text{ eV, electrons can only be excited up to } n = 3.$$

$$\text{Number of emitted spectral lines: } N = \frac{n(n - 1)}{2} = \frac{3(3 - 1)}{2} = 3$$

$$\Delta E = E_{n_i} - E_{n_f} = 13.6 \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \text{ eV}$$

$$\Delta E = \frac{hc}{\lambda} \implies \lambda = \frac{hc}{\Delta E}$$

$$\Delta E = 13.6 \left( \frac{1}{1^2} - \frac{1}{4^2} \right)$$

$$\Delta E = 13.6 \left( \frac{15}{16} \right) = 12.75 \text{ eV}$$

$$hc = (4 \times 10^{-15} \text{ eV}\cdot\text{s}) \times (3 \times 10^8 \text{ m/s})$$

$$hc = 12 \times 10^{-7} \text{ eV}\cdot\text{m}$$

$$hc = 12 \times 10^{-7} \times 10^9 \text{ eV}\cdot\text{nm} = 1200 \text{ eV}\cdot\text{nm}$$

$$\lambda = \frac{1200 \text{ eV}\cdot\text{nm}}{12.75 \text{ eV}}$$

$$\lambda \approx 94.117 \text{ nm}$$

We need to find the energy released when an electron in a hydrogen-like atom with $$Z = 4$$ jumps from the $$4^{th}$$ to the $$2^{nd}$$ energy state.

The energy of the $$n^{th}$$ level in a hydrogen-like atom is given by $$E_n = -\dfrac{Rch \cdot Z^2}{n^2}$$, where $$Rch = 13.6$$ eV. The energy released during a transition from level $$n_2$$ to level $$n_1$$ can be calculated by $$\Delta E = Rch \cdot Z^2 \left(\dfrac{1}{n_1^2} - \dfrac{1}{n_2^2}\right)$$. Substituting $$n_1 = 2$$ and $$n_2 = 4$$ leads to $$\Delta E = 13.6 \times 4^2 \times \left(\dfrac{1}{4} - \dfrac{1}{16}\right) = 13.6 \times 16 \times \dfrac{3}{16} = 13.6 \times 3 = 40.8$$ eV.

The correct answer is Option D: 40.8 eV.

In photoelectric effect,

Stopping potential is related to maximum kinetic energy by:

$$eV_0=K_{\max}$$

and

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

where
$$\nu$$ = frequency of incident light
$$\phi$$= work function of the metal.

Since stopping potential depends on $$⁡K_{\max}$$​, it depends on frequency (or wavelength), not on intensity/power of light.

Therefore, Statement I is correct.

Also,

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

and

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

So maximum kinetic energy depends on wavelength of incident light.

Therefore, Statement II is also correct.

We need to find the speed of an electron in the 3rd orbit of hydrogen, given that the speed in the 7th orbit is $$3.6 \times 10^6$$ m/s.

Speed in Bohr's orbit:

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

$$v_n = \frac{v_0}{n}$$

where $$v_0$$ is a constant (speed in the first orbit).

This means speed is inversely proportional to the orbit number: $$v \propto \frac{1}{n}$$.

Using the ratio:

$$\frac{v_3}{v_7} = \frac{7}{3}$$

$$v_3 = v_7 \times \frac{7}{3} = 3.6 \times 10^6 \times \frac{7}{3} = 8.4 \times 10^6$$ m/s

The speed of the electron in the 3rd orbit is $$8.4 \times 10^6$$ m/s.

The correct answer is Option 4: $$8.4 \times 10^6$$ m/s.

In Bohr's model, the angular momentum of an electron in the $$n^{th}$$ orbit is:

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

For the first orbit: $$L_1 = \frac{h}{2\pi} = L$$ (given as L)

For the second orbit: $$L_2 = \frac{2h}{2\pi} = 2L$$

Change in angular momentum: $$\Delta L = L_2 - L_1 = 2L - L = L$$

The correct answer is Option 3: L.

The de Broglie wavelength: $$\lambda = \frac{h}{\sqrt{2mE}}$$

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

If KE becomes $$E/4$$:

$$\lambda' = \frac{h}{\sqrt{2m(E/4)}} = \frac{h}{\frac{1}{2}\sqrt{2mE}} = 2 \times \frac{h}{\sqrt{2mE}} = 2\lambda$$

The new wavelength is $$2\lambda$$.

This matches option 4.

The energy of a hydrogen-like ion in state $$n$$ is:

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

For He$$^+$$: $$Z = 2$$. The first excited state means $$n = 2$$.

$$E_2 = -\frac{13.6 \times 4}{4} = -13.6 \text{ eV}$$

Given that the radius of the electron's second stationary orbit in Bohr's atom is R, we note that in Bohr's model the radius of the n-th orbit is proportional to n^2: $$ r_n = r_0 \cdot n^2 $$ where $$r_0$$ is the radius of the first orbit.

For n = 2, this gives $$ r_2 = r_0 \times 4 = R $$, and thus $$ r_0 = \frac{R}{4} $$.

Substituting into the expression for the third orbit (n = 3) yields $$ r_3 = r_0 \times 9 = \frac{R}{4} \times 9 = \frac{9R}{4} = 2.25R $$. Therefore, the radius of the third orbit is 2.25R.

We need the ratio of slopes of stopping potential vs frequency plots for Gold and Aluminium.

State the photoelectric equation.

Einstein's photoelectric equation: $$eV_0 = h\nu - \phi$$

$$V_0 = \dfrac{h}{e}\nu - \dfrac{\phi}{e}$$

Identify the slope.

The slope of $$V_0$$ vs $$\nu$$ plot is $$\dfrac{h}{e}$$, which is a universal constant independent of the material.

Therefore, the ratio of slopes = $$\dfrac{h/e}{h/e} = 1$$.

This corresponds to Option 3.

Solution :

Energy of photon emitted is equal to the difference between the energy levels.

Using,

E = hc/λ

Given :

h = 6.62 × 10⁻³⁴ J s

c = 3 × 10⁸ m/s

λ = 124.1 nm = 124.1 × 10⁻⁹ m

Substituting,

E = (6.62 × 10⁻³⁴ × 3 × 10⁸)/(124.1 × 10⁻⁹)

E ≈ 1.6 × 10⁻¹⁸ J

Now,

1 eV = 1.6 × 10⁻¹⁹ J

Therefore,

E = (1.6 × 10⁻¹⁸)/(1.6 × 10⁻¹⁹)

= 10 eV

From the figure, transition from 0 eV to −10 eV has energy difference of 10 eV.

Hence, transition D will emit photon of wavelength 124.1 nm.

Final Answer :

D

Find the ratio $$\frac{r_2}{r_1}$$ where $$r_1$$ is the radius of the 2nd orbit of $$He^+$$ and $$r_2$$ is the radius of the 4th orbit of $$Be^{3+}$$.

$$r_n = a_0 \cdot \frac{n^2}{Z}$$

where $$a_0$$ is the Bohr radius, $$n$$ is the principal quantum number, and $$Z$$ is the atomic number.

$$r_1 = a_0 \cdot \frac{2^2}{2} = a_0 \cdot \frac{4}{2} = 2a_0$$

$$r_2 = a_0 \cdot \frac{4^2}{4} = a_0 \cdot \frac{16}{4} = 4a_0$$

$$\frac{r_2}{r_1} = \frac{4a_0}{2a_0} = 2$$

So $$x : 1 = 2 : 1$$, giving $$x = 2$$.

The answer is $$\boxed{2}$$.

Solution :

De-Broglie wavelength is given by :

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

For α-particle :

Mass,

$$m_\alpha = 4m_p$$

Charge,

$$q_\alpha = 2e$$

Therefore,

$$\lambda_\alpha = \frac{h}{\sqrt{2(4m_p)(2e)V}}$$

$$= \frac{h}{\sqrt{16m_peV}}$$

$$= \frac{h}{4\sqrt{m_peV}}$$

For proton :

$$\lambda_p = \frac{h}{\sqrt{2m_peV}}$$

Comparing with :

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

we get,

$$m = 8$$

Final Answer :

$$8$$

We have a photon of energy 12.75 eV incident on a hydrogen atom in the ground state. The energy levels of hydrogen are given by

The ground state energy is $$E_1 = -13.6$$ eV. After absorbing 12.75 eV, the energy becomes

Now finding the excited state,

The angular momentum in the $$n$$-th orbit by Bohr's postulate is

Substituting $$h = 4.14 \times 10^{-15}$$ eVs,

Comparing with $$\frac{x}{\pi} \times 10^{-17}$$ eVs, we get $$x = 828$$. So, the answer is $$828$$.

When hydrogen atoms absorb light and subsequently emit radiation of 6 different wavelengths, we use the formula for number of spectral lines:

$$\text{Number of lines} = \frac{n(n-1)}{2}$$

Setting this equal to 6:

$$\frac{n(n-1)}{2} = 6 \implies n(n-1) = 12 \implies n = 4$$

So the hydrogen atoms are excited from the ground state ($$n = 1$$) to $$n = 4$$.

Energy required:

$$E = 13.6\left(\frac{1}{1^2} - \frac{1}{4^2}\right) = 13.6\left(1 - \frac{1}{16}\right) = 13.6 \times \frac{15}{16}$$

$$E = \frac{204}{16} = 12.75 \text{ eV}$$

Frequency of incident light:

$$f = \frac{E}{h} = \frac{12.75}{4.25 \times 10^{-15}} = 3 \times 10^{15} \text{ Hz}$$

Therefore, $$x = 3$$.

The atom absorbs a photon of wavelength 500 nm and emits a photon of wavelength 600 nm.

Energy absorbed: $$E_1 = \frac{hc}{\lambda_1} = \frac{6.6 \times 10^{-34} \times 3 \times 10^8}{500 \times 10^{-9}}$$

$$E_1 = \frac{19.8 \times 10^{-26}}{5 \times 10^{-7}} = 3.96 \times 10^{-19} \text{ J}$$

Energy emitted: $$E_2 = \frac{hc}{\lambda_2} = \frac{6.6 \times 10^{-34} \times 3 \times 10^8}{600 \times 10^{-9}}$$

$$E_2 = \frac{19.8 \times 10^{-26}}{6 \times 10^{-7}} = 3.30 \times 10^{-19} \text{ J}$$

Net energy absorbed:

$$\Delta E = E_1 - E_2 = (3.96 - 3.30) \times 10^{-19} = 0.66 \times 10^{-19} \text{ J}$$

Converting to eV: $$\Delta E = \frac{0.66 \times 10^{-19}}{1.6 \times 10^{-19}} = 0.4125 \text{ eV}$$

$$\Delta E = 4125 \times 10^{-4} \text{ eV}$$

Therefore $$n = \mathbf{4125}$$.

$$n_A = 2, \quad n_B = 3, \quad n_C = 4$$

$$\frac{1}{\lambda_1} = R \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R \left( \frac{9 - 4}{36} \right) = \frac{5R}{36}$$

$$\lambda_1 = \frac{36}{5R}$$

$$\frac{1}{\lambda_2} = R \left( \frac{1}{3^2} - \frac{1}{4^2} \right) = R \left( \frac{16 - 9}{144} \right) = \frac{7R}{144}$$

$$\lambda_2 = \frac{144}{7R}$$

$$\frac{\lambda_1}{\lambda_2} = \frac{\frac{36}{5R}}{\frac{144}{7R}}$$

$$\frac{\lambda_1}{\lambda_2} = \frac{36}{5} \times \frac{7}{144}$$

$$\frac{\lambda_1}{\lambda_2} = \frac{7}{5 \times 4} = \frac{7}{20}$$

$$\frac{7}{20} = \frac{7}{4n}$$

$$4n = 20$$

$$n = 5$$

Solution :

For hydrogen atom,

$$\frac{1}{\lambda} = R\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$$

Transition 1 :

$$n_2 = 3,\ n_1 = 1$$

Therefore,

$$\frac{1}{\lambda_1} = R\left(1 - \frac{1}{9}\right)$$

$$= R \times \frac{8}{9}$$

Transition 2 :

$$n_2 = 2,\ n_1 = 1$$

Therefore,

$$\frac{1}{\lambda_2} = R\left(1 - \frac{1}{4}\right)$$

$$= R \times \frac{3}{4}$$

Now,

$$\frac{\lambda_1}{\lambda_2}=\frac{\frac{1}{R(8/9)}}{\frac{1}{R(3/4)}}$$

$$=\frac{3/4}{8/9}$$

$$=\frac{3}{4}\times\frac{9}{8}$$

$$=\frac{27}{32}$$

Given,

$$\frac{\lambda_1}{\lambda_2} = \frac{x}{32}$$

Therefore,

$$x = 27$$

Final Answer :

$$27$$

We need to find the lowest wavelength of the Balmer series, given that the lowest wavelength of the Lyman series is 917 angstroms.

The wavelength of spectral lines in the hydrogen spectrum is given by:

where $$R$$ is the Rydberg constant, $$n_1$$ is the lower energy level, and $$n_2$$ is the upper energy level ($$n_2 > n_1$$).

Find the lowest wavelength (series limit) for each series.

The lowest wavelength in any series occurs when $$n_2 \to \infty$$, giving the series limit.

Lyman series ($$n_1 = 1$$): The series limit is:

So $$\lambda_L = \frac{1}{R} = 917$$ angstroms (given).

Balmer series ($$n_1 = 2$$): The series limit is:

So $$\lambda_B = \frac{4}{R}$$.

Since $$\lambda_L = \frac{1}{R}$$, we have $$R = \frac{1}{\lambda_L}$$. Substituting:

The lowest wavelength of the Balmer series is 3668 angstroms.

The binding energy of the ground state electron in a hydrogen atom is 13.6 eV, and we are asked to find the energy required to remove the electron from the second excited state of $$\text{Li}^{2+}$$. Since $$\text{Li}^{2+}$$ is a hydrogen-like ion with $$Z = 3$$ and the second excited state corresponds to $$n = 3$$ (ground state $$n=1$$, first excited $$n=2$$, second excited $$n=3$$), we use the formula for the binding energy in the $$n$$-th orbit of hydrogen-like atoms:

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

Substituting $$Z = 3$$ and $$n = 3$$ gives

$$E_3 = \frac{13.6 \times 3^2}{3^2} = \frac{13.6 \times 9}{9} = 13.6 \text{ eV}$$

Next, to express this energy in the form $$x \times 10^{-1}$$ eV, we note that

$$13.6 \text{ eV} = x \times 10^{-1} \text{ eV}$$

From the above,

$$x = \frac{13.6}{10^{-1}} = \frac{13.6}{0.1} = 136$$

Therefore, the value of $$x$$ is 136.

The radius of the $$n$$th orbit in a hydrogen-like atom is given by

$$r_n = \frac{n^2 a_0}{Z}$$

where $$a_0$$ is the Bohr radius and $$Z$$ is the atomic number.

For Li$$^{++}$$, we have $$Z = 3$$ and $$n = 5$$, so

$$r_5 = \frac{5^2 \times a_0}{3} = \frac{25 \times 0.51}{3}\;\mathring{A}$$

$$r_5 = \frac{12.75}{3} = 4.25\;\mathring{A} = 4.25 \times 10^{-10}\;\text{m} = 425 \times 10^{-12}\;\text{m}$$

Hence, the radius of the fifth orbit of Li$$^{++}$$ is $$425 \times 10^{-12}$$ m. So, the answer is $$425$$.

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 are given that the wavelength of radiation emitted when an electron jumps from the second excited state to the first excited state of the hydrogen atom is $$\lambda_0$$ and we need to find the wavelength when the electron jumps from the third excited state to the second orbit.

First, the second excited state corresponds to $$n = 3$$ and the first excited state to $$n = 2$$, while the third excited state is $$n = 4$$ and the second orbit is $$n = 2$$.

For the transition $$n = 3 \to n = 2$$, $$\frac{1}{\lambda_0} = R\left(\frac{1}{2^2} - \frac{1}{3^2}\right) = R\left(\frac{1}{4} - \frac{1}{9}\right) = R \cdot \frac{5}{36}.$$

For the transition $$n = 4 \to n = 2$$, $$\frac{1}{\lambda} = R\left(\frac{1}{2^2} - \frac{1}{4^2}\right) = R\left(\frac{1}{4} - \frac{1}{16}\right) = R \cdot \frac{3}{16}.$$

Taking the ratio gives $$\frac{\lambda_0}{\lambda} = \frac{R \cdot \tfrac{3}{16}}{R \cdot \tfrac{5}{36}} = \frac{3}{16} \times \frac{36}{5} = \frac{108}{80} = \frac{27}{20},$$ so $$\lambda = \frac{20}{27}\,\lambda_0.$$

Comparing this with the given expression $$\lambda = \frac{20}{x}\,\lambda_0$$ shows that $$x = 27\,. $$

Given

• Moseley's Law relation between frequency ($\nu$) and atomic number ($Z$).
• Graph: $\nu^n$ vs $Z$ is a straight line.

Step 1: Core Formula

According to Moseley's Law for characteristic X-rays:

$$\sqrt{\nu} = a(Z - b)$$

Where $a$ and $b$ are constants.

Step 2: Identifying the Exponent

For the graph to be a straight line (linear) when plotted against $Z$:

$$\nu^n \propto Z$$

Rewriting Moseley's Law:

$$\nu^{1/2} = aZ - ab$$

Comparing the power of $\nu$ in the equation to the power $n$ in the graph:

$$n = \frac{1}{2}$$

Final Answer

$$\boxed{n = \frac{1}{2}}$$

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.

Given: $$U = \frac{1}{2}m\omega^2 r^2$$, where $$\omega$$ is constant.

The force is: $$F = -\frac{dU}{dr} = -m\omega^2 r$$ (directed toward the origin).

For a circular orbit of radius $$r$$, this force provides the centripetal force:

$$ m\omega^2 r = \frac{mv^2}{r} \implies v = \omega r $$

Applying Bohr's quantization condition for angular momentum:

$$ mvr = n\hbar $$

$$ m(\omega r)r = n\hbar $$

$$ m\omega r^2 = n\hbar $$

Solving for $$r$$:

$$ r^2 = \frac{n\hbar}{m\omega} \implies r = \sqrt{\frac{n\hbar}{m\omega}} $$

Therefore, $$r \propto \sqrt{n}$$.

The correct answer is $$\sqrt{n}$$.

We need to find the wavelength of the second line of the Paschen series of the hydrogen atom, given that the first line has wavelength 720 nm.

The Paschen series corresponds to transitions to $$n = 3$$, and the Rydberg formula for hydrogen is $$\frac{1}{\lambda} = R\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$$.

For the first line of the Paschen series, corresponding to a transition from $$n_2 = 4$$ to $$n_1 = 3$$, $$\frac{1}{\lambda_1} = R\left(\frac{1}{9} - \frac{1}{16}\right) = R\left(\frac{16 - 9}{144}\right) = \frac{7R}{144}$$, while for the second line, corresponding to a transition from $$n_2 = 5$$ to $$n_1 = 3$$, $$\frac{1}{\lambda_2} = R\left(\frac{1}{9} - \frac{1}{25}\right) = R\left(\frac{25 - 9}{225}\right) = \frac{16R}{225}$$.

Taking the ratio of these expressions gives $$\frac{\lambda_2}{\lambda_1} = \frac{7R/144}{16R/225} = \frac{7 \times 225}{16 \times 144} = \frac{1575}{2304}$$, and hence $$\lambda_2 = 720 \times \frac{1575}{2304} = 720 \times \frac{1575}{2304}$$. Simplifying $$\frac{720}{2304} = \frac{5}{16}$$ yields $$\lambda_2 = \frac{5 \times 1575}{16} = \frac{7875}{16} = 492.1875 \text{ nm}$$, which rounds to the nearest integer as $$\lambda_2 \approx 492$$ nm.

The answer is 492.

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.

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: $$\dfrac{\lambda_0}{\left(1 + \dfrac{eE_0 t}{mv_0}\right)}$$.

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 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.

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 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.