(i) Number of electrons in 1 gram:
Mass of 1 electron $m_e = 9.10939 \times 10^{-31} \text{ kg} = 9.10939 \times 10^{-28} \text{ g}$.
Number of electrons = $\frac{\text{Total Mass}}{\text{Mass of 1 e}^-} = \frac{1 \text{ g}}{9.10939 \times 10^{-28} \text{ g}}$.

(ii) Mass and Charge of 1 mole of electrons:
1 mole = $6.022 \times 10^{23}$ electrons ($N_A$).
Mass: $M = N_A \times m_e = (6.022 \times 10^{23}) \times (9.10939 \times 10^{-31} \text{ kg})$
$= 5.48 \times 10^{-7} \text{ kg}$.
Charge: Charge of 1 $e^- = 1.602 \times 10^{-19} \text{ C}$.
$Q = N_A \times e = (6.022 \times 10^{23}) \times (1.602 \times 10^{-19})$
$= 9.65 \times 10^4 \text{ C}$ (1 Faraday).

(i) Electrons in 1 mole of Methane ($CH_4$):
1 molecule $CH_4$ has: $6(C) + 4 \times 1(H) = 10$ electrons.
1 mole $CH_4$ has $6.022 \times 10^{23}$ molecules.
Total electrons = $10 \times 6.022 \times 10^{23} = \mathbf{6.022 \times 10^{24}}$.

(ii) Neutrons in 7 mg of $^{14}C$:
Isotope $^{14}_6C$ has $14 – 6 = 8$ neutrons.
Atomic mass = 14 g/mol. Mass given = 7 mg = $7 \times 10^{-3}$ g.
Moles of $^{14}C = \frac{7 \times 10^{-3}}{14} = 0.5 \times 10^{-3}$ mol.
Number of atoms = $0.5 \times 10^{-3} \times 6.022 \times 10^{23} = 3.011 \times 10^{20}$.
(a) Total Neutrons = $8 \times 3.011 \times 10^{20} = \mathbf{2.4088 \times 10^{21}}$.
(b) Total Mass of Neutrons = $N \times m_n = (2.4088 \times 10^{21}) \times (1.675 \times 10^{-27} \text{ kg}) = \mathbf{4.03 \times 10^{-6} \text{ kg}}$.

(iii) Protons in 34 mg of $NH_3$:
1 molecule $NH_3$ ($^{14}N + 3 \times ^1H$) has $7 + 3(1) = 10$ protons.
Molar mass $NH_3 = 17$ g/mol. Mass = 34 mg = $34 \times 10^{-3}$ g.
Moles = $\frac{34 \times 10^{-3}}{17} = 2 \times 10^{-3}$ mol.
Number of molecules = $2 \times 10^{-3} \times 6.022 \times 10^{23} = 1.2044 \times 10^{21}$.
(a) Total Protons = $10 \times 1.2044 \times 10^{21} = \mathbf{1.2044 \times 10^{22}}$.
(b) Total Mass = $(1.2044 \times 10^{22}) \times (1.6726 \times 10^{-27} \text{ kg}) = \mathbf{2.01 \times 10^{-5} \text{ kg}}$.
Effect of P/T: No, the number of protons and mass depend only on the amount of substance (mass), which is constant regardless of T and P.

(i) Z = 17 (Chlorine): A = 35. Symbol: $\mathbf{^{35}_{17}Cl}$.

(ii) Z = 92 (Uranium): A = 233. Symbol: $\mathbf{^{233}_{92}U}$.

(iii) Z = 4 (Beryllium): A = 9. Symbol: $\mathbf{^9_4Be}$.

Given: $\lambda = 580 \text{ nm} = 580 \times 10^{-9} \text{ m}$. Speed of light $c = 3 \times 10^8 \text{ m/s}$.

(i) Frequency ($\nu$):
$\nu = \frac{c}{\lambda} = \frac{3 \times 10^8}{580 \times 10^{-9}} = \frac{300}{580} \times 10^{15} = \mathbf{5.17 \times 10^{14} \text{ Hz}}$.

(ii) Wavenumber ($\bar{\nu}$):
$\bar{\nu} = \frac{1}{\lambda} = \frac{1}{580 \times 10^{-9}} = \mathbf{1.72 \times 10^6 \text{ m}^{-1}}$.

Planck’s constant $h = 6.626 \times 10^{-34} \text{ J s}$.

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

(ii) Wavelength $\lambda = 0.50 \text{ \AA} = 0.50 \times 10^{-10} \text{ m}$:
$E = \frac{hc}{\lambda} = \frac{(6.626 \times 10^{-34})(3 \times 10^8)}{0.50 \times 10^{-10}} = \frac{1.988 \times 10^{-25}}{0.5 \times 10^{-10}} = \mathbf{3.98 \times 10^{-15} \text{ J}}$.

Given Period $T = 2.0 \times 10^{-10} \text{ s}$.

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

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

Wavenumber ($\bar{\nu}$):
$\bar{\nu} = \frac{1}{\lambda} = \frac{1}{0.06} = \mathbf{16.66 \text{ m}^{-1}}$.

Given $\lambda = 4000 \text{ pm} = 4000 \times 10^{-12} \text{ m} = 4 \times 10^{-9} \text{ m}$.
Total Energy $E_{total} = 1 \text{ J}$.

Energy of 1 photon $E = \frac{hc}{\lambda} = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{4 \times 10^{-9}} = \frac{1.988 \times 10^{-25}}{4 \times 10^{-9}} = 4.97 \times 10^{-17} \text{ J}$.

Given: $\lambda = 4 \times 10^{-7} \text{ m}$. Work Function $W_0 = 2.13 \text{ eV}$.
$1 \text{ eV} = 1.602 \times 10^{-19} \text{ J}$.

(i) Energy of Photon (eV):
$E = \frac{hc}{\lambda} = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{4 \times 10^{-7}} = 4.97 \times 10^{-19} \text{ J}$.
In eV: $\frac{4.97 \times 10^{-19}}{1.602 \times 10^{-19}} \approx \mathbf{3.10 \text{ eV}}$.

(ii) Kinetic Energy ($K_{max}$):
$K_{max} = E – W_0 = 3.10 \text{ eV} – 2.13 \text{ eV} = \mathbf{0.97 \text{ eV}}$.
In Joules: $0.97 \times 1.602 \times 10^{-19} = 1.55 \times 10^{-19} \text{ J}$.

(iii) Velocity of Photoelectron:
$K_{max} = \frac{1}{2}m_e v^2 \Rightarrow v = \sqrt{\frac{2 K_{max}}{m_e}}$.
$v = \sqrt{\frac{2 \times 1.55 \times 10^{-19}}{9.11 \times 10^{-31}}} = \sqrt{0.34 \times 10^{12}} = \sqrt{3.4 \times 10^{11}} \approx \mathbf{5.84 \times 10^5 \text{ m/s}}$.

Wavelength $\lambda = 242 \text{ nm} = 242 \times 10^{-9} \text{ m}$.
Energy per atom $E = \frac{hc}{\lambda} = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{242 \times 10^{-9}} = 8.21 \times 10^{-19} \text{ J}$.

Energy per mole (Ionisation Energy):
$E_{mol} = E \times N_A = 8.21 \times 10^{-19} \times 6.022 \times 10^{23} = 49.4 \times 10^4 \text{ J/mol}$.
$E_{mol} = \mathbf{494 \text{ kJ mol}^{-1}}$.

Power $P = 25 \text{ Watt} = 25 \text{ J/s}$. Wavelength $\lambda = 0.57 \mu\text{m} = 0.57 \times 10^{-6} \text{ m}$.
Energy of 1 quantum $E = \frac{hc}{\lambda} = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{0.57 \times 10^{-6}} = 3.487 \times 10^{-19} \text{ J}$.

Emitted with zero velocity implies Incident Energy = Work Function ($h\nu = W_0$).
$\lambda = 6800 \text{ \AA} = 6800 \times 10^{-10} \text{ m} = 6.8 \times 10^{-7} \text{ m}$.

Threshold Frequency ($\nu_0$):
$\nu_0 = \frac{c}{\lambda} = \frac{3 \times 10^8}{6.8 \times 10^{-7}} = \mathbf{4.41 \times 10^{14} \text{ Hz}}$.

Work Function ($W_0$):
$W_0 = h\nu_0 = 6.626 \times 10^{-34} \times 4.41 \times 10^{14} = \mathbf{2.92 \times 10^{-19} \text{ J}}$.

Rydberg Formula: $\frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} – \frac{1}{n_2^2} \right)$.
$R_H = 109677 \text{ cm}^{-1}$. $n_1 = 2, n_2 = 4$ (Emission).

$\frac{1}{\lambda} = 109677 \left( \frac{1}{2^2} – \frac{1}{4^2} \right) = 109677 \left( \frac{1}{4} – \frac{1}{16} \right)$
$\frac{1}{\lambda} = 109677 \left( \frac{3}{16} \right) \approx 20564.4 \text{ cm}^{-1}$.
$\lambda = \frac{1}{20564.4} \text{ cm} = 4.86 \times 10^{-5} \text{ cm} = \mathbf{486 \text{ nm}}$. (Blue-Green line of Balmer series).

Energy of $n^{th}$ orbit: $E_n = \frac{-2.18 \times 10^{-18}}{n^2} \text{ J}$.
Ionisation means transition from $n \to \infty$ ($E_\infty = 0$).

From n=5:
$\Delta E = E_\infty – E_5 = 0 – \left( \frac{-2.18 \times 10^{-18}}{5^2} \right) = \frac{2.18 \times 10^{-18}}{25} = \mathbf{8.72 \times 10^{-20} \text{ J}}$.

From n=1 (Ionization Enthalpy):
$\Delta E = \frac{2.18 \times 10^{-18}}{1^2} = 2.18 \times 10^{-18} \text{ J}$.

Comparison: Energy required from $n=5$ is much less (about 25 times less) than from ground state.

Formula for max number of lines: $\frac{n(n-1)}{2}$ where $n$ is excited state and ground state is 1.
Here $n = 6$.
Number of lines = $\frac{6(6-1)}{2} = \frac{30}{2} = \mathbf{15}$.

(Transitions: $6\to5, 6\to4, \dots, 2\to1$).

(i) Energy:
$E_n = \frac{E_1}{n^2}$. Given $E_1 = -2.18 \times 10^{-18} \text{ J}$.
$E_5 = \frac{-2.18 \times 10^{-18}}{25} = \mathbf{-8.72 \times 10^{-20} \text{ J}}$.

(ii) Radius:
$r_n = 0.0529 \text{ nm} \times n^2$ (for H atom).
$r_5 = 0.0529 \times 5^2 = 0.0529 \times 25 = \mathbf{1.3225 \text{ nm}}$.

Balmer Series: $n_1 = 2$.
Longest wavelength $\lambda$ corresponds to smallest energy gap ($\Delta E$), which is the transition from the nearest level ($n_2 = 3$).

$\bar{\nu} = R_H \left( \frac{1}{2^2} – \frac{1}{3^2} \right) = 109677 \left( \frac{1}{4} – \frac{1}{9} \right)$.
$\bar{\nu} = 109677 \left( \frac{5}{36} \right) \approx \mathbf{15233 \text{ cm}^{-1}}$.

Given Ground state energy $E_1 = -2.18 \times 10^{-11} \text{ ergs} = -2.18 \times 10^{-18} \text{ J}$.

Energy required ($n=1 \to n=5$):
$\Delta E = E_5 – E_1 = E_1(\frac{1}{25}) – E_1 = E_1(\frac{1}{25} – 1) = -E_1(\frac{24}{25})$.
$\Delta E = 2.18 \times 10^{-18} \times 0.96 = \mathbf{2.09 \times 10^{-18} \text{ J}}$.

Wavelength emitted ($n=5 \to n=1$):
Energy released is the same magnitude $\Delta E = 2.09 \times 10^{-18} \text{ J}$.
$\lambda = \frac{hc}{\Delta E} = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{2.09 \times 10^{-18}}$
$\lambda = \frac{1.988 \times 10^{-25}}{2.09 \times 10^{-18}} \approx 9.51 \times 10^{-8} \text{ m} = \mathbf{951 \text{ \AA}}$.

Energy at $n=2$: $E_2 = \frac{-2.18 \times 10^{-18}}{2^2} = -5.45 \times 10^{-19} \text{ J}$.
Energy required to remove ($n \to \infty$) is $\Delta E = 0 – E_2 = \mathbf{5.45 \times 10^{-19} \text{ J}}$.

Longest Wavelength:
$\lambda = \frac{hc}{\Delta E} = \frac{1.988 \times 10^{-25}}{5.45 \times 10^{-19}} = 3.647 \times 10^{-7} \text{ m}$.
In cm: $\mathbf{3.647 \times 10^{-5} \text{ cm}}$.

De Broglie Equation: $\lambda = \frac{h}{mv}$.
$h = 6.626 \times 10^{-34} \text{ J s}$.
$m_e = 9.11 \times 10^{-31} \text{ kg}$.
$v = 2.05 \times 10^7 \text{ m/s}$.

Given: Mass $m = 9.1 \times 10^{-31} \text{ kg}$. K.E. $K = 3.0 \times 10^{-25} \text{ J}$.
$K = \frac{1}{2}mv^2 \implies v = \sqrt{\frac{2K}{m}}$.
$v = \sqrt{\frac{2 \times 3.0 \times 10^{-25}}{9.1 \times 10^{-31}}} = \sqrt{0.659 \times 10^6} \approx 811.8 \text{ m/s}$.

De Broglie Wavelength: $\lambda = \frac{h}{mv}$.
$\lambda = \frac{6.626 \times 10^{-34}}{9.1 \times 10^{-31} \times 811.8} = \frac{6.626 \times 10^{-34}}{7387.4 \times 10^{-31}}$.
$\lambda = \frac{6.626}{7.387} \times 10^{-6} \approx \mathbf{8.96 \times 10^{-7} \text{ m}}$.

Isoelectronic means having the same number of electrons.

• $Na^+ (Z=11)$: $11 – 1 = 10 e^-$.
• $K^+ (Z=19)$: $19 – 1 = 18 e^-$.
• $Mg^{2+} (Z=12)$: $12 – 2 = 10 e^-$.
• $Ca^{2+} (Z=20)$: $20 – 2 = 18 e^-$.
• $S^{2-} (Z=16)$: $16 + 2 = 18 e^-$.
• $Ar (Z=18)$: $18 e^-$.

Group 1 (10 electrons): $Na^+, Mg^{2+}$.
Group 2 (18 electrons): $K^+, Ca^{2+}, S^{2-}, Ar$.

(i) Ions Configuration:
(a) $H^-$ (2e): $1s^2$.
(b) $Na^+$ (10e): $1s^2 2s^2 2p^6$.
(c) $O^{2-}$ (10e): $1s^2 2s^2 2p^6$.
(d) $F^-$ (10e): $1s^2 2s^2 2p^6$.

(ii) Atomic Numbers from outermost shell:
(a) $3s^1$: Period 3, Group 1 $\rightarrow$ Na ($Z=11$).
(b) $2p^3$: $1s^2 2s^2 2p^3 \rightarrow$ Total 7 $\rightarrow$ N ($Z=7$).
(c) $3p^5$: $[Ne] 3s^2 3p^5 \rightarrow 10 + 2 + 5 = 17 \rightarrow$ Cl ($Z=17$).

(iii) Atom Identification:
(a) $[He] 2s^1 \rightarrow Li$.
(b) $[Ne] 3s^2 3p^3 \rightarrow P$.
(c) $[Ar] 4s^2 3d^1 \rightarrow Sc$.

Orbital types correspond to $l$ values:
$s \to 0, p \to 1, d \to 2, f \to 3, g \to 4$.
Since $l$ can range from $0$ to $n-1$, we need $n-1 \ge 4 \implies n \ge 5$.
Lowest value of n is 5.

For a 3d orbital:
Principal quantum number $\mathbf{n = 3}$.
Azimuthal quantum number $\mathbf{l = 2}$ (since d-orbital).
Magnetic quantum number $\mathbf{m_l}$: Integers from $-l$ to $+l$.
Possible values: $\mathbf{-2, -1, 0, +1, +2}$.

(i) Protons: In a neutral atom, protons = electrons = 29.

(ii) Configuration: $Z=29$ is Copper (Cu).
Expected: $[Ar] 4s^2 3d^9$.
Actual (Stability rule): $\mathbf{[Ar] 4s^1 3d^{10}}$.

$H_2^+$: 2 protons, but charge +1 means 1 electron lost.
Total e = $2(1) – 1 = \mathbf{1}$.

$H_2$: Neutral. Total e = $2(1) = \mathbf{2}$.

$O_2^+$: $O_2$ has $16$ e. Charge +1 means 1 lost.
Total e = $16 – 1 = \mathbf{15}$.

(i) For n=3:
Possible $l$: 0, 1, 2.
Possible $m_l$:
For $l=0 (3s): 0$.
For $l=1 (3p): -1, 0, +1$.
For $l=2 (3d): -2, -1, 0, +1, +2$.

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

(iii) Possible orbitals:
1p: No ($l=1$ not allowed for $n=1$).
2s: Possible ($n=2, l=0$).
2p: Possible ($n=2, l=1$).
3f: No ($l=3$ not allowed for $n=3$).

(a) $n=1, l=0 \rightarrow \mathbf{1s}$.
(b) $n=3, l=1 \rightarrow \mathbf{3p}$.
(c) $n=4, l=2 \rightarrow \mathbf{4d}$.
(d) $n=4, l=3 \rightarrow \mathbf{4f}$.

(a) $n=0$: Not Possible ($n \ge 1$).
(b) $n=1, l=0, m=0, s=-1/2$: Possible.
(c) $n=1, l=1$: Not Possible ($l$ must be $< n$).
(d) $n=2, l=1, m=0$: Possible.
(e) $n=3, l=3$: Not Possible ($l$ must be $< n$).
(f) $n=3, l=1, m=0$: Possible.

(a) $n=4, m_s = -1/2$:
Total electrons in $n=4$ shell = $2n^2 = 32$.
Half have spin $-1/2$. Answer: $\mathbf{16}$.

(b) $n=3, l=0$:
This specifies the $3s$ orbital. Maximum capacity of an s-orbital is $\mathbf{2}$ electrons.

Bohr’s quantization condition: $mvr = \frac{nh}{2\pi}$.
Rearranging: $2\pi r = \frac{nh}{mv}$.
De Broglie wavelength: $\lambda = \frac{h}{mv}$.
Substituting $\lambda$: $2\pi r = n \lambda$.
Conclusion: The circumference ($2\pi r$) is an integral multiple ($n$) of the wavelength ($\lambda$).

Wavenumber $\bar{\nu} = R Z^2 (\frac{1}{n_1^2} – \frac{1}{n_2^2})$.
For $He^+ (Z=2)$, transition $4 \to 2$:
$\bar{\nu}_{He} = R (2^2) (\frac{1}{2^2} – \frac{1}{4^2}) = 4R (\frac{1}{4} – \frac{1}{16}) = 4R (\frac{3}{16}) = \frac{3R}{4}$.

For Hydrogen ($Z=1$), we need $\bar{\nu}_H = \frac{3R}{4}$.
$\bar{\nu}_H = R (1^2) (\frac{1}{n_1^2} – \frac{1}{n_2^2}) = \frac{3R}{4} \implies \frac{1}{n_1^2} – \frac{1}{n_2^2} = \frac{3}{4}$.
By inspection, $n_1 = 1, n_2 = 2$ gives $1 – 1/4 = 3/4$.
Transition: n = 2 to n = 1 (Lyman Series).

Energy of H ground state ($E_H$) = $-2.18 \times 10^{-18}$ J.
For hydrogen-like ion, $E_n = E_H \times \frac{Z^2}{n^2}$.
For $He^+ (Z=2)$ ground state ($n=1$):
$E_{He} = (-2.18 \times 10^{-18}) \times \frac{2^2}{1^2} = -8.72 \times 10^{-18} \text{ J}$.
Ionization Energy = $E_\infty – E_1 = 0 – (-8.72 \times 10^{-18}) = \mathbf{8.72 \times 10^{-18} \text{ J}}$.

Diameter of C atom $= 0.15 \text{ nm} = 1.5 \times 10^{-10} \text{ m}$.
Length of scale $= 20 \text{ cm} = 0.2 \text{ m}$.
Number of atoms = $\frac{\text{Total Length}}{\text{Diameter}} = \frac{0.2}{1.5 \times 10^{-10}} = \mathbf{1.33 \times 10^9}$.

Number of atoms $N = 2 \times 10^8$. Length $L = 2.4 \text{ cm}$.
Diameter $d = L/N = \frac{2.4}{2 \times 10^8} = 1.2 \times 10^{-8} \text{ cm}$.
Radius $r = d/2 = 0.6 \times 10^{-8} \text{ cm} = 0.06 \text{ nm} = \mathbf{0.06 \text{ nm}}$.

(a) Radius in pm:
Diameter $= 2.6 \text{ \AA} = 2.6 \times 10^{-10} \text{ m}$.
Radius $= 1.3 \times 10^{-10} \text{ m} = 130 \times 10^{-12} \text{ m} = \mathbf{130 \text{ pm}}$.

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

Charge $Q = 2.5 \times 10^{-16} \text{ C}$.
Charge of electron $e = 1.602 \times 10^{-19} \text{ C}$.
Number $n = Q/e = \frac{2.5 \times 10^{-16}}{1.602 \times 10^{-19}} = 1.56 \times 10^3 = \mathbf{1560 \text{ electrons}}$.

Charge on drop $Q = -1.282 \times 10^{-18} \text{ C}$.
Number $n = \frac{-1.282 \times 10^{-18}}{-1.602 \times 10^{-19}} = \frac{12.82}{1.602} \approx \mathbf{8 \text{ electrons}}$.

Heavy atoms (Gold) have a large positive nuclear charge, providing strong repulsion to alpha particles, causing large deflections.
If light atoms (Aluminium) are used, their nuclei have small positive charge and mass. They would not cause significant deflection or back-scattering of heavy alpha particles. The alpha particles would mostly pass through or even push the light nuclei, failing to demonstrate the presence of a dense, heavy nucleus clearly.

Standard notation is $\mathbf{^A_Z X}$.
$Z$ (Atomic Number) is fixed for an element (Br is always 35).
$A$ (Mass Number) can vary (isotopes).
$\mathbf{^{79}_{35}Br}$ is correct because it specifies Z=35 and A=79.
$\mathbf{^{79}Br}$ is acceptable because ‘Br’ implies Z=35.
$\mathbf{^{35}_{79}Br}$ is wrong because Z cannot be 79 for Br.
$\mathbf{^{35}Br}$ is wrong because ’35’ usually denotes A, and an atom of mass 35 cannot be Bromine (Br has mass ~80).

Given: Mass Number $A = 81$.
Neutrons ($n$) are 31.7% more than protons ($p$).
Equation 1: $n = p + \frac{31.7}{100}p = 1.317 p$.
Equation 2: $n + p = A = 81$.

Atomic Number $Z = 35$: This corresponds to Bromine ($Br$).
Symbol: $\mathbf{^{81}_{35}Br}$.

Given: $A = 37$. Charge = -1.
Charge -1 means electrons $e = p + 1$.
Neutrons ($n$) are 11.1% more than electrons ($e$).
$n = e + 0.111 e = 1.111 e = 1.111 (p+1)$.

Atomic Number $Z = 17$: Chlorine ($Cl$).
Symbol: $\mathbf{^{37}_{17}Cl^-}$.

Given: $A = 56$. Charge = +3.
Charge +3 means electrons $e = p – 3$.
Neutrons ($n$) are 30.4% more than electrons.
$n = 1.304 e = 1.304 (p-3)$.

Atomic Number $Z = 26$: Iron ($Fe$).
Symbol: $\mathbf{^{56}_{26}Fe^{3+}}$.

Order of increasing frequency ($\nu$):
Radio Waves < Microwaves < Visible Light < X-rays < Cosmic Rays.

Correct Order:
(c) FM Radio $\rightarrow$ (a) Microwave Oven $\rightarrow$ (b) Amber Light $\rightarrow$ (e) X-rays $\rightarrow$ (d) Cosmic Rays.

$\lambda = 337.1 \text{ nm} = 337.1 \times 10^{-9} \text{ m}$. No. of photons $N = 5.6 \times 10^{24}$.
Energy of 1 photon $E = \frac{hc}{\lambda} = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{337.1 \times 10^{-9}} = 5.89 \times 10^{-19} \text{ J}$.

Total Energy $E_{total} = N \times E = 5.6 \times 10^{24} \times 5.89 \times 10^{-19} = \mathbf{3.3 \times 10^6 \text{ J}}$.

Note: Power requires time. If this is a pulse, energy is calculated. If “produced” means per second, Power = $3.3 \times 10^6$ Watts.

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

(a) Frequency ($\nu$): $\frac{c}{\lambda} = \frac{3 \times 10^8}{6.16 \times 10^{-7}} = \mathbf{4.87 \times 10^{14} \text{ Hz}}$.

(b) Distance in 30s: $d = c \times t = 3 \times 10^8 \times 30 = \mathbf{9.0 \times 10^9 \text{ m}}$.

(c) Energy of Quantum ($E$): $h\nu = 6.626 \times 10^{-34} \times 4.87 \times 10^{14} = \mathbf{3.23 \times 10^{-19} \text{ J}}$.

(d) Number of Quanta for 2 J: $N = \frac{2}{3.23 \times 10^{-19}} = \mathbf{6.2 \times 10^{18}}$.

Total Energy received $E_{tot} = 3.15 \times 10^{-18} \text{ J}$.
$\lambda = 600 \text{ nm} = 6 \times 10^{-7} \text{ m}$.

Energy per photon $E = \frac{hc}{\lambda} = \frac{1.988 \times 10^{-25}}{6 \times 10^{-7}} = 3.31 \times 10^{-19} \text{ J}$.

Duration $t = 2 \text{ ns}$ (irrelevant for total energy if N is given).
$N = 2.5 \times 10^{15}$ photons. Frequency $\nu$ needed.
The question implies we calculate energy if frequency was given, or assume standard source. Wait, text might imply calculating Power? Or Energy? If no frequency given, let’s assume standard value or variable $\nu$. Or maybe use uncertainty principle? “Duration nearly in nano second range”. Usually frequency $\nu = 1/t = 1/(2 \times 10^{-9}) = 0.5 \times 10^9$ Hz? Energy $E = N h \nu = 2.5 \times 10^{15} \times 6.626 \times 10^{-34} \times 0.5 \times 10^9 \approx \mathbf{8.28 \times 10^{-10} \text{ J}}$.

$\lambda_1 = 589 \text{ nm}$. $\lambda_2 = 589.6 \text{ nm}$.

Frequency:
$\nu_1 = \frac{3 \times 10^8}{589 \times 10^{-9}} = \mathbf{5.093 \times 10^{14} \text{ Hz}}$.
$\nu_2 = \frac{3 \times 10^8}{589.6 \times 10^{-9}} = \mathbf{5.088 \times 10^{14} \text{ Hz}}$.

Energy Difference ($\Delta E$):
$\Delta E = h(\nu_1 – \nu_2) = 6.626 \times 10^{-34} \times (5.093 – 5.088) \times 10^{14}$
$= 6.626 \times 10^{-34} \times 0.005 \times 10^{14} = \mathbf{3.31 \times 10^{-22} \text{ J}}$.

Work Function $W_0 = 1.9 \text{ eV} = 1.9 \times 1.602 \times 10^{-19} = 3.04 \times 10^{-19} \text{ J}$.

(a) Threshold Wavelength $\lambda_0$:
$\lambda_0 = \frac{hc}{W_0} = \frac{1.988 \times 10^{-25}}{3.04 \times 10^{-19}} = 6.54 \times 10^{-7} \text{ m} = \mathbf{654 \text{ nm}}$.

(b) Threshold Frequency $\nu_0$:
$\nu_0 = \frac{c}{\lambda_0} = \frac{3 \times 10^8}{6.54 \times 10^{-7}} = \mathbf{4.59 \times 10^{14} \text{ Hz}}$.

Kinetic Energy with 500 nm:
$E = \frac{hc}{500 \times 10^{-9}} = 3.98 \times 10^{-19} \text{ J}$.
$K.E. = E – W_0 = (3.98 – 3.04) \times 10^{-19} = \mathbf{9.4 \times 10^{-20} \text{ J}}$.
Velocity $v = \sqrt{2K/m} = \mathbf{4.54 \times 10^5 \text{ m/s}}$.

Eq: $h(\nu – \nu_0) = \frac{1}{2}mv^2$.
Using pairs of $(\nu, K.E.)$. Let’s solve for $h$ using slope.
$K_1 = \frac{1}{2}m(2.55 \times 10^5)^2$. $K_2 = \frac{1}{2}m(4.35 \times 10^5)^2$.
$\nu_1 = c/500\text{nm}$. $\nu_2 = c/450\text{nm}$.
$h(\nu_2 – \nu_1) = K_2 – K_1$.
Solving gives $h \approx \mathbf{6.6 \times 10^{-34} \text{ J s}}$.
Threshold wavelength $\lambda_0 \approx \mathbf{540 \text{ nm}}$.

Stopping Potential $V_0 = 0.35 \text{ V} \implies K.E. = 0.35 \text{ eV}$.
$\lambda = 256.7 \text{ nm}$.
$E = \frac{1240 \text{ eV nm}}{256.7 \text{ nm}} \approx 4.83 \text{ eV}$.
Work Function $W_0 = E – K.E. = 4.83 – 0.35 = \mathbf{4.48 \text{ eV}}$.

Photon $\lambda = 150 \text{ pm} = 150 \times 10^{-12} \text{ m}$.
$E_{photon} = \frac{hc}{\lambda} = \frac{1.988 \times 10^{-25}}{1.5 \times 10^{-10}} = 1.325 \times 10^{-15} \text{ J}$.
K.E. of electron $= \frac{1}{2}mv^2 = 0.5 \times 9.11 \times 10^{-31} \times (1.5 \times 10^7)^2 = 1.025 \times 10^{-16} \text{ J}$.
Binding Energy $W_0 = E – K.E. = 13.25 \times 10^{-16} – 1.025 \times 10^{-16} = \mathbf{1.22 \times 10^{-15} \text{ J}}$.
In eV: $\approx \mathbf{7.6 \times 10^3 \text{ eV}}$.

$\nu = 3.29 \times 10^{15} (\frac{1}{3^2} – \frac{1}{n^2})$.
Observed $\lambda = 1285 \text{ nm}$. $\nu = c/\lambda = 2.33 \times 10^{14} \text{ Hz}$.
$2.33 \times 10^{14} = 3.29 \times 10^{15} (\frac{1}{9} – \frac{1}{n^2})$.
$0.0708 = 0.111 – 1/n^2 \implies 1/n^2 = 0.0403 \implies n^2 \approx 25 \implies \mathbf{n=5}$.
Region: **Infrared** (Paschen series).

Radius $r_n = 0.0529 n^2 \text{ nm}$.
Start: $1.3225 = 0.0529 n_1^2 \implies n_1^2 = 25 \implies n_1 = 5$.
End: $211.6 \text{ pm} = 0.2116 \text{ nm}$. $0.2116 = 0.0529 n_2^2 \implies n_2^2 = 4 \implies n_2 = 2$.
Transition: $5 \to 2$ (**Balmer Series**, Visible).
Wavelength calculation: $\bar{\nu} = 109677 (\frac{1}{4} – \frac{1}{25}) = 23032 \text{ cm}^{-1}$.
$\lambda = 434 \text{ nm}$ (Blue-Violet).

$v = 1.6 \times 10^6 \text{ m/s}$.
$\lambda = \frac{h}{mv} = \frac{6.626 \times 10^{-34}}{9.11 \times 10^{-31} \times 1.6 \times 10^6}$.
$\lambda = \mathbf{4.55 \times 10^{-10} \text{ m}} = \mathbf{455 \text{ pm}}$.

$\lambda = 800 \text{ pm} = 8 \times 10^{-10} \text{ m}$. Mass $m_n = 1.675 \times 10^{-27} \text{ kg}$.
$v = \frac{h}{m\lambda} = \frac{6.626 \times 10^{-34}}{1.675 \times 10^{-27} \times 8 \times 10^{-10}} = \mathbf{494 \text{ m/s}}$.

$v = 2.19 \times 10^6 \text{ m/s}$.
$\lambda = \frac{h}{mv} = \frac{6.626 \times 10^{-34}}{9.11 \times 10^{-31} \times 2.19 \times 10^6} = \mathbf{3.32 \times 10^{-10} \text{ m}}$.

Mass $m = 0.1 \text{ kg}$. Velocity $v = 4.37 \times 10^5 \text{ m/s}$.
$\lambda = \frac{6.626 \times 10^{-34}}{0.1 \times 4.37 \times 10^5} = \mathbf{1.52 \times 10^{-38} \text{ m}}$.
This wavelength is excessively small, confirming macroscopic objects do not show wave nature.

$\Delta x = 0.002 \text{ nm} = 2 \times 10^{-12} \text{ m}$.
$\Delta p \ge \frac{h}{4\pi \Delta x} = \frac{6.626 \times 10^{-34}}{12.56 \times 2 \times 10^{-12}} = \mathbf{2.64 \times 10^{-23} \text{ kg m/s}}$.
Actual momentum given: $p = \frac{h}{4\pi \times 0.05 \text{ nm}}$.
Since $\Delta p$ is comparable to or larger than $p$, the momentum cannot be defined precisely. Yes, there is a problem.

Rule: Higher $(n+l)$ means higher energy. If $(n+l)$ same, higher $n$ is higher energy.
1. $4d$ ($n+l = 6$)
2. $3d$ ($n+l = 5$)
3. $4p$ ($n+l = 5$)
4. $3d$ ($n+l = 5$)
5. $3p$ ($n+l = 4$)
6. $4p$ ($n+l = 5$)
Order:
$5 (3p) < 2 (3d) = 4 (3d) < 3 (4p) = 6 (4p) < 1 (4d)$.

Shielding effect increases as $n$ increases.
Outer electrons are shielded by inner electrons.
4p electron is furthest from the nucleus and most shielded, thus experiences the **lowest** effective nuclear charge.

Closer to nucleus = higher $Z_{eff}$.
(i) 2s > 3s.
(ii) 4d > 4f ($d$ penetrates more than $f$).
(iii) 3p > 3d.

Al ($Z=13$) vs Si ($Z=14$).
Both have electron in $3p$.
Silicon has higher nuclear charge (+14) compared to Aluminum (+13).
Si electron experiences more effective nuclear charge.

(a) P (15): $[Ne] 3s^2 3p^3$. 3 unpaired.
(b) Si (14): $[Ne] 3s^2 3p^2$. 2 unpaired.
(c) Cr (24): $[Ar] 4s^1 3d^5$. 6 unpaired.
(d) Fe (26): $[Ar] 4s^2 3d^6$. 4 unpaired.
(e) Kr (36): Full shell. 0 unpaired.

(a) Subshells ($n=4$):
$l = 0, 1, 2, 3$ ($4s, 4p, 4d, 4f$). Total 4 subshells.

(b) Electrons with $m_s = -1/2$:
Total electrons in $n=4$ is $2n^2 = 32$.
Half have spin $-1/2$. Answer: 16.