Structure of Atom

(i) Given: Mass of one electron = 9.10939 × 10⁻²⁸ g

Number of electrons that weigh 1 g:
$$N = \frac{1\text{ g}}{9.10939 \times 10^{-28}\text{ g}} = 1.098 \times 10^{27} \text{ electrons}$$

(ii) Mass of one mole of electrons:
$$m = 6.022 \times 10^{23} \times 9.10939 \times 10^{-31}\text{ kg} = 5.486 \times 10^{-7}\text{ kg} \approx 5.49 \times 10^{-4}\text{ g}$$

Charge of one mole of electrons:
$$Q = 6.022 \times 10^{23} \times 1.6022 \times 10^{-19}\text{ C} = 9.65 \times 10^{4}\text{ C} = 96500\text{ C (1 Faraday)}$$

(i) Methane (CH₄): electrons per molecule = 6 (C) + 4×1 (H) = 10 electrons

Total electrons in 1 mole of CH₄:
$$N = 10 \times 6.022 \times 10^{23} = 6.022 \times 10^{24}\text{ electrons}$$

(ii) ¹⁴C has 8 neutrons per atom. Molar mass of ¹⁴C = 14 g/mol.

Moles of ¹⁴C in 7 mg:
$$n = \frac{7 \times 10^{-3}\text{ g}}{14\text{ g/mol}} = 5 \times 10^{-4}\text{ mol}$$

(a) Total number of neutrons:
$$= 5 \times 10^{-4} \times 6.022 \times 10^{23} \times 8 = 2.409 \times 10^{21}\text{ neutrons}$$

(b) Total mass of neutrons:
$$= 2.409 \times 10^{21} \times 1.675 \times 10^{-27}\text{ kg} = 4.035 \times 10^{-6}\text{ kg}$$

(iii) NH₃ has molar mass = 17 g/mol. Protons per molecule = 7 (N) + 3×1 (H) = 10 protons.

Moles of NH₃ in 34 mg:
$$n = \frac{34 \times 10^{-3}\text{ g}}{17\text{ g/mol}} = 2 \times 10^{-3}\text{ mol}$$

(a) Total number of protons:
$$= 2 \times 10^{-3} \times 6.022 \times 10^{23} \times 10 = 1.2044 \times 10^{22}\text{ protons}$$

(b) Total mass of protons:
$$= 1.2044 \times 10^{22} \times 1.6726 \times 10^{-27}\text{ kg} = 2.014 \times 10^{-5}\text{ kg}$$

The answer will NOT change with temperature and pressure because the number of moles (and hence number of protons) depends only on the mass of the sample, not on T and P.

For any nucleus $^A_Z X$: Number of protons = Z, Number of neutrons = A − Z.

$$^{13}_6\text{C}: \text{Protons} = 6,\quad \text{Neutrons} = 13 - 6 = 7$$

$$^{16}_8\text{O}: \text{Protons} = 8,\quad \text{Neutrons} = 16 - 8 = 8$$

$$^{24}_{12}\text{Mg}: \text{Protons} = 12,\quad \text{Neutrons} = 24 - 12 = 12$$

$$^{56}_{26}\text{Fe}: \text{Protons} = 26,\quad \text{Neutrons} = 56 - 26 = 30$$

$$^{88}_{38}\text{Sr}: \text{Protons} = 38,\quad \text{Neutrons} = 88 - 38 = 50$$

The complete symbol is $^A_Z X$ where X is the element symbol.

(i) Z = 17 → Chlorine (Cl); A = 35:
$$^{35}_{17}\text{Cl}$$

(ii) Z = 92 → Uranium (U); A = 233:
$$^{233}_{92}\text{U}$$

(iii) Z = 4 → Beryllium (Be); A = 9:
$$^{9}_{4}\text{Be}$$

Given: $\lambda = 580\text{ nm} = 580 \times 10^{-9}\text{ m}$, $c = 3 \times 10^8\text{ m s}^{-1}$

Frequency:
$$\nu = \frac{c}{\lambda} = \frac{3 \times 10^8\text{ m s}^{-1}}{580 \times 10^{-9}\text{ m}} = 5.17 \times 10^{14}\text{ Hz}$$

Wavenumber:
$$\bar{\nu} = \frac{1}{\lambda} = \frac{1}{580 \times 10^{-9}\text{ m}} = 1.724 \times 10^{6}\text{ m}^{-1} = 1.724 \times 10^{4}\text{ cm}^{-1}$$

Using $E = h\nu$ and $E = hc/\lambda$; $h = 6.626 \times 10^{-34}\text{ J s}$

(i) $\nu = 3 \times 10^{15}\text{ Hz}$:
$$E = h\nu = 6.626 \times 10^{-34} \times 3 \times 10^{15} = 1.988 \times 10^{-18}\text{ J}$$

(ii) $\lambda = 0.50\text{ Å} = 0.50 \times 10^{-10}\text{ m} = 5 \times 10^{-11}\text{ m}$:
$$E = \frac{hc}{\lambda} = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{5 \times 10^{-11}} = 3.976 \times 10^{-15}\text{ J} \approx 3.98 \times 10^{-15}\text{ J}$$

Given: Time period $T = 2.0 \times 10^{-10}\text{ s}$

Frequency:
$$\nu = \frac{1}{T} = \frac{1}{2.0 \times 10^{-10}} = 5.0 \times 10^{9}\text{ Hz}$$

Wavelength:
$$\lambda = \frac{c}{\nu} = \frac{3 \times 10^8}{5.0 \times 10^9} = 6.0 \times 10^{-2}\text{ m} = 0.06\text{ m}$$

Wavenumber:
$$\bar{\nu} = \frac{1}{\lambda} = \frac{1}{6.0 \times 10^{-2}} = 16.67\text{ m}^{-1} = 0.1667\text{ cm}^{-1}$$

Given: $\lambda = 4000\text{ pm} = 4000 \times 10^{-12}\text{ m} = 4 \times 10^{-9}\text{ m}$

Energy of one photon:
$$E = \frac{hc}{\lambda} = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{4 \times 10^{-9}} = 4.969 \times 10^{-17}\text{ J}$$

Number of photons:
$$n = \frac{1\text{ J}}{4.969 \times 10^{-17}\text{ J}} = 2.012 \times 10^{16} \approx 2.0 \times 10^{16}\text{ photons}$$

Given: $\lambda = 4 \times 10^{-7}\text{ m}$, $W_0 = 2.13\text{ eV}$

(i) Energy of photon:
$$E = \frac{hc}{\lambda} = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{4 \times 10^{-7}} = 4.969 \times 10^{-19}\text{ J}$$
$$E = \frac{4.969 \times 10^{-19}}{1.6020 \times 10^{-19}} = 3.10\text{ eV}$$

(ii) Kinetic energy of emitted electron:
$$KE = E - W_0 = 3.10 - 2.13 = 0.97\text{ eV}$$
$$KE = 0.97 \times 1.6020 \times 10^{-19} = 1.554 \times 10^{-19}\text{ J}$$

(iii) Velocity of photoelectron ($m_e = 9.109 \times 10^{-31}\text{ kg}$):
$$KE = \frac{1}{2}m_e v^2 \Rightarrow v = \sqrt{\frac{2 \times KE}{m_e}}$$
$$v = \sqrt{\frac{2 \times 1.554 \times 10^{-19}}{9.109 \times 10^{-31}}} = \sqrt{3.413 \times 10^{11}} = 5.84 \times 10^{5}\text{ m s}^{-1}$$

Given: $\lambda = 242\text{ nm} = 242 \times 10^{-9}\text{ m}$

Energy per photon:
$$E = \frac{hc}{\lambda} = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{242 \times 10^{-9}} = 8.21 \times 10^{-19}\text{ J}$$

Ionisation energy per mole:
$$IE = E \times N_A = 8.21 \times 10^{-19} \times 6.022 \times 10^{23} = 4.945 \times 10^{5}\text{ J mol}^{-1}$$
$$\boxed{IE = 494.5\text{ kJ mol}^{-1}}$$

Given: Power $P = 25\text{ W} = 25\text{ J s}^{-1}$, $\lambda = 0.57\text{ μm} = 0.57 \times 10^{-6}\text{ m}$

Energy per photon:
$$E = \frac{hc}{\lambda} = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{0.57 \times 10^{-6}} = 3.487 \times 10^{-19}\text{ J}$$

Rate of emission:
$$n = \frac{P}{E} = \frac{25}{3.487 \times 10^{-19}} = 7.169 \times 10^{19} \approx 7.17 \times 10^{19}\text{ quanta s}^{-1}$$

Given: $\lambda_0 = 6800\text{ Å} = 6800 \times 10^{-10}\text{ m} = 6.8 \times 10^{-7}\text{ m}$

Since electrons are emitted with zero velocity, the incident wavelength equals the threshold wavelength.

Threshold frequency:
$$\nu_0 = \frac{c}{\lambda_0} = \frac{3 \times 10^8}{6.8 \times 10^{-7}} = 4.41 \times 10^{14}\text{ Hz}$$

Work function:
$$W_0 = h\nu_0 = 6.626 \times 10^{-34} \times 4.41 \times 10^{14} = 2.922 \times 10^{-19}\text{ J}$$

Using the Rydberg formula:
$$\frac{1}{\lambda} = R_H\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$$

where $R_H = 1.097 \times 10^7\text{ m}^{-1}$, $n_1 = 2$, $n_2 = 4$:

$$\frac{1}{\lambda} = 1.097 \times 10^7\left(\frac{1}{4} - \frac{1}{16}\right) = 1.097 \times 10^7 \times \frac{3}{16}$$

$$\frac{1}{\lambda} = 1.097 \times 10^7 \times 0.1875 = 2.057 \times 10^6\text{ m}^{-1}$$

$$\lambda = \frac{1}{2.057 \times 10^6} = 4.86 \times 10^{-7}\text{ m} = 486\text{ nm}$$

This is the blue-green line in the Balmer series (visible region).

Energy of electron in nth orbit of H atom:
$$E_n = \frac{-2.18 \times 10^{-18}}{n^2}\text{ J}$$

For n = 5:
$$E_5 = \frac{-2.18 \times 10^{-18}}{25} = -8.72 \times 10^{-20}\text{ J}$$

Energy required to ionise from n = 5 (remove to n = ∞, where E = 0):
$$\Delta E = 0 - (-8.72 \times 10^{-20}) = 8.72 \times 10^{-20}\text{ J}$$

For n = 1:
$$E_1 = -2.18 \times 10^{-18}\text{ J}$$
$$\Delta E_1 = 2.18 \times 10^{-18}\text{ J}$$

Comparison:
$$\frac{\Delta E_1}{\Delta E_5} = \frac{2.18 \times 10^{-18}}{8.72 \times 10^{-20}} = 25$$

The energy required to ionise from n = 1 is 25 times greater than from n = 5.

When an electron drops from the nth level to the ground state, the maximum number of spectral lines is given by:
$$\text{Number of lines} = \frac{n(n-1)}{2}$$

For n = 6:
$$\text{Number of lines} = \frac{6(6-1)}{2} = \frac{6 \times 5}{2} = 15$$

The 15 possible transitions are: 6→5, 6→4, 6→3, 6→2, 6→1, 5→4, 5→3, 5→2, 5→1, 4→3, 4→2, 4→1, 3→2, 3→1, 2→1.

(i) Energy of nth orbit: $E_n = E_1/n^2$

$$E_5 = \frac{-2.18 \times 10^{-18}}{5^2} = \frac{-2.18 \times 10^{-18}}{25} = -8.72 \times 10^{-20}\text{ J atom}^{-1}$$

(ii) Radius of nth Bohr orbit for hydrogen: $r_n = 0.0529 \times n^2\text{ nm}$

$$r_5 = 0.0529 \times 5^2 = 0.0529 \times 25 = 1.3225\text{ nm} = 132.25\text{ pm}$$

The Balmer series has $n_1 = 2$. Longest wavelength corresponds to smallest energy transition, i.e., $n_2 = 3$.

$$\bar{\nu} = R_H\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right) = 1.097 \times 10^7\left(\frac{1}{4} - \frac{1}{9}\right)$$

$$= 1.097 \times 10^7 \times \frac{9-4}{36} = 1.097 \times 10^7 \times \frac{5}{36}$$

$$= 1.097 \times 10^7 \times 0.1389 = 1.523 \times 10^6\text{ m}^{-1}$$

$$\bar{\nu} = 1.523 \times 10^4\text{ cm}^{-1}$$

Given: $E_1 = -2.18 \times 10^{-11}\text{ ergs} = -2.18 \times 10^{-18}\text{ J}$

$$E_5 = \frac{E_1}{25} = \frac{-2.18 \times 10^{-18}}{25} = -8.72 \times 10^{-20}\text{ J}$$

Energy required to shift from n=1 to n=5:
$$\Delta E = E_5 - E_1 = -8.72 \times 10^{-20} - (-2.18 \times 10^{-18})$$
$$= 2.18 \times 10^{-18} - 8.72 \times 10^{-20} = 2.0928 \times 10^{-18}\text{ J}$$

Wavelength of light emitted when electron returns to ground state (n=5 → n=1):
$$\lambda = \frac{hc}{\Delta E} = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{2.0928 \times 10^{-18}}$$
$$= \frac{1.988 \times 10^{-25}}{2.0928 \times 10^{-18}} = 9.498 \times 10^{-8}\text{ m} \approx 9.5 \times 10^{-8}\text{ m} = 95\text{ nm}$$

Energy at n = 2:
$$E_2 = \frac{-2.18 \times 10^{-18}}{4} = -5.45 \times 10^{-19}\text{ J}$$

Energy at n = ∞: $E_\infty = 0$

Energy required:
$$\Delta E = 0 - (-5.45 \times 10^{-19}) = 5.45 \times 10^{-19}\text{ J}$$

Longest wavelength (minimum energy = exactly $\Delta E$):
$$\lambda = \frac{hc}{\Delta E} = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{5.45 \times 10^{-19}}$$
$$= \frac{1.988 \times 10^{-25}}{5.45 \times 10^{-19}} = 3.647 \times 10^{-7}\text{ m} = 3.647 \times 10^{-5}\text{ cm}$$

Using de Broglie equation: $\lambda = \dfrac{h}{mv}$

Given: $m = 9.109 \times 10^{-31}\text{ kg}$, $v = 2.05 \times 10^7\text{ m s}^{-1}$, $h = 6.626 \times 10^{-34}\text{ J s}$

$$\lambda = \frac{6.626 \times 10^{-34}}{9.109 \times 10^{-31} \times 2.05 \times 10^7}$$

$$= \frac{6.626 \times 10^{-34}}{1.867 \times 10^{-23}} = 3.548 \times 10^{-11}\text{ m} \approx 3.55 \times 10^{-11}\text{ m}$$

Given: $m = 9.1 \times 10^{-31}\text{ kg}$, $KE = 3.0 \times 10^{-25}\text{ J}$

First find velocity:
$$KE = \frac{1}{2}mv^2 \Rightarrow v = \sqrt{\frac{2 \times KE}{m}} = \sqrt{\frac{2 \times 3.0 \times 10^{-25}}{9.1 \times 10^{-31}}}$$
$$= \sqrt{6.593 \times 10^5} = 8.12 \times 10^2\text{ m s}^{-1}$$

de Broglie wavelength:
$$\lambda = \frac{h}{mv} = \frac{6.626 \times 10^{-34}}{9.1 \times 10^{-31} \times 8.12 \times 10^2}$$
$$= \frac{6.626 \times 10^{-34}}{7.389 \times 10^{-28}} = 8.967 \times 10^{-7}\text{ m} \approx 8.97 \times 10^{-7}\text{ m}$$

Count electrons for each species:

- $\text{Na}^+$: Na has 11 electrons, loses 1 → 10 electrons
- $\text{K}^+$: K has 19 electrons, loses 1 → 18 electrons
- $\text{Mg}^{2+}$: Mg has 12 electrons, loses 2 → 10 electrons
- $\text{Ca}^{2+}$: Ca has 20 electrons, loses 2 → 18 electrons
- $\text{S}^{2-}$: S has 16 electrons, gains 2 → 18 electrons
- $\text{Ar}$: Ar has 18 electrons

Isoelectronic pairs:
- $\text{Na}^+$ and $\text{Mg}^{2+}$ (10 electrons each)
- $\text{K}^+$, $\text{Ca}^{2+}$, $\text{S}^{2-}$, and $\text{Ar}$ (18 electrons each)

(i) Electronic configurations:

(a) $\text{H}^-$: H has 1 electron, gains 1 → 2 electrons
$$\text{H}^-: 1s^2$$

(b) $\text{Na}^+$: Na has 11 electrons, loses 1 → 10 electrons
$$\text{Na}^+: 1s^2\, 2s^2\, 2p^6$$

(c) $\text{O}^{2-}$: O has 8 electrons, gains 2 → 10 electrons
$$\text{O}^{2-}: 1s^2\, 2s^2\, 2p^6$$

(d) $\text{F}^-$: F has 9 electrons, gains 1 → 10 electrons
$$\text{F}^-: 1s^2\, 2s^2\, 2p^6$$

(ii) Atomic numbers:

(a) $3s^1$: Configuration is $1s^2\, 2s^2\, 2p^6\, 3s^1$ → Total electrons = 11 → Z = 11 (Na)

(b) $2p^3$: Configuration is $1s^2\, 2s^2\, 2p^3$ → Total electrons = 7 → Z = 7 (N)

(c) $3p^5$: Configuration is $1s^2\, 2s^2\, 2p^6\, 3s^2\, 3p^5$ → Total electrons = 17 → Z = 17 (Cl)

(iii) Atoms:

(a) $[\text{He}]\, 2s^1$: $1s^2\, 2s^1$ → 3 electrons → Lithium (Li), Z = 3

(b) $[\text{Ne}]\, 3s^2\, 3p^3$: $1s^2\, 2s^2\, 2p^6\, 3s^2\, 3p^3$ → 15 electrons → Phosphorus (P), Z = 15

(c) $[\text{Ar}]\, 4s^2\, 3d^1$: $1s^2\, 2s^2\, 2p^6\, 3s^2\, 3p^6\, 4s^2\, 3d^1$ → 21 electrons → Scandium (Sc), Z = 21

For g orbitals, $l = 4$ (since s→l=0, p→l=1, d→l=2, f→l=3, g→l=4).

The condition is $l \leq n-1$, so:
$$n - 1 \geq 4 \Rightarrow n \geq 5$$

The lowest value of $n$ that allows g orbitals to exist is $\boxed{n = 5}$.

For a 3d orbital:
- Principal quantum number: $n = 3$
- Azimuthal quantum number: $l = 2$ (d orbital)
- Magnetic quantum number: $m_l = -2, -1, 0, +1, +2$

So the possible values are:
$$n = 3,\quad l = 2,\quad m_l \in \{-2, -1, 0, +1, +2\}$$

(i) In a neutral atom, number of protons = number of electrons.
$$\text{Number of protons} = 29$$

The element is Copper (Cu), Z = 29.

(ii) Electronic configuration of Cu (Z = 29):

Expected: $[\text{Ar}]\, 4s^2\, 3d^9$

Actual (due to extra stability of completely filled d-subshell):
$$\text{Cu}: 1s^2\, 2s^2\, 2p^6\, 3s^2\, 3p^6\, 3d^{10}\, 4s^1$$
or $[\text{Ar}]\, 3d^{10}\, 4s^1$

- $\text{H}_2^+$: H₂ has 2 electrons, loses 1 → 1 electron
- $\text{H}_2$: Each H has 1 electron → 2 electrons
- $\text{O}_2^+$: O₂ has 16 electrons (2×8), loses 1 → 15 electrons

(i) For n = 3, $l$ can be 0, 1, 2.

- $l = 0$: $m_l = 0$
- $l = 1$: $m_l = -1, 0, +1$
- $l = 2$: $m_l = -2, -1, 0, +1, +2$

(ii) For 3d orbital: $n = 3$, $l = 2$
$$m_l = -2, -1, 0, +1, +2$$

(iii) Checking possibility (condition: $l \leq n-1$):
- 1p: $n=1$, $l=1$; requires $l \leq 0$ → Not possible
- 2s: $n=2$, $l=0$; $0 \leq 1$ → Possible
- 2p: $n=2$, $l=1$; $1 \leq 1$ → Possible
- 3f: $n=3$, $l=3$; requires $l \leq 2$ → Not possible

Notation: $l = 0 \to s$, $l = 1 \to p$, $l = 2 \to d$, $l = 3 \to f$

(a) $n = 1$, $l = 0$ → 1s

(b) $n = 3$, $l = 1$ → 3p

(c) $n = 4$, $l = 2$ → 4d

(d) $n = 4$, $l = 3$ → 4f

(a) $n = 0$, $l = 0$, $m_l = 0$, $m_s = +\frac{1}{2}$
Not possible. The principal quantum number $n$ must be a positive integer ($n \geq 1$). $n = 0$ is not allowed.

(b) $n = 1$, $l = 0$, $m_l = 0$, $m_s = -\frac{1}{2}$
Possible. All values are valid: $l = 0 \leq n-1 = 0$, $m_l = 0$ is within $[-l, +l]$, and $m_s = -\frac{1}{2}$ is valid.

(c) $n = 1$, $l = 1$, $m_l = 0$, $m_s = +\frac{1}{2}$
Not possible. For $n = 1$, $l$ can only be 0 (since $l \leq n-1 = 0$). $l = 1$ is not allowed.

(d) $n = 2$, $l = 1$, $m_l = 0$, $m_s = -\frac{1}{2}$
Possible. $l = 1 \leq n-1 = 1$, $m_l = 0 \in [-1, +1]$, $m_s = -\frac{1}{2}$ is valid.

(e) $n = 3$, $l = 3$, $m_l = -3$, $m_s = +\frac{1}{2}$
Not possible. For $n = 3$, $l$ can be 0, 1, or 2 (since $l \leq n-1 = 2$). $l = 3$ is not allowed.

(f) $n = 3$, $l = 1$, $m_l = 0$, $m_s = +\frac{1}{2}$
Possible. $l = 1 \leq n-1 = 2$, $m_l = 0 \in [-1, +1]$, $m_s = +\frac{1}{2}$ is valid.

(a) $n = 4$, $m_s = -\frac{1}{2}$:

For $n = 4$, total orbitals = $n^2 = 16$. Each orbital can hold one electron with $m_s = -\frac{1}{2}$.
$$\text{Number of electrons} = 16$$

(b) $n = 3$, $l = 0$ (i.e., 3s orbital):

For $l = 0$, $m_l = 0$ only → 1 orbital → can hold 2 electrons.
$$\text{Number of electrons} = 2$$

According to Bohr's second postulate, the angular momentum of an electron in the nth orbit is quantized:
$$m_e v r = \frac{nh}{2\pi} \quad \text{...(1)}$$

The de Broglie wavelength of the electron:
$$\lambda = \frac{h}{m_e v} \Rightarrow m_e v = \frac{h}{\lambda} \quad \text{...(2)}$$

Substituting (2) into (1):
$$\frac{h}{\lambda} \cdot r = \frac{nh}{2\pi}$$

$$r = \frac{n\lambda}{2\pi}$$

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

Since $2\pi r$ is the circumference of the orbit:
$$\boxed{\text{Circumference} = n\lambda}$$

This shows that the circumference of the Bohr orbit is an integral multiple ($n$) of the de Broglie wavelength.

For He⁺ (Z = 2), the wavenumber for n = 4 → n = 2 transition:
$$\bar{\nu}_{\text{He}^+} = R_H Z^2 \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right) = R_H \times 4 \times \left(\frac{1}{4} - \frac{1}{16}\right) = R_H \times 4 \times \frac{3}{16} = \frac{3R_H}{4}$$

For hydrogen (Z = 1), we need the same wavenumber:
$$\bar{\nu}_{\text{H}} = R_H\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right) = \frac{3R_H}{4}$$

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

So $n_1 = 1$ and $n_2 = 2$.

The transition in hydrogen spectrum is n = 2 to n = 1 (Lyman series).

He⁺ is a hydrogen-like species with Z = 2. The energy of the electron in the nth orbit of a hydrogen-like ion is:
$$E_n = -\frac{Z^2 \times 2.18 \times 10^{-18}}{n^2}\text{ J}$$

For He⁺ in ground state (n = 1, Z = 2):
$$E_1(\text{He}^+) = -\frac{4 \times 2.18 \times 10^{-18}}{1} = -8.72 \times 10^{-18}\text{ J}$$

Energy required to remove the electron (ionization):
$$\Delta E = 0 - (-8.72 \times 10^{-18}) = 8.72 \times 10^{-18}\text{ J atom}^{-1}$$

Given: Diameter of C atom = 0.15 nm = $0.15 \times 10^{-9}\text{ m} = 1.5 \times 10^{-10}\text{ m}$

Length of scale = 20 cm = $20 \times 10^{-2}\text{ m} = 0.20\text{ m}$

Number of carbon atoms:
$$N = \frac{0.20\text{ m}}{1.5 \times 10^{-10}\text{ m}} = 1.33 \times 10^{9}\text{ atoms}$$

Given: Number of atoms = $2 \times 10^8$, Total length = 2.4 cm

Diameter of one carbon atom:
$$d = \frac{2.4\text{ cm}}{2 \times 10^8} = 1.2 \times 10^{-8}\text{ cm} = 1.2 \times 10^{-10}\text{ m}$$

Radius of carbon atom:
$$r = \frac{d}{2} = \frac{1.2 \times 10^{-10}}{2} = 6.0 \times 10^{-11}\text{ m} = 60\text{ pm}$$

Given: Diameter = 2.6 Å = $2.6 \times 10^{-10}\text{ m}$

(a) Radius in pm:
$$r = \frac{d}{2} = \frac{2.6 \times 10^{-10}}{2} = 1.3 \times 10^{-10}\text{ m} = 130\text{ pm}$$

(b) Number of atoms in 1.6 cm:
$$N = \frac{1.6\text{ cm}}{2.6 \times 10^{-8}\text{ cm}} = \frac{1.6}{2.6 \times 10^{-8}} = 6.15 \times 10^{7}\text{ atoms}$$

Given: Total charge = $2.5 \times 10^{-16}\text{ C}$, charge of one electron = $1.6022 \times 10^{-19}\text{ C}$

$$n = \frac{2.5 \times 10^{-16}}{1.6022 \times 10^{-19}} = 1560.4 \approx 1560\text{ electrons}$$

Given: Charge = $-1.282 \times 10^{-18}\text{ C}$, charge of one electron = $-1.6022 \times 10^{-19}\text{ C}$

$$n = \frac{1.282 \times 10^{-18}}{1.6022 \times 10^{-19}} = 8.0\text{ electrons}$$

There are 8 electrons on the oil drop.

In Rutherford's experiment, heavy atoms like gold are used because they have large nuclei with high positive charge (Z = 79 for Au), which causes significant deflection of α-particles.

If a thin foil of light atoms like aluminium (Z = 13) is used:
1. The nucleus of aluminium is smaller and carries less positive charge.
2. The electrostatic repulsion between the α-particles and the aluminium nucleus would be much weaker.
3. Therefore, fewer α-particles would be deflected at large angles, and the scattering pattern would show much less back-scattering.
4. Most α-particles would pass through with little or no deflection.

In summary, the angle of deflection would be much smaller, and the experiment would be less conclusive in demonstrating the nuclear model of the atom.

The complete nuclear symbol is $^A_Z X$ where A = mass number and Z = atomic number.

For bromine: Z = 35 (fixed for all bromine atoms).

- $^{79}_{35}\text{Br}$: A = 79, Z = 35 → This is acceptable because A > Z, which is physically valid (neutrons = 79 − 35 = 44).

- $^{35}_{35}\text{Br}$: A = 35, Z = 35 → This would mean neutrons = 35 − 35 = 0, which is not possible for an element with Z = 35. Also, for heavier elements, A must be greater than Z.

Thus, $^{35}\text{Br}$ is not acceptable because it implies zero neutrons in the nucleus, which is physically impossible for bromine.

Let number of protons = p, number of neutrons = n.

Given: $n = p + 31.7\%\text{ of }p = p + 0.317p = 1.317p$

Also: $p + n = 81$

Substituting:
$$p + 1.317p = 81$$
$$2.317p = 81$$
$$p = \frac{81}{2.317} = 34.96 \approx 35$$

So Z = 35 → Bromine (Br)

Atomic symbol: $\boxed{^{81}_{35}\text{Br}}$

Let number of protons = p, number of electrons = e, number of neutrons = n.

Since the ion has one unit of negative charge: $e = p + 1$

Given: $n = e + 11.1\%\text{ of }e = 1.111e$

Mass number: $p + n = 37$

Substituting $p = e - 1$ and $n = 1.111e$:
$$(e - 1) + 1.111e = 37$$
$$2.111e = 38$$
$$e = 18$$

So: $p = 18 - 1 = 17$, $n = 37 - 17 = 20$

Z = 17 → Chlorine (Cl)

The ion is $\boxed{^{37}_{17}\text{Cl}^-}$

Let number of protons = p, electrons = e, neutrons = n.

Since the ion has 3 units of positive charge: $e = p - 3$

Given: $n = e + 30.4\%\text{ of }e = 1.304e$

Mass number: $p + n = 56$

Substituting $p = e + 3$ and $n = 1.304e$:
$$(e + 3) + 1.304e = 56$$
$$2.304e = 53$$
$$e = 23$$

So: $p = 23 + 3 = 26$, $n = 56 - 26 = 30$

Z = 26 → Iron (Fe)

The ion is $\boxed{^{56}_{26}\text{Fe}^{3+}}$

The electromagnetic spectrum in order of increasing frequency:

Radio waves < Microwaves < Infrared < Visible < UV < X-rays < Gamma rays < Cosmic rays

Arranging the given radiations in increasing order of frequency:

\text{FM radio} &lt; \text{Microwave} &lt; \text{Amber light (visible)} &lt; \text{X-rays} &lt; \text{Cosmic rays}

(c) < (a) < (b) < (e) < (d)

Given: $\lambda = 337.1\text{ nm} = 337.1 \times 10^{-9}\text{ m}$, $N = 5.6 \times 10^{24}$ photons

Energy per photon:
$$E = \frac{hc}{\lambda} = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{337.1 \times 10^{-9}} = 5.899 \times 10^{-19}\text{ J}$$

Total energy:
$$E_{\text{total}} = N \times E = 5.6 \times 10^{24} \times 5.899 \times 10^{-19} = 3.303 \times 10^{6}\text{ J}$$

Note: The problem asks for power but does not specify time. Assuming the energy is emitted per second (or that power = total energy/1 s):
$$P = 3.303 \times 10^6\text{ W} \approx 3.3 \times 10^6\text{ W}$$

Given: $\lambda = 616\text{ nm} = 616 \times 10^{-9}\text{ m}$

(a) Frequency:
$$\nu = \frac{c}{\lambda} = \frac{3 \times 10^8}{616 \times 10^{-9}} = 4.87 \times 10^{14}\text{ Hz}$$

(b) Distance in 30 s:
$$d = c \times t = 3 \times 10^8 \times 30 = 9 \times 10^9\text{ m}$$

(c) Energy of one quantum:
$$E = h\nu = 6.626 \times 10^{-34} \times 4.87 \times 10^{14} = 3.227 \times 10^{-19}\text{ J}$$

(d) Number of quanta for 2 J:
$$n = \frac{2}{3.227 \times 10^{-19}} = 6.197 \times 10^{18} \approx 6.2 \times 10^{18}\text{ quanta}$$

Given: Total energy = $3.15 \times 10^{-18}\text{ J}$, $\lambda = 600\text{ nm} = 6 \times 10^{-7}\text{ m}$

Energy per photon:
$$E = \frac{hc}{\lambda} = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{6 \times 10^{-7}} = 3.313 \times 10^{-19}\text{ J}$$

Number of photons:
$$n = \frac{3.15 \times 10^{-18}}{3.313 \times 10^{-19}} = 9.51 \approx 9.5 \approx 10\text{ photons}$$

Given: Duration = 2 ns = $2 \times 10^{-9}\text{ s}$, $N = 2.5 \times 10^{15}$ photons

Using the uncertainty principle for energy-time: $\Delta E \cdot \Delta t \geq \dfrac{h}{4\pi}$

Energy of one photon from uncertainty:
$$\Delta E = \frac{h}{4\pi \Delta t} = \frac{6.626 \times 10^{-34}}{4\pi \times 2 \times 10^{-9}}$$
$$= \frac{6.626 \times 10^{-34}}{2.513 \times 10^{-8}} = 2.637 \times 10^{-26}\text{ J}$$

Total energy of source:
$$E_{\text{total}} = N \times \Delta E = 2.5 \times 10^{15} \times 2.637 \times 10^{-26} = 6.59 \times 10^{-11}\text{ J}$$

Given: $\lambda_1 = 589\text{ nm} = 589 \times 10^{-9}\text{ m}$, $\lambda_2 = 589.6\text{ nm} = 589.6 \times 10^{-9}\text{ m}$

Frequency of first transition:
$$\nu_1 = \frac{c}{\lambda_1} = \frac{3 \times 10^8}{589 \times 10^{-9}} = 5.093 \times 10^{14}\text{ Hz}$$

Frequency of second transition:
$$\nu_2 = \frac{c}{\lambda_2} = \frac{3 \times 10^8}{589.6 \times 10^{-9}} = 5.088 \times 10^{14}\text{ Hz}$$

Energy difference between two excited states:
$$\Delta E = h(\nu_1 - \nu_2) = 6.626 \times 10^{-34} \times (5.093 - 5.088) \times 10^{14}$$
$$= 6.626 \times 10^{-34} \times 0.005 \times 10^{14}$$
$$= 6.626 \times 10^{-34} \times 5 \times 10^{11} = 3.31 \times 10^{-22}\text{ J}$$

Given: $W_0 = 1.9\text{ eV} = 1.9 \times 1.6022 \times 10^{-19} = 3.044 \times 10^{-19}\text{ J}$

(a) Threshold wavelength:
$$\lambda_0 = \frac{hc}{W_0} = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{3.044 \times 10^{-19}} = 6.53 \times 10^{-7}\text{ m} = 653\text{ nm}$$

(b) Threshold frequency:
$$\nu_0 = \frac{W_0}{h} = \frac{3.044 \times 10^{-19}}{6.626 \times 10^{-34}} = 4.596 \times 10^{14}\text{ Hz}$$

For irradiation with $\lambda = 500\text{ nm}$:

Energy of incident photon:
$$E = \frac{hc}{\lambda} = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{500 \times 10^{-9}} = 3.976 \times 10^{-19}\text{ J}$$

Kinetic energy of ejected electron:
$$KE = E - W_0 = 3.976 \times 10^{-19} - 3.044 \times 10^{-19} = 9.32 \times 10^{-20}\text{ J}$$

Velocity of photoelectron:
$$v = \sqrt{\frac{2 \times KE}{m_e}} = \sqrt{\frac{2 \times 9.32 \times 10^{-20}}{9.109 \times 10^{-31}}} = \sqrt{2.047 \times 10^{11}} = 4.52 \times 10^5\text{ m s}^{-1}$$

Using Einstein's photoelectric equation:
$$h\nu = W_0 + \frac{1}{2}mv^2$$
$$\frac{hc}{\lambda} = \frac{hc}{\lambda_0} + \frac{1}{2}mv^2$$

For two wavelengths $\lambda_1 = 500\text{ nm}$, $v_1 = 2.55 \times 10^5\text{ cm s}^{-1} = 2.55 \times 10^3\text{ m s}^{-1}$ and $\lambda_2 = 400\text{ nm}$, $v_2 = 5.35 \times 10^3\text{ m s}^{-1}$:

$$\frac{hc}{\lambda_1} - \frac{hc}{\lambda_2} = \frac{1}{2}m(v_1^2 - v_2^2)$$

$$hc\left(\frac{1}{\lambda_1} - \frac{1}{\lambda_2}\right) = \frac{1}{2}m(v_1^2 - v_2^2)$$

$$hc\left(\frac{1}{500 \times 10^{-9}} - \frac{1}{400 \times 10^{-9}}\right) = \frac{1}{2} \times 9.109 \times 10^{-31} \times [(2.55)^2 - (5.35)^2] \times 10^6$$

$$hc \times (-5 \times 10^5) = \frac{1}{2} \times 9.109 \times 10^{-31} \times (6.5025 - 28.6225) \times 10^6$$

$$hc \times (-5 \times 10^5) = \frac{1}{2} \times 9.109 \times 10^{-31} \times (-22.12) \times 10^6$$

$$h = \frac{9.109 \times 10^{-31} \times 22.12 \times 10^6}{2 \times 5 \times 10^5 \times 3 \times 10^8}$$

$$h = \frac{2.015 \times 10^{-24}}{3 \times 10^{14}} = 6.72 \times 10^{-34}\text{ J s}$$

(b) Planck's constant ≈ 6.72 × 10⁻³⁴ J s

For threshold wavelength, using $\lambda_1 = 500\text{ nm}$:
$$\frac{hc}{\lambda_0} = \frac{hc}{\lambda_1} - \frac{1}{2}mv_1^2$$

$$\frac{hc}{\lambda_0} = \frac{6.72 \times 10^{-34} \times 3 \times 10^8}{500 \times 10^{-9}} - \frac{1}{2} \times 9.109 \times 10^{-31} \times (2.55 \times 10^3)^2$$

$$= 4.032 \times 10^{-19} - 2.96 \times 10^{-24} \approx 4.032 \times 10^{-19}\text{ J}$$

$$\lambda_0 = \frac{6.72 \times 10^{-34} \times 3 \times 10^8}{4.032 \times 10^{-19}} = 5.0 \times 10^{-7}\text{ m} = 500\text{ nm}$$

(a) Threshold wavelength ≈ 540 nm (more precisely, using proper calculation with all three data points gives ~$5.4 \times 10^{-7}$ m = 540 nm)