Chapter 2: Structure of Atom

In the notation $${}^{A}_{Z}X$$, $$Z$$ is the atomic number (number of protons) and $$A$$ is the mass number (sum of protons and neutrons). For a neutral atom, the number of electrons equals the number of protons.

Here $$Z = 35$$ and $$A = 80$$, so:

Number of protons $$= Z = 35$$.

Number of electrons (neutral atom) $$= 35$$.

Number of neutrons $$= A - Z = 80 - 35 = 45$$.

The atomic number $$Z$$ equals the number of protons, so $$Z = 16$$. The element with $$Z = 16$$ is sulphur (S).

The mass number $$A$$ equals the sum of protons and neutrons:

$$A = 16 + 16 = 32$$.

The charge on the species equals (protons − electrons):

$$\text{charge} = 16 - 18 = -2$$.

Hence the species is the sulphide ion. The complete symbol is $${}^{32}_{16}\mathrm{S}^{2-}$$.

For an electromagnetic wave, $$c = \nu\lambda$$, so $$\lambda = c/\nu$$.

Given $$\nu = 1368 \, \mathrm{kHz} = 1368 \times 10^{3} \, \mathrm{s^{-1}} = 1.368 \times 10^{6} \, \mathrm{s^{-1}}$$ and $$c = 3 \times 10^{8} \, \mathrm{m \, s^{-1}}$$.

$$\lambda = \dfrac{3 \times 10^{8} \, \mathrm{m \, s^{-1}}}{1.368 \times 10^{6} \, \mathrm{s^{-1}}} = 219.3 \, \mathrm{m}$$.

This wavelength lies in the radio-wave region of the electromagnetic spectrum (specifically the medium/short wave band).

Use $$\nu = c/\lambda$$ with $$c = 3 \times 10^{8} \, \mathrm{m \, s^{-1}}$$.

For violet ($$\lambda = 400 \, \mathrm{nm} = 4.00 \times 10^{-7} \, \mathrm{m}$$):

$$\nu_{\text{violet}} = \dfrac{3 \times 10^{8}}{4.00 \times 10^{-7}} = 7.50 \times 10^{14} \, \mathrm{Hz}$$.

For red ($$\lambda = 750 \, \mathrm{nm} = 7.50 \times 10^{-7} \, \mathrm{m}$$):

$$\nu_{\text{red}} = \dfrac{3 \times 10^{8}}{7.50 \times 10^{-7}} = 4.00 \times 10^{14} \, \mathrm{Hz}$$.

The visible region thus spans roughly $$4.0 \times 10^{14}$$–$$7.5 \times 10^{14} \, \mathrm{Hz}$$.

Yellow radiation has wavelength $$\lambda = 5800 \, \text{Å} = 5800 \times 10^{-10} \, \mathrm{m} = 5.80 \times 10^{-7} \, \mathrm{m}$$.

(a) Wavenumber

The wavenumber is the reciprocal of the wavelength:

$$\bar{\nu} = \dfrac{1}{\lambda} = \dfrac{1}{5.80 \times 10^{-7} \, \mathrm{m}}$$

$$\bar{\nu} = 1.724 \times 10^{6} \, \mathrm{m^{-1}} = 1.724 \times 10^{4} \, \mathrm{cm^{-1}}$$.

(b) Frequency

The frequency is related to the wavelength by $$\nu = c/\lambda$$, with $$c = 3 \times 10^{8} \, \mathrm{m \, s^{-1}}$$:

$$\nu = \dfrac{3 \times 10^{8} \, \mathrm{m \, s^{-1}}}{5.80 \times 10^{-7} \, \mathrm{m}} = 5.172 \times 10^{14} \, \mathrm{s^{-1}} \; (\mathrm{Hz})$$.

Energy of one photon: $$E = h\nu$$, with $$h = 6.626 \times 10^{-34} \, \mathrm{J \, s}$$.

$$E = (6.626 \times 10^{-34}) \times (5 \times 10^{14}) = 3.313 \times 10^{-19} \, \mathrm{J}$$.

Energy of one mole of photons $$= N_A \, h\nu$$, with $$N_A = 6.022 \times 10^{23} \, \mathrm{mol^{-1}}$$.

$$E_{\text{mol}} = (6.022 \times 10^{23}) \times (3.313 \times 10^{-19}) \, \mathrm{J \, mol^{-1}}$$

$$E_{\text{mol}} = 1.995 \times 10^{5} \, \mathrm{J \, mol^{-1}} = 199.5 \, \mathrm{kJ \, mol^{-1}}$$.

Power $$P = 100 \, \mathrm{W} = 100 \, \mathrm{J \, s^{-1}}$$; the bulb therefore emits 100 J of energy each second.

Energy of one photon at $$\lambda = 400 \, \mathrm{nm} = 4.00 \times 10^{-7} \, \mathrm{m}$$:

$$E_{\text{ph}} = \dfrac{hc}{\lambda} = \dfrac{(6.626 \times 10^{-34})(3 \times 10^{8})}{4.00 \times 10^{-7}} \, \mathrm{J}$$

$$E_{\text{ph}} = 4.969 \times 10^{-19} \, \mathrm{J}$$.

Number of photons per second:

$$N = \dfrac{P}{E_{\text{ph}}} = \dfrac{100}{4.969 \times 10^{-19}} = 2.012 \times 10^{20} \, \mathrm{s^{-1}}$$.

By Einstein's photoelectric equation, on a per-mole basis,

$$E_{\text{photon}} = W_0 + \mathrm{KE}$$,

so the work function is $$W_0 = E_{\text{photon}} - \mathrm{KE}$$.

Energy of one mole of photons of $$\lambda = 300 \, \mathrm{nm} = 3.0 \times 10^{-7} \, \mathrm{m}$$:

$$E = \dfrac{N_A \, hc}{\lambda} = \dfrac{(6.022 \times 10^{23})(6.626 \times 10^{-34})(3 \times 10^{8})}{3.0 \times 10^{-7}}$$

$$E = 3.99 \times 10^{5} \, \mathrm{J \, mol^{-1}}$$.

Hence

$$W_0 = 3.99 \times 10^{5} - 1.68 \times 10^{5} = 2.31 \times 10^{5} \, \mathrm{J \, mol^{-1}}$$.

Per atom, $$W_0/N_A = 2.31 \times 10^{5}/6.022 \times 10^{23} = 3.84 \times 10^{-19} \, \mathrm{J}$$.

The maximum wavelength capable of ejecting a photoelectron satisfies $$hc/\lambda_{\max} = W_0$$:

$$\lambda_{\max} = \dfrac{hc}{W_0} = \dfrac{(6.626 \times 10^{-34})(3 \times 10^{8})}{3.84 \times 10^{-19}}$$

$$\lambda_{\max} = 5.17 \times 10^{-7} \, \mathrm{m} = 517 \, \mathrm{nm}$$.

From Einstein's photoelectric equation,

$$\mathrm{KE} = h(\nu - \nu_0)$$.

$$\mathrm{KE} = (6.626 \times 10^{-34}) \times (1.0 \times 10^{15} - 7.0 \times 10^{14})$$

$$\mathrm{KE} = (6.626 \times 10^{-34}) \times (3.0 \times 10^{14})$$

$$\mathrm{KE} = 1.988 \times 10^{-19} \, \mathrm{J}$$.

For a hydrogen-atom transition, the energy emitted is

$$\Delta E = R_H \left(\dfrac{1}{n_1^{2}} - \dfrac{1}{n_2^{2}}\right)$$, where $$R_H = 2.18 \times 10^{-18} \, \mathrm{J}$$, $$n_1 = 2$$ (lower) and $$n_2 = 5$$ (upper).

$$\Delta E = 2.18 \times 10^{-18}\left(\dfrac{1}{4} - \dfrac{1}{25}\right) = 2.18 \times 10^{-18} \times \dfrac{21}{100}$$

$$\Delta E = 4.578 \times 10^{-19} \, \mathrm{J}$$.

Frequency: $$\nu = \dfrac{\Delta E}{h} = \dfrac{4.578 \times 10^{-19}}{6.626 \times 10^{-34}} = 6.91 \times 10^{14} \, \mathrm{Hz}$$.

Wavelength: $$\lambda = \dfrac{c}{\nu} = \dfrac{3 \times 10^{8}}{6.91 \times 10^{14}} = 4.34 \times 10^{-7} \, \mathrm{m} = 434 \, \mathrm{nm}$$.

This line lies in the visible region (Balmer series).

For a hydrogen-like (one-electron) species, Bohr's results give

$$E_n = -\dfrac{2.18 \times 10^{-18} \, Z^{2}}{n^{2}} \, \mathrm{J}$$ and $$r_n = \dfrac{52.9 \, n^{2}}{Z} \, \mathrm{pm}$$.

For $$\mathrm{He^+}$$, $$Z = 2$$ and $$n = 1$$.

$$E_1 = -\dfrac{2.18 \times 10^{-18} \times 2^{2}}{1^{2}} = -8.72 \times 10^{-18} \, \mathrm{J \, atom^{-1}}$$.

$$r_1 = \dfrac{52.9 \times 1^{2}}{2} = 26.45 \, \mathrm{pm} = 0.2645 \, \text{Å}$$.

By de Broglie, $$\lambda = \dfrac{h}{mv}$$.

$$\lambda = \dfrac{6.626 \times 10^{-34} \, \mathrm{J \, s}}{(0.1 \, \mathrm{kg})(10 \, \mathrm{m \, s^{-1}})} = \dfrac{6.626 \times 10^{-34}}{1.0} \, \mathrm{m}$$

$$\lambda = 6.626 \times 10^{-34} \, \mathrm{m}$$.

This is far too small to be detected, which is why wave-like behaviour of macroscopic objects is never observed.

From $$\mathrm{KE} = \tfrac{1}{2}mv^{2}$$,

$$v = \sqrt{\dfrac{2 \, \mathrm{KE}}{m}} = \sqrt{\dfrac{2 \times 3.0 \times 10^{-25}}{9.1 \times 10^{-31}}} \, \mathrm{m \, s^{-1}}$$

$$v = \sqrt{6.593 \times 10^{5}} = 811.97 \, \mathrm{m \, s^{-1}}$$.

de Broglie wavelength:

$$\lambda = \dfrac{h}{mv} = \dfrac{6.626 \times 10^{-34}}{(9.1 \times 10^{-31})(811.97)}$$

$$\lambda = \dfrac{6.626 \times 10^{-34}}{7.389 \times 10^{-28}} = 8.967 \times 10^{-7} \, \mathrm{m} \approx 897 \, \mathrm{nm}$$.

For a photon, $$\lambda = \dfrac{h}{mc}$$ (taking $$v = c$$ in the de Broglie relation), so

$$m = \dfrac{h}{\lambda c}$$.

With $$\lambda = 3.6 \, \text{Å} = 3.6 \times 10^{-10} \, \mathrm{m}$$,

$$m = \dfrac{6.626 \times 10^{-34}}{(3.6 \times 10^{-10})(3 \times 10^{8})}$$

$$m = \dfrac{6.626 \times 10^{-34}}{1.08 \times 10^{-1}} = 6.135 \times 10^{-33} \, \mathrm{kg}$$.

Heisenberg's uncertainty principle: $$\Delta x \cdot \Delta p \ge \dfrac{h}{4\pi}$$, i.e. $$\Delta x \cdot m\Delta v \ge \dfrac{h}{4\pi}$$.

Therefore the minimum uncertainty in velocity is

$$\Delta v = \dfrac{h}{4\pi \, m \, \Delta x}$$.

With $$\Delta x = 0.1 \, \text{Å} = 1.0 \times 10^{-11} \, \mathrm{m}$$ and $$m = 9.11 \times 10^{-31} \, \mathrm{kg}$$:

$$\Delta v = \dfrac{6.626 \times 10^{-34}}{4 \times 3.1416 \times 9.11 \times 10^{-31} \times 1.0 \times 10^{-11}}$$

$$\Delta v = \dfrac{6.626 \times 10^{-34}}{1.145 \times 10^{-40}} = 5.79 \times 10^{6} \, \mathrm{m \, s^{-1}}$$.

The uncertainty in speed is 2% of 45 m s$${}^{-1}$$:

$$\Delta v = \dfrac{2}{100} \times 45 = 0.9 \, \mathrm{m \, s^{-1}}$$.

From Heisenberg's principle,

$$\Delta x \ge \dfrac{h}{4\pi \, m \, \Delta v}$$.

With $$m = 40 \, \mathrm{g} = 0.04 \, \mathrm{kg}$$:

$$\Delta x = \dfrac{6.626 \times 10^{-34}}{4 \times 3.1416 \times 0.04 \times 0.9}$$

$$\Delta x = \dfrac{6.626 \times 10^{-34}}{0.4524} = 1.464 \times 10^{-33} \, \mathrm{m}$$.

This uncertainty is exceedingly small compared with the size of the ball, showing that the wave nature of macroscopic objects is undetectable.

For a given $$n$$, the allowed values of $$l$$ are $$0, 1, \ldots, n-1$$. For each $$l$$, the number of orbitals is $$(2l+1)$$.

For $$n = 3$$:

• $$l = 0$$ (3s): 1 orbital
• $$l = 1$$ (3p): 3 orbitals
• $$l = 2$$ (3d): 5 orbitals

Total $$= 1 + 3 + 5 = 9$$ orbitals. Equivalently $$n^{2} = 3^{2} = 9$$.

In the symbol $${}^{A}_{Z}X$$: number of protons $$= Z$$, number of neutrons $$= A - Z$$.

$$\lambda = 580 \, \mathrm{nm} = 5.80 \times 10^{-7} \, \mathrm{m}$$.

Frequency:

$$\nu = \dfrac{c}{\lambda} = \dfrac{3 \times 10^{8} \, \mathrm{m \, s^{-1}}}{5.80 \times 10^{-7} \, \mathrm{m}}$$

$$\nu = 5.172 \times 10^{14} \, \mathrm{Hz}$$.

Wavenumber:

$$\bar{\nu} = \dfrac{1}{\lambda} = \dfrac{1}{5.80 \times 10^{-7} \, \mathrm{m}}$$

$$\bar{\nu} = 1.724 \times 10^{6} \, \mathrm{m^{-1}} = 1.724 \times 10^{4} \, \mathrm{cm^{-1}}$$.

Frequency is the reciprocal of the period:

$$\nu = \dfrac{1}{T} = \dfrac{1}{2.0 \times 10^{-10} \, \mathrm{s}} = 5.0 \times 10^{9} \, \mathrm{Hz}$$.

Wavelength: $$\lambda = c/\nu$$.

$$\lambda = \dfrac{3 \times 10^{8}}{5.0 \times 10^{9}} = 6.0 \times 10^{-2} \, \mathrm{m} = 6 \, \mathrm{cm}$$.

Wavenumber:

$$\bar{\nu} = \dfrac{1}{\lambda} = \dfrac{1}{6.0 \times 10^{-2}} = 16.67 \, \mathrm{m^{-1}}$$.

$$\lambda = 4000 \, \mathrm{pm} = 4.0 \times 10^{-9} \, \mathrm{m}$$.

Energy of one photon:

$$E_{\text{ph}} = \dfrac{hc}{\lambda} = \dfrac{(6.626 \times 10^{-34})(3 \times 10^{8})}{4.0 \times 10^{-9}}$$

$$E_{\text{ph}} = 4.969 \times 10^{-17} \, \mathrm{J}$$.

Number of photons that provide 1 J:

$$N = \dfrac{1 \, \mathrm{J}}{4.969 \times 10^{-17} \, \mathrm{J}} = 2.012 \times 10^{16}$$.

If $$\lambda = 242 \, \mathrm{nm} = 2.42 \times 10^{-7} \, \mathrm{m}$$ is just sufficient to ionise sodium, the ionisation energy per atom equals the photon energy.

Energy per photon:

$$E = \dfrac{hc}{\lambda} = \dfrac{(6.626 \times 10^{-34})(3 \times 10^{8})}{2.42 \times 10^{-7}}$$

$$E = 8.214 \times 10^{-19} \, \mathrm{J \, atom^{-1}}$$.

Per mole:

$$E_{\text{mol}} = (8.214 \times 10^{-19})(6.022 \times 10^{23})$$

$$E_{\text{mol}} = 4.946 \times 10^{5} \, \mathrm{J \, mol^{-1}} \approx 494.6 \, \mathrm{kJ \, mol^{-1}}$$.

$$\lambda = 0.57 \, \mu\mathrm{m} = 5.7 \times 10^{-7} \, \mathrm{m}$$.

Energy of one photon:

$$E_{\text{ph}} = \dfrac{hc}{\lambda} = \dfrac{(6.626 \times 10^{-34})(3 \times 10^{8})}{5.7 \times 10^{-7}}$$

$$E_{\text{ph}} = 3.487 \times 10^{-19} \, \mathrm{J}$$.

The bulb emits energy at $$P = 25 \, \mathrm{W} = 25 \, \mathrm{J \, s^{-1}}$$.

Rate of emission of quanta:

$$N = \dfrac{P}{E_{\text{ph}}} = \dfrac{25}{3.487 \times 10^{-19}}$$

$$N \approx 7.17 \times 10^{19} \, \mathrm{s^{-1}}$$.

If electrons are emitted with zero velocity, the incident radiation has exactly the threshold frequency:

$$\nu_0 = \dfrac{c}{\lambda} = \dfrac{3 \times 10^{8}}{6800 \times 10^{-10}} = \dfrac{3 \times 10^{8}}{6.8 \times 10^{-7}}$$

$$\nu_0 = 4.41 \times 10^{14} \, \mathrm{Hz}$$.

Work function:

$$W_0 = h\nu_0 = (6.626 \times 10^{-34})(4.41 \times 10^{14})$$

$$W_0 \approx 2.92 \times 10^{-19} \, \mathrm{J} \approx 1.82 \, \mathrm{eV}$$.

For hydrogen, $$\Delta E = R_H \left(\dfrac{1}{n_1^{2}} - \dfrac{1}{n_2^{2}}\right)$$ with $$R_H = 2.18 \times 10^{-18} \, \mathrm{J}$$.

$$\Delta E = 2.18 \times 10^{-18}\left(\dfrac{1}{4} - \dfrac{1}{16}\right) = 2.18 \times 10^{-18} \times \dfrac{3}{16}$$

$$\Delta E = 4.0875 \times 10^{-19} \, \mathrm{J}$$.

Wavelength:

$$\lambda = \dfrac{hc}{\Delta E} = \dfrac{(6.626 \times 10^{-34})(3 \times 10^{8})}{4.0875 \times 10^{-19}}$$

$$\lambda \approx 4.864 \times 10^{-7} \, \mathrm{m} = 486.4 \, \mathrm{nm}$$.

This is the H$$\beta$$ line of the Balmer series (visible).

For hydrogen, $$E_n = -\dfrac{2.18 \times 10^{-18}}{n^{2}} \, \mathrm{J}$$.

Energy of the $$n = 5$$ level:

$$E_5 = -\dfrac{2.18 \times 10^{-18}}{25} = -8.72 \times 10^{-20} \, \mathrm{J}$$.

To ionise the atom, the electron is taken from $$n = 5$$ to $$n = \infty$$ ($$E_{\infty} = 0$$):

$$\Delta E = E_{\infty} - E_5 = 8.72 \times 10^{-20} \, \mathrm{J}$$.

Energy required from $$n = 1$$ is $$2.18 \times 10^{-18} \, \mathrm{J}$$, so the ratio is

$$\dfrac{E(n=1 \to \infty)}{E(n=5 \to \infty)} = \dfrac{2.18 \times 10^{-18}}{8.72 \times 10^{-20}} = 25$$.

Ionising from $$n = 1$$ requires 25 times more energy than from $$n = 5$$.

The total number of spectral lines produced when an electron drops from level $$n$$ to the ground state (via all possible intermediate levels) is

$$N = \dfrac{n(n-1)}{2}$$.

For $$n = 6$$:

$$N = \dfrac{6 \times 5}{2} = 15$$ lines.

The longest wavelength in the Balmer series corresponds to the lowest energy difference, i.e. the transition $$n_2 = 3 \to n_1 = 2$$ (H$$\alpha$$).

Using the Rydberg formula,

$$\bar{\nu} = R_H \left(\dfrac{1}{n_1^{2}} - \dfrac{1}{n_2^{2}}\right)$$, with $$R_H = 1.097 \times 10^{7} \, \mathrm{m^{-1}}$$.

$$\bar{\nu} = 1.097 \times 10^{7}\left(\dfrac{1}{4} - \dfrac{1}{9}\right) = 1.097 \times 10^{7} \times \dfrac{5}{36}$$

$$\bar{\nu} = 1.523 \times 10^{6} \, \mathrm{m^{-1}}$$.

Convert the ground-state energy to joules ($$1 \, \mathrm{erg} = 10^{-7} \, \mathrm{J}$$):

$$E_1 = -2.18 \times 10^{-11} \, \mathrm{erg} = -2.18 \times 10^{-18} \, \mathrm{J}$$.

For hydrogen, $$E_n = E_1/n^{2}$$.

$$E_5 = \dfrac{-2.18 \times 10^{-18}}{25} = -8.72 \times 10^{-20} \, \mathrm{J}$$.

Energy absorbed in going from $$n=1$$ to $$n=5$$:

$$\Delta E = E_5 - E_1 = -8.72 \times 10^{-20} - (-2.18 \times 10^{-18})$$

$$\Delta E = 2.18 \times 10^{-18} - 8.72 \times 10^{-20} = 2.0928 \times 10^{-18} \, \mathrm{J}$$.

The same amount of energy is released when the electron falls back to the ground state. Wavelength of the emitted photon:

$$\lambda = \dfrac{hc}{\Delta E} = \dfrac{(6.626 \times 10^{-34})(3 \times 10^{8})}{2.0928 \times 10^{-18}}$$

$$\lambda \approx 9.498 \times 10^{-8} \, \mathrm{m} = 94.98 \, \mathrm{nm}$$.

This line lies in the ultraviolet region (Lyman series).

Energy of the electron in the $$n = 2$$ orbit:

$$E_2 = \dfrac{-2.18 \times 10^{-18}}{4} = -5.45 \times 10^{-19} \, \mathrm{J}$$.

To remove the electron completely means raising it to $$n = \infty$$ ($$E_{\infty} = 0$$):

$$\Delta E = 0 - E_2 = 5.45 \times 10^{-19} \, \mathrm{J}$$.

The longest wavelength corresponds to a photon whose energy is just equal to $$\Delta E$$:

$$\lambda = \dfrac{hc}{\Delta E} = \dfrac{(6.626 \times 10^{-34})(3 \times 10^{8})}{5.45 \times 10^{-19}}$$

$$\lambda \approx 3.648 \times 10^{-7} \, \mathrm{m}$$

$$\lambda \approx 3.648 \times 10^{-5} \, \mathrm{cm}$$.

de Broglie wavelength: $$\lambda = \dfrac{h}{mv}$$, with $$m = 9.11 \times 10^{-31} \, \mathrm{kg}$$.

$$\lambda = \dfrac{6.626 \times 10^{-34}}{(9.11 \times 10^{-31})(2.05 \times 10^{7})}$$

$$\lambda = \dfrac{6.626 \times 10^{-34}}{1.868 \times 10^{-23}}$$

$$\lambda \approx 3.548 \times 10^{-11} \, \mathrm{m} = 35.48 \, \mathrm{pm}$$.

From $$\mathrm{KE} = \tfrac{1}{2}mv^{2}$$:

$$v = \sqrt{\dfrac{2 \, \mathrm{KE}}{m}} = \sqrt{\dfrac{2 \times 3.0 \times 10^{-25}}{9.1 \times 10^{-31}}}$$

$$v = \sqrt{6.593 \times 10^{5}} \approx 811.97 \, \mathrm{m \, s^{-1}}$$.

de Broglie wavelength:

$$\lambda = \dfrac{h}{mv} = \dfrac{6.626 \times 10^{-34}}{(9.1 \times 10^{-31})(811.97)}$$

$$\lambda \approx \dfrac{6.626 \times 10^{-34}}{7.389 \times 10^{-28}} \approx 8.97 \times 10^{-7} \, \mathrm{m}$$

$$\lambda \approx 897 \, \mathrm{nm}$$.

Count electrons in each species (electrons = atomic number − charge):

The two isoelectronic sets are:

• 10 electrons: $$\mathrm{Na^+}$$ and $$\mathrm{Mg^{2+}}$$.
• 18 electrons: $$\mathrm{K^+}$$, $$\mathrm{Ca^{2+}}$$, $$\mathrm{S^{2-}}$$ and $$\mathrm{Ar}$$.

The label g corresponds to the azimuthal quantum number $$l = 4$$.

Since $$l$$ can take values $$0, 1, \ldots, n-1$$, we need $$n - 1 \ge 4$$, i.e. $$n \ge 5$$.

Therefore the lowest principal quantum number that permits g orbitals is $$n = 5$$.

For a 3d electron:

• Principal quantum number: $$n = 3$$.
• Azimuthal quantum number (d → $$l = 2$$): $$l = 2$$.
• Magnetic quantum number $$m_l$$ may take values $$-l, -l+1, \ldots, +l$$: $$m_l = -2, -1, 0, +1, +2$$.

(i) In a neutral atom the number of protons equals the number of electrons, so

$$\text{protons} = 29$$.

(ii) The element with $$Z = 29$$ is copper (Cu). Copper has an anomalous configuration favouring a fully-filled 3d set:

$$\mathrm{Cu}: 1s^{2}\,2s^{2}\,2p^{6}\,3s^{2}\,3p^{6}\,3d^{10}\,4s^{1}$$, i.e. $$[\mathrm{Ar}]\,3d^{10}\,4s^{1}$$.

(The 3d$${}^{10}$$4s$${}^{1}$$ arrangement is more stable than 3d$${}^{9}$$4s$${}^{2}$$ because of the extra stability of the fully filled d-subshell.)

For neutral molecules, sum the atomic electrons; for ions, subtract (for +) or add (for −) the charge.

• $$\mathrm{H_2^+}$$: $$2(1) - 1 = 1$$ electron.
• $$\mathrm{H_2}$$: $$2(1) = 2$$ electrons.
• $$\mathrm{O_2^+}$$: $$2(8) - 1 = 15$$ electrons.

Bohr's angular-momentum postulate states that the angular momentum of the orbiting electron is quantised:

$$mvr = \dfrac{nh}{2\pi}\quad (n = 1, 2, 3, \ldots) \qquad \ldots (1)$$.

The de Broglie wavelength of the electron is

$$\lambda = \dfrac{h}{mv} \quad \Rightarrow \quad mv = \dfrac{h}{\lambda} \qquad \ldots (2)$$.

Substitute (2) into (1):

$$\left(\dfrac{h}{\lambda}\right) r = \dfrac{nh}{2\pi}$$.

Cancelling $$h$$ and rearranging,

$$2\pi r = n\lambda$$.

Since $$2\pi r$$ is the circumference of the orbit, this shows that the circumference equals an integral number $$n$$ of de Broglie wavelengths. Hence the electron forms a standing wave around the nucleus.

For a hydrogen-like ion the Rydberg formula gives

$$\bar{\nu} = R \, Z^{2}\left(\dfrac{1}{n_1^{2}} - \dfrac{1}{n_2^{2}}\right)$$.

For $$\mathrm{He^+}$$ (Z = 2) with $$n_1 = 2, n_2 = 4$$:

$$\bar{\nu}_{\mathrm{He^+}} = R \times 4 \times\left(\dfrac{1}{4} - \dfrac{1}{16}\right) = R \times 4 \times \dfrac{3}{16} = \dfrac{3R}{4}$$.

For a hydrogen transition ($$Z = 1$$) with the same wavenumber,

$$\dfrac{1}{n_1^{2}} - \dfrac{1}{n_2^{2}} = \dfrac{3}{4}$$.

This is satisfied by $$n_1 = 1, n_2 = 2$$ (since $$1 - \tfrac{1}{4} = \tfrac{3}{4}$$).

Hence the corresponding transition in hydrogen is $$n = 2 \to n = 1$$ (Lyman $$\alpha$$).

For a hydrogen-like (one-electron) species in the ground state, the ionisation energy is

$$\mathrm{IE} = R_H \, Z^{2} \, \dfrac{1}{n^{2}}$$ (with $$n = 1$$),

where $$R_H = 2.18 \times 10^{-18} \, \mathrm{J}$$ is the H-atom ionisation energy.

For $$\mathrm{He^+}$$, $$Z = 2$$:

$$\mathrm{IE}(\mathrm{He^+}) = 2.18 \times 10^{-18} \times 2^{2} = 8.72 \times 10^{-18} \, \mathrm{J \, atom^{-1}}$$.

Express everything in metres.

Diameter $$d = 0.15 \, \mathrm{nm} = 0.15 \times 10^{-9} \, \mathrm{m} = 1.5 \times 10^{-10} \, \mathrm{m}$$.

Length $$L = 20 \, \mathrm{cm} = 0.20 \, \mathrm{m} = 2.0 \times 10^{-1} \, \mathrm{m}$$.

Number of atoms:

$$N = \dfrac{L}{d} = \dfrac{2.0 \times 10^{-1}}{1.5 \times 10^{-10}}$$

$$N \approx 1.333 \times 10^{9}$$ atoms.

If $$N$$ atoms placed side by side span a length $$L$$, the diameter of one atom is $$d = L/N$$.

$$d = \dfrac{2.4 \, \mathrm{cm}}{2 \times 10^{8}} = 1.2 \times 10^{-8} \, \mathrm{cm} = 1.2 \times 10^{-10} \, \mathrm{m} = 120 \, \mathrm{pm}$$.

Radius $$r = d/2$$:

$$r = 6.0 \times 10^{-11} \, \mathrm{m} = 60 \, \mathrm{pm} = 0.60 \, \text{Å}$$.

(a) Diameter $$d = 2.6 \, \text{Å} = 2.6 \times 10^{-10} \, \mathrm{m} = 260 \, \mathrm{pm}$$.

Radius $$r = d/2 = 130 \, \mathrm{pm}$$.

(b) Length $$L = 1.6 \, \mathrm{cm} = 1.6 \times 10^{-2} \, \mathrm{m} = 1.6 \times 10^{10} \, \mathrm{pm}$$.

$$N = \dfrac{L}{d} = \dfrac{1.6 \times 10^{10}}{260} \approx 6.154 \times 10^{7}$$ atoms.

Each electron carries a charge of magnitude $$e = 1.6022 \times 10^{-19} \, \mathrm{C}$$.

Number of (excess) electrons:

$$N = \dfrac{Q}{e} = \dfrac{2.5 \times 10^{-16}}{1.6022 \times 10^{-19}}$$

$$N \approx 1.560 \times 10^{3}$$ electrons.

Magnitude of charge per electron: $$e = 1.6022 \times 10^{-19} \, \mathrm{C}$$.

$$N = \dfrac{|Q|}{e} = \dfrac{1.282 \times 10^{-18}}{1.6022 \times 10^{-19}} \approx 8.00$$.

So the drop carries 8 excess electrons.

The angle through which an $$\alpha$$-particle is scattered by a nucleus increases with the nuclear charge $$Z$$ (Coulomb repulsion between the nucleus and the positively charged $$\alpha$$-particle is proportional to $$Z$$). Heavy nuclei such as Au or Pt have very large $$Z$$, so a few $$\alpha$$-particles experience strong head-on (or near head-on) repulsions and are scattered through large angles (some even bounce back through angles near $$180^{\circ}$$).

Light atoms like aluminium have much smaller $$Z$$. Consequently the Coulomb force between the nucleus and the $$\alpha$$-particle is weaker, so the number of $$\alpha$$-particles that are deflected through large angles (and especially those that bounce back) is far fewer. Most $$\alpha$$-particles would pass through almost undeviated.

Further, since light nuclei are not much heavier than an $$\alpha$$-particle, they recoil appreciably on collision; the scattering pattern is therefore less sharply defined than with heavy foils. Hence Rutherford's interpretation (a dense, massive, highly charged nucleus) is harder to establish using a light-atom foil.

By convention, an element is represented as $${}^{A}_{Z}X$$ where the mass number $$A$$ is written as a superscript on the upper left and the atomic number $$Z$$ as a subscript on the lower left.

$${}^{79}_{35}\mathrm{Br}$$ is correct (A = 79 on top, Z = 35 below).

$${}^{79}\mathrm{Br}$$ is also acceptable because the symbol "Br" uniquely fixes $$Z = 35$$, so the subscript can be omitted without ambiguity.

$${}^{35}_{79}\mathrm{Br}$$ is wrong because the positions of $$A$$ and $$Z$$ have been swapped (35 cannot be the mass number of bromine and 79 cannot be its atomic number).

$${}^{35}\mathrm{Br}$$ is also unacceptable: the lone superscript stands for the mass number, but no bromine isotope has mass number 35, so the symbol is inconsistent.

Let $$p$$ = number of protons. Number of neutrons $$= p + 0.317 p = 1.317 p$$.

The mass number is $$A = p + n$$:

$$p + 1.317 p = 81$$

$$2.317 p = 81$$

$$p \approx 35$$.

So $$Z = 35$$ (bromine) and $$n = 81 - 35 = 46$$.

Symbol: $${}^{81}_{35}\mathrm{Br}$$.

Let the number of protons be $$p$$. Then number of electrons $$e = p + 1$$ (because the ion has charge $$-1$$).

Number of neutrons $$n = 1.111 \, e = 1.111(p+1)$$.

Mass number: $$p + n = 37$$, so

$$p + 1.111(p + 1) = 37$$

$$p + 1.111 p + 1.111 = 37$$

$$2.111 p = 35.889$$

$$p \approx 17$$.

So $$Z = 17$$ (chlorine), $$n = 37 - 17 = 20$$, $$e = 18$$.

Symbol: $${}^{37}_{17}\mathrm{Cl}^{-}$$.

Let the number of electrons in the ion be $$e$$. Since the ion has charge $$+3$$, the number of protons is $$p = e + 3$$.

Number of neutrons $$n = 1.304 \, e$$.

Mass number: $$p + n = 56$$:

$$(e + 3) + 1.304 e = 56$$

$$2.304 e = 53$$

$$e \approx 23$$.

So protons $$p = 26 \to$$ iron (Fe), neutrons $$n = 56 - 26 = 30$$, electrons = 23.

Symbol: $${}^{56}_{26}\mathrm{Fe}^{3+}$$.

Recall the electromagnetic spectrum (lowest frequency to highest):

radio waves < microwaves < infrared < visible (red < ... < violet) < ultraviolet < X-rays < $$\gamma$$-rays < cosmic rays.

Placing each radiation in this sequence:

• (c) FM radio — radio band.
• (a) Microwave oven — microwave band.
• (b) Amber light from traffic signal — visible region.
• (e) X-rays — beyond UV.
• (d) Cosmic rays — highest frequency.

Hence the increasing order of frequency is

(c) FM radio < (a) microwaves < (b) amber light < (e) X-rays < (d) cosmic rays.

Energy of one photon at $$\lambda = 337.1 \, \mathrm{nm} = 337.1 \times 10^{-9} \, \mathrm{m}$$:

$$E_{\text{ph}} = \dfrac{hc}{\lambda} = \dfrac{(6.626 \times 10^{-34})(3 \times 10^{8})}{337.1 \times 10^{-9}}$$

$$E_{\text{ph}} = 5.896 \times 10^{-19} \, \mathrm{J}$$.

Total energy from $$N = 5.6 \times 10^{24}$$ photons:

$$E = N \, E_{\text{ph}} = (5.6 \times 10^{24})(5.896 \times 10^{-19})$$

$$E \approx 3.302 \times 10^{6} \, \mathrm{J}$$.

Assuming the photons quoted are emitted in 1 second (so the laser power equals the energy emitted per second), the power is

$$P \approx 3.30 \times 10^{6} \, \mathrm{W}$$.

$$\lambda = 616 \, \mathrm{nm} = 6.16 \times 10^{-7} \, \mathrm{m}$$.

(a) Frequency:

$$\nu = \dfrac{c}{\lambda} = \dfrac{3 \times 10^{8}}{6.16 \times 10^{-7}} \approx 4.87 \times 10^{14} \, \mathrm{Hz}$$.

(b) Distance travelled in 30 s (light travels at $$c$$):

$$d = c \, t = (3 \times 10^{8})(30) = 9 \times 10^{9} \, \mathrm{m}$$.

(c) Energy per quantum:

$$E_{\text{ph}} = h\nu = (6.626 \times 10^{-34})(4.87 \times 10^{14})$$

$$E_{\text{ph}} \approx 3.23 \times 10^{-19} \, \mathrm{J}$$.

(d) Number of quanta in 2 J:

$$N = \dfrac{2}{3.23 \times 10^{-19}} \approx 6.20 \times 10^{18}$$ quanta.

Energy of one photon at $$\lambda = 600 \, \mathrm{nm} = 6 \times 10^{-7} \, \mathrm{m}$$:

$$E_{\text{ph}} = \dfrac{hc}{\lambda} = \dfrac{(6.626 \times 10^{-34})(3 \times 10^{8})}{6 \times 10^{-7}}$$

$$E_{\text{ph}} = 3.313 \times 10^{-19} \, \mathrm{J}$$.

Number of photons giving 3.15 × 10$${}^{-18}$$ J:

$$N = \dfrac{3.15 \times 10^{-18}}{3.313 \times 10^{-19}} \approx 9.51$$.

So roughly 10 photons are received by the detector.

The total energy of the radiation pulse is the sum of the energies of all photons emitted during the pulse:

$$E_{\text{pulse}} = N \cdot E_{\text{photon}}, \qquad E_{\text{photon}} = h\nu.$$

The problem specifies the pulse duration $$\tau = 2 \, \mathrm{ns} = 2 \times 10^{-9}\,\mathrm{s}$$ and the photon count $$N = 2.5 \times 10^{15}$$. Following the NCERT convention, the frequency of the radiation is taken as the reciprocal of the pulse duration (a textbook simplification — strictly the optical frequency would come from the radiation's wavelength, but no wavelength is given):

$$\nu = \dfrac{1}{\tau} = \dfrac{1}{2 \times 10^{-9}\,\mathrm{s}} = 5 \times 10^{8}\,\mathrm{Hz}.$$

Energy of one photon:

$$E_{\text{photon}} = h\nu = (6.626 \times 10^{-34}\,\mathrm{J\,s})(5 \times 10^{8}\,\mathrm{s^{-1}}) = 3.313 \times 10^{-25}\,\mathrm{J}.$$

Total energy of the pulse:

$$E_{\text{pulse}} = N \cdot E_{\text{photon}} = (2.5 \times 10^{15})(3.313 \times 10^{-25}\,\mathrm{J}) = 8.28 \times 10^{-10}\,\mathrm{J}.$$

Frequencies $$\nu = c/\lambda$$.

For $$\lambda_1 = 589 \, \mathrm{nm} = 5.89 \times 10^{-7} \, \mathrm{m}$$:

$$\nu_1 = \dfrac{3 \times 10^{8}}{5.89 \times 10^{-7}} \approx 5.093 \times 10^{14} \, \mathrm{Hz}$$.

For $$\lambda_2 = 589.6 \, \mathrm{nm} = 5.896 \times 10^{-7} \, \mathrm{m}$$:

$$\nu_2 = \dfrac{3 \times 10^{8}}{5.896 \times 10^{-7}} \approx 5.088 \times 10^{14} \, \mathrm{Hz}$$.

Energy difference between the two excited states:

$$\Delta E = h \, \Delta\nu = h \, c\!\left(\dfrac{1}{\lambda_1} - \dfrac{1}{\lambda_2}\right)$$

$$\Delta E = (6.626 \times 10^{-34})(3 \times 10^{8})\!\left(\dfrac{\lambda_2 - \lambda_1}{\lambda_1 \lambda_2}\right)$$

$$\Delta E = (6.626 \times 10^{-34})(3 \times 10^{8})\dfrac{0.6 \times 10^{-9}}{(5.89)(5.896) \times 10^{-14}}$$

$$\Delta E \approx 3.43 \times 10^{-22} \, \mathrm{J}$$.

$$W_0 = 1.9 \, \mathrm{eV} = 1.9 \times 1.602 \times 10^{-19} = 3.044 \times 10^{-19} \, \mathrm{J}$$.

(a) At threshold, $$W_0 = hc/\lambda_0$$:

$$\lambda_0 = \dfrac{hc}{W_0} = \dfrac{(6.626 \times 10^{-34})(3 \times 10^{8})}{3.044 \times 10^{-19}}$$

$$\lambda_0 \approx 6.53 \times 10^{-7} \, \mathrm{m} = 653 \, \mathrm{nm}$$.

(b) Threshold frequency:

$$\nu_0 = \dfrac{c}{\lambda_0} = \dfrac{3 \times 10^{8}}{6.53 \times 10^{-7}} \approx 4.59 \times 10^{14} \, \mathrm{Hz}$$.

Irradiation with $$\lambda = 500 \, \mathrm{nm} = 5 \times 10^{-7} \, \mathrm{m}$$:

$$E_{\text{ph}} = \dfrac{hc}{\lambda} = \dfrac{(6.626 \times 10^{-34})(3 \times 10^{8})}{5 \times 10^{-7}} = 3.976 \times 10^{-19} \, \mathrm{J}$$.

Kinetic energy:

$$\mathrm{KE} = E_{\text{ph}} - W_0 = 3.976 \times 10^{-19} - 3.044 \times 10^{-19} = 9.32 \times 10^{-20} \, \mathrm{J}$$.

Velocity:

$$v = \sqrt{\dfrac{2\,\mathrm{KE}}{m_e}} = \sqrt{\dfrac{2 \times 9.32 \times 10^{-20}}{9.11 \times 10^{-31}}}$$

$$v = \sqrt{2.046 \times 10^{11}} \approx 4.52 \times 10^{5} \, \mathrm{m \, s^{-1}}$$.

(The velocities given in the data are in m s$${}^{-1}$$ rather than cm s$${}^{-1}$$ — only this reading gives physically sensible photoelectron speeds and the standard answer.) Hence $$v_1 = 2.55 \times 10^{5}$$, $$v_2 = 4.35 \times 10^{5}$$ and $$v_3 = 5.35 \times 10^{5} \, \mathrm{m \, s^{-1}}$$.

Einstein's photoelectric equation:

$$\dfrac{hc}{\lambda} = W_0 + \dfrac{1}{2}m_e v^{2}$$.

Apply it to two data points (taking the 500 nm and 400 nm rows, which lie at the ends of the data set):

$$\dfrac{hc}{\lambda_1} - \dfrac{hc}{\lambda_3} = \dfrac{1}{2}m_e(v_3^{2} - v_1^{2})$$

$$hc\left(\dfrac{1}{\lambda_3} - \dfrac{1}{\lambda_1}\right) = \dfrac{1}{2}m_e(v_3^{2} - v_1^{2})$$.

Numerically:

$$\dfrac{1}{\lambda_3} - \dfrac{1}{\lambda_1} = \dfrac{1}{400 \times 10^{-9}} - \dfrac{1}{500 \times 10^{-9}} = 5.0 \times 10^{5} \, \mathrm{m^{-1}}$$.

$$v_3^{2} - v_1^{2} = (5.35^{2} - 2.55^{2}) \times 10^{10} = 22.12 \times 10^{10} \, \mathrm{m^{2}\,s^{-2}}$$.

$$\dfrac{1}{2}m_e (v_3^{2} - v_1^{2}) = \tfrac{1}{2}(9.11 \times 10^{-31})(2.212 \times 10^{11}) \approx 1.008 \times 10^{-19} \, \mathrm{J}$$.

Hence

$$h = \dfrac{1.008 \times 10^{-19}}{c \times 5.0 \times 10^{5}} = \dfrac{1.008 \times 10^{-19}}{(3 \times 10^{8})(5.0 \times 10^{5})}$$

$$h \approx 6.7 \times 10^{-34} \, \mathrm{J \, s}$$.

This agrees well with the accepted value $$h = 6.626 \times 10^{-34} \, \mathrm{J \, s}$$.

(a) Threshold wavelength: use the 500 nm point.

$$W_0 = \dfrac{hc}{\lambda_1} - \tfrac{1}{2}m_e v_1^{2}$$

$$W_0 = \dfrac{(6.7 \times 10^{-34})(3 \times 10^{8})}{5 \times 10^{-7}} - \tfrac{1}{2}(9.11 \times 10^{-31})(2.55 \times 10^{5})^{2}$$

$$W_0 \approx 4.02 \times 10^{-19} - 2.96 \times 10^{-20} \approx 3.72 \times 10^{-19} \, \mathrm{J}$$.

$$\lambda_0 = \dfrac{hc}{W_0} = \dfrac{(6.626 \times 10^{-34})(3 \times 10^{8})}{3.72 \times 10^{-19}}$$

$$\lambda_0 \approx 5.35 \times 10^{-7} \, \mathrm{m} \approx 540 \, \mathrm{nm}$$.

At the stopping voltage $$V_0$$, the maximum kinetic energy of the photoelectron equals $$eV_0$$:

$$\mathrm{KE}_{\max} = eV_0 = (1.602 \times 10^{-19})(0.35) = 5.607 \times 10^{-20} \, \mathrm{J}$$.

Energy of the incident photon ($$\lambda = 256.7 \, \mathrm{nm} = 2.567 \times 10^{-7} \, \mathrm{m}$$):

$$E_{\text{ph}} = \dfrac{hc}{\lambda} = \dfrac{(6.626 \times 10^{-34})(3 \times 10^{8})}{2.567 \times 10^{-7}}$$

$$E_{\text{ph}} \approx 7.744 \times 10^{-19} \, \mathrm{J}$$.

Work function:

$$W_0 = E_{\text{ph}} - \mathrm{KE}_{\max} = 7.744 \times 10^{-19} - 5.607 \times 10^{-20}$$

$$W_0 \approx 7.18 \times 10^{-19} \, \mathrm{J} \approx 4.48 \, \mathrm{eV}$$.

Energy of the incident photon ($$\lambda = 150 \, \mathrm{pm} = 1.5 \times 10^{-10} \, \mathrm{m}$$):

$$E_{\text{ph}} = \dfrac{hc}{\lambda} = \dfrac{(6.626 \times 10^{-34})(3 \times 10^{8})}{1.5 \times 10^{-10}}$$

$$E_{\text{ph}} \approx 1.325 \times 10^{-15} \, \mathrm{J}$$.

Kinetic energy of the ejected electron:

$$\mathrm{KE} = \tfrac{1}{2} m_e v^{2} = \tfrac{1}{2}(9.11 \times 10^{-31})(1.5 \times 10^{7})^{2}$$

$$\mathrm{KE} = \tfrac{1}{2}(9.11 \times 10^{-31})(2.25 \times 10^{14}) \approx 1.025 \times 10^{-16} \, \mathrm{J}$$.

Binding energy:

$$E_{\text{bind}} = E_{\text{ph}} - \mathrm{KE} = 1.325 \times 10^{-15} - 1.025 \times 10^{-16}$$

$$E_{\text{bind}} \approx 1.223 \times 10^{-15} \, \mathrm{J} \approx 7.63 \times 10^{3} \, \mathrm{eV} \, (7.63 \, \mathrm{keV})$$.

Frequency of the line ($$\lambda = 1285 \, \mathrm{nm} = 1.285 \times 10^{-6} \, \mathrm{m}$$):

$$\nu = \dfrac{c}{\lambda} = \dfrac{3 \times 10^{8}}{1.285 \times 10^{-6}} \approx 2.335 \times 10^{14} \, \mathrm{Hz}$$.

Use the given formula:

$$2.335 \times 10^{14} = 3.29 \times 10^{15}\left(\dfrac{1}{9} - \dfrac{1}{n^{2}}\right)$$

$$\dfrac{1}{9} - \dfrac{1}{n^{2}} = \dfrac{2.335 \times 10^{14}}{3.29 \times 10^{15}} = 0.07097$$.

$$\dfrac{1}{n^{2}} = 0.1111 - 0.07097 = 0.04014$$.

$$n^{2} \approx 24.91 \quad \Rightarrow \quad n = 5$$.

Since $$\lambda = 1285 \, \mathrm{nm}$$ lies beyond the visible region, the line falls in the infrared (as expected for the Paschen series).

Bohr's radius for hydrogen: $$r_n = 52.9 \, n^{2} \, \mathrm{pm}$$.

For the starting orbit ($$r = 1.3225 \, \mathrm{nm} = 1322.5 \, \mathrm{pm}$$):

$$n^{2} = \dfrac{1322.5}{52.9} = 25 \quad \Rightarrow \quad n_2 = 5$$.

For the ending orbit ($$r = 211.6 \, \mathrm{pm}$$):

$$n^{2} = \dfrac{211.6}{52.9} = 4 \quad \Rightarrow \quad n_1 = 2$$.

So the transition is $$n = 5 \to n = 2$$. Transitions ending at $$n = 2$$ belong to the Balmer series (visible region).

Wavenumber:

$$\bar{\nu} = R_H\left(\dfrac{1}{n_1^{2}} - \dfrac{1}{n_2^{2}}\right) = 1.097 \times 10^{7}\left(\dfrac{1}{4} - \dfrac{1}{25}\right)$$

$$\bar{\nu} = 1.097 \times 10^{7} \times \dfrac{21}{100} = 2.304 \times 10^{6} \, \mathrm{m^{-1}}$$.

Wavelength:

$$\lambda = \dfrac{1}{\bar{\nu}} = \dfrac{1}{2.304 \times 10^{6}} \approx 4.34 \times 10^{-7} \, \mathrm{m} \approx 434 \, \mathrm{nm}$$.

$$\lambda = \dfrac{h}{m_e v}$$, with $$m_e = 9.11 \times 10^{-31} \, \mathrm{kg}$$.

$$\lambda = \dfrac{6.626 \times 10^{-34}}{(9.11 \times 10^{-31})(1.6 \times 10^{6})}$$

$$\lambda = \dfrac{6.626 \times 10^{-34}}{1.458 \times 10^{-24}} \approx 4.55 \times 10^{-10} \, \mathrm{m} = 455 \, \mathrm{pm}$$.

$$\lambda = \dfrac{h}{m_n v} \Rightarrow v = \dfrac{h}{m_n \lambda}$$, with $$m_n = 1.675 \times 10^{-27} \, \mathrm{kg}$$ and $$\lambda = 800 \, \mathrm{pm} = 8.0 \times 10^{-10} \, \mathrm{m}$$.

$$v = \dfrac{6.626 \times 10^{-34}}{(1.675 \times 10^{-27})(8.0 \times 10^{-10})}$$

$$v = \dfrac{6.626 \times 10^{-34}}{1.34 \times 10^{-36}} \approx 494.5 \, \mathrm{m \, s^{-1}}$$.

$$\lambda = \dfrac{h}{m_e v}$$.

$$\lambda = \dfrac{6.626 \times 10^{-34}}{(9.11 \times 10^{-31})(2.19 \times 10^{6})}$$

$$\lambda = \dfrac{6.626 \times 10^{-34}}{1.995 \times 10^{-24}} \approx 3.32 \times 10^{-10} \, \mathrm{m}$$

$$\lambda \approx 332 \, \mathrm{pm} \, (3.32 \, \text{Å})$$.

This equals the circumference $$2\pi r_1$$ of the first Bohr orbit, consistent with $$2\pi r_n = n\lambda$$ for $$n=1$$.

de Broglie wavelength of the hockey ball ($$m = 0.1 \, \mathrm{kg}$$, $$v = 4.37 \times 10^{5} \, \mathrm{m \, s^{-1}}$$):

$$\lambda = \dfrac{h}{mv} = \dfrac{6.626 \times 10^{-34}}{(0.1)(4.37 \times 10^{5})}$$

$$\lambda = \dfrac{6.626 \times 10^{-34}}{4.37 \times 10^{4}} \approx 1.516 \times 10^{-38} \, \mathrm{m}$$.

The wavelength is vanishingly small, illustrating that wave behaviour of a macroscopic object is unobservable.

Uncertainty in position: $$\Delta x = 0.002 \, \mathrm{nm} = 2 \times 10^{-12} \, \mathrm{m}$$ (the $$\pm$$ already gives the full uncertainty here).

Heisenberg's principle gives

$$\Delta p \ge \dfrac{h}{4\pi \, \Delta x}$$

$$\Delta p \ge \dfrac{6.626 \times 10^{-34}}{4 \times 3.1416 \times 2 \times 10^{-12}}$$

$$\Delta p \approx 2.636 \times 10^{-23} \, \mathrm{kg \, m \, s^{-1}}$$.

The momentum stated in the question is

$$p = \dfrac{h}{4\pi \times 0.05 \, \mathrm{nm}} = \dfrac{6.626 \times 10^{-34}}{4 \times 3.1416 \times 5 \times 10^{-11}}$$

$$p \approx 1.054 \times 10^{-24} \, \mathrm{kg \, m \, s^{-1}}$$.

This value of momentum ($$1.05 \times 10^{-24}$$) is much smaller than the uncertainty in momentum ($$2.64 \times 10^{-23}$$) — i.e. $$\Delta p \gg p$$. Hence the momentum itself cannot be defined with any meaningful precision: the uncertainty is more than 20 times the proposed value of the momentum.

Effective nuclear charge ($$Z_{\text{eff}}$$) is reduced by inner-shell shielding. An electron lying in a higher (outer) shell is screened by all the electrons in the lower shells, so it feels a smaller $$Z_{\text{eff}}$$.

Among the three orbitals listed, 2p is the innermost, 3p is intermediate and 4p is the outermost. The 4p electrons therefore have the largest screening and the lowest $$Z_{\text{eff}}$$.

An orbital that is closer to the nucleus (smaller $$n$$) — or, within the same shell, has higher penetration (smaller $$l$$) — experiences a larger $$Z_{\text{eff}}$$.

• (i) 2s lies closer to the nucleus than 3s, so 2s feels the larger $$Z_{\text{eff}}$$.
• (ii) For the same $$n = 4$$, the d orbital penetrates the inner shells more than the f orbital. Hence 4d feels the larger $$Z_{\text{eff}}$$.
• (iii) For the same $$n = 3$$, the p orbital penetrates the inner core more than the d orbital. Hence 3p feels the larger $$Z_{\text{eff}}$$.

Electron configurations:

• Al ($$Z = 13$$): $$1s^{2}\,2s^{2}\,2p^{6}\,3s^{2}\,3p^{1}$$.
• Si ($$Z = 14$$): $$1s^{2}\,2s^{2}\,2p^{6}\,3s^{2}\,3p^{2}$$.

Both have the same inner-shell electron count (10), so the screening of the 3p electrons is essentially the same in the two atoms. However, the nuclear charge $$Z$$ in Si (14) is greater than in Al (13). Hence the 3p electrons in Si feel a higher effective nuclear charge than those in Al.