Atomic Structure (Classical)

Question 1

(a) Derivation [10]

Force balance (Coulomb = centripetal): $\frac{1}{4\pi\varepsilon_0}\frac{Ze^2}{r^2} = \frac{m_e v^2}{r} \implies m_e v^2 = \frac{Ze^2}{4\pi\varepsilon_0 r}\quad(1)\ \textbf{[2]}$

Quantization: $m_e v r = \dfrac{nh}{2\pi} \implies v = \dfrac{nh}{2\pi m_e r}\quad(2)\ \textbf{[1]}$

Substitute (2) into (1): $m_e\left(\frac{nh}{2\pi m_e r}\right)^2 = \frac{Ze^2}{4\pi\varepsilon_0 r}\ \textbf{[2]}$ $\frac{n^2 h^2}{4\pi^2 m_e r} = \frac{Ze^2}{4\pi\varepsilon_0}\implies r_n = \frac{\varepsilon_0 h^2 n^2}{\pi m_e e^2 Z}\ \textbf{[2]}$

Energy: $E = KE + PE = \tfrac12 m_e v^2 - \dfrac{Ze^2}{4\pi\varepsilon_0 r}$. Using (1), $\tfrac12 m_e v^2 = \dfrac{Ze^2}{8\pi\varepsilon_0 r}$, so $E = \frac{Ze^2}{8\pi\varepsilon_0 r} - \frac{Ze^2}{4\pi\varepsilon_0 r} = -\frac{Ze^2}{8\pi\varepsilon_0 r}\ \textbf{[2]}$ Insert $r_n$: $E_n = -\frac{Ze^2}{8\pi\varepsilon_0}\cdot\frac{\pi m_e e^2 Z}{\varepsilon_0 h^2 n^2} = -\frac{m_e e^4 Z^2}{8\varepsilon_0^2 h^2 n^2}\ \textbf{[1]}$

(b) Numerical check [4]

$E_1 = -\frac{(9.109\times10^{-31})(1.602\times10^{-19})^4}{8(8.854\times10^{-12})^2(6.626\times10^{-34})^2}$ Numerator $= 9.109\times10^{-31}\times6.586\times10^{-76} = 6.000\times10^{-106}$ [1] Denominator $= 8\times7.838\times10^{-23}\times4.390\times10^{-67}=2.754\times10^{-88}$ [1] $E_1 = -2.179\times10^{-18}$ J [1] $= -2.179\times10^{-18}/1.602\times10^{-19} = -13.6$ eV ✓ [1]

(c) Virial theorem [6]

$KE = \tfrac12 m_e v^2 = \dfrac{Ze^2}{8\pi\varepsilon_0 r}$ (positive) [1] $PE = -\dfrac{Ze^2}{4\pi\varepsilon_0 r} = -2\,KE$ [2] $E = KE + PE = KE - 2KE = -KE$ Therefore $E/KE = -1$... let me state carefully: $E = -KE$, so $\dfrac{E}{KE} = -1$; and $KE=-E$ means $E = -KE$. The commonly-quoted ratio $E:KE = -1$; here we prove: $\frac{E}{KE} = \frac{-KE}{KE} = -1,\qquad \frac{E}{PE} = \frac{-KE}{-2KE} = +\tfrac12\ \textbf{[2]}$ (Accept the virial relations $E=-KE=\tfrac12 PE$.) The requested $E/PE=+\tfrac12$ is proven. [1] Note: full-mark answer states $E=-KE$ and $E=\tfrac12 PE$, consistent with the virial theorem $\langle KE\rangle = -\tfrac12\langle PE\rangle$.

Question 2

(a) Rydberg derivation [5]

Photon energy for transition $n_2\to n_1$: $\Delta E = E_{n_2}-E_{n_1} = \frac{m_e e^4 Z^2}{8\varepsilon_0^2 h^2}\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)\ \textbf{[2]}$ $\frac{1}{\lambda} = \frac{\Delta E}{hc} = \underbrace{\frac{m_e e^4}{8\varepsilon_0^2 h^3 c}}_{R_H}\,Z^2\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)\ \textbf{[2]}$ $R_H = \dfrac{m_e e^4}{8\varepsilon_0^2 h^3 c} \approx 1.097\times10^7\ \mathrm{m^{-1}}$ [1]

(b) Balmer computations [6]

Series limit ($n_1=2,\ n_2\to\infty$): $\dfrac1\lambda = R_H(\tfrac14 - 0)= 2.743\times10^6\ \mathrm{m^{-1}}$ $\lambda = 3.646\times10^{-7}$ m $= 364.6$ nm — UV/near-UV edge [2]

$H_\alpha$ ($3\to2$): $\dfrac1\lambda = R_H(\tfrac14-\tfrac19)=1.097\times10^7\times0.13889 = 1.524\times10^6$ $\lambda = 6.563\times10^{-7}$ m $= 656.3$ nm — visible (red) [3] (1 mark each: setup, value, region) [1]

(c) Code + hand check [6]

RH = 1.097e7  # m^-1
def line_wavelength(n1, n2, Z=1):
    inv_lambda = RH * Z**2 * (1/n1**2 - 1/n2**2)
    return 1e9 / inv_lambda   # nm
 
for n2 in range(2, 6):          # Lyman: n1 = 1
    print(n2, round(line_wavelength(1, n2), 2))

[4] (correct formula, nm conversion, loop over Lyman lines)

Hand check, shortest Lyman line = series limit $n_2\to\infty$: $\dfrac1\lambda = R_H(1-0)=1.097\times10^7 \implies \lambda = 91.2$ nm. [2] (For $n_2=5$ line the code gives $\approx 94.97$ nm; series limit 91.2 nm confirms shortest.)

(d) He⁺ transition [5]

For $\mathrm{He}^+$, $Z=2$, so $1/\lambda \propto Z^2 = 4$. [1] $H_\alpha$: $\tfrac1\lambda \propto 1\cdot(\tfrac14-\tfrac19)=\tfrac{5}{36}=0.13889$. Need He⁺ transition with $4\left(\tfrac{1}{n_1^2}-\tfrac{1}{n_2^2}\right)=0.13889$, i.e. $\tfrac{1}{n_1^2}-\tfrac{1}{n_2^2}=0.03472=\tfrac{5}{144}$. [2] Try $n_1=4,\ n_2=6$: $\tfrac1{16}-\tfrac1{36}=0.0625-0.02778=0.03472$ ✓ [2] So the $\mathbf{6\to4}$ transition in $\mathrm{He}^+$ gives the same wavelength as $H_\alpha$ (indeed all He⁺ even-level transitions map onto H because $Z^2/n^2$ scaling doubles the quantum numbers).

Question 3

(a) Abundance [5]

Let $x$ = fraction of $^{35}$Cl. [1] $34.969x + 36.966(1-x) = 35.453$ [2] $36.966 - 1.997x = 35.453 \implies 1.997x = 1.513 \implies x = 0.7576$ [1] $^{35}\mathrm{Cl} = 75.76\%,\quad ^{37}\mathrm{Cl} = 24.24\%$ [1]

(b) Definitions + examples [5]

• Isotopes: same $Z$, different $A$ (different neutrons). Example: $^{14}_6\mathrm{C}$ & $^{13}_6\mathrm{C}$ (also $^{14}_7\mathrm{N}$ & $^{15}_7\mathrm{N}$). [1+1]
• Isobars: same $A$, different $Z$. Example: $^{14}_6\mathrm{C}$ & $^{14}_7\mathrm{N}$. [1]
• Isotones: same number of neutrons $N=A-Z$. Example: $^{13}_6\mathrm{C}$ ($N=7$) & $^{14}_7\mathrm{N}$ ($N=7$); also $^{15}_7\mathrm{N}$ ($N=8$) & $^{16}_8\mathrm{O}$ ($N=8$). [1+1]

(c) Limitations [8]

Three limitations (2 marks each, ½ concept + ½ observation, up to 6):

1. Fails for multi-electron atoms — cannot predict spectra of He, Li, etc. (no electron–electron repulsion). [2]
2. No fine structure / spectral line splitting — closely spaced lines seen under high resolution unexplained; also cannot explain Zeeman/Stark effects (splitting in magnetic/electric fields). [2]
3. Violates Heisenberg's uncertainty principle — assumes definite orbit (fixed $r$ and $v$ simultaneously); also gives no intensity of spectral lines and treats angular momentum incorrectly ($n\hbar$ vs $\sqrt{l(l+1)}\hbar$). [2]

Connection to Rutherford instability [2]: Rutherford's classical model has an accelerating (orbiting) electron which, by Maxwell's electrodynamics, must continuously radiate energy, spiral inward, and collapse into the nucleus in $\sim10^{-11}$ s — the atom would not be stable and would emit a continuous spectrum. Bohr's quantized stationary states were introduced precisely to forbid this radiation (postulate: no energy loss in allowed orbits), but this was an ad hoc fix without theoretical justification — itself a conceptual limitation resolved only by quantum mechanics.

[
  {"claim":"Bohr ground-state energy of H equals -13.6 eV to 3 sig figs",
   "code":"me=9.109e-31; e=1.602e-19; eps0=8.854e-12; h=6.626e-34; E=-me*e**4/(8*eps0**2*h**2)/e; result = round(E,1)==-13.6"},
  {"claim":"Balmer series limit wavelength ~= 364.6 nm",
   "code":"RH=1.097e7; lam=1/(RH*(1/2**2))*1e9; result = abs(lam-364.6)<0.5"},
  {"claim":"H-alpha (3->2) wavelength ~= 656.3 nm",
   "code":"RH=1.097e7; lam=1/(RH*(1/4-1/9))*1e9; result = abs(lam-656.3)<0.5"},
  {"claim":"Chlorine 35-Cl abundance ~ 75.76 percent",
   "code":"x=(36.966-35.453)/(36.966-34.969); result = abs(x*100-75.76)<0.1"},
  {"claim":"He+ 6->4 transition matches H-alpha scaling: 4*(1/16-1/36)=5/36",
   "code":"lhs=Rational(4)*(Rational(1,16)-Rational(1,36)); result = lhs==Rational(5,36)"}
]