Atomic Structure (Classical)

Q1 (10 marks)

(a) Let abundance of $^{25}$X be $x$. Peak heights ∝ abundances: $^{24}$X $= 7.85x$, $^{26}$X $= 1.101x$. Sum $= 7.85x + x + 1.101x = 9.951x = 1 \Rightarrow x = 0.10049$. (1)

• $^{25}$X: $0.1005 \approx 10.05\%$
• $^{24}$X: $7.85\times0.10049 = 0.7889 \approx 78.9\%$ (1.5)
• $^{26}$X: $1.101\times0.10049 = 0.1106 \approx 11.06\%$ (1.5)

(b) Average mass $= 24(0.7889) + 25(0.1005) + 26(0.1106)$ (use masses = mass numbers) $= 18.933 + 2.512 + 2.876 = 24.32$ u. (3) (This is Mg — Z = 12.)

(c) Neutrons $= A - Z$, with $Z=12$: (1)

• $^{24}$X: 12 neutrons; $^{25}$X: 13; $^{26}$X: 14. (1)

Isotones have equal neutron number. $^{26}$X (14 n) and the hypothetical $^{27}$Y (14 n) both have 14 neutrons → they are isotones. (1)

Q2 (10 marks)

(a) Rydberg-type energy scaling: $\dfrac{1}{\lambda} = R Z^2\left(\dfrac{1}{n_1^2}-\dfrac{1}{n_2^2}\right)$. For H Lyman-α: $Z=1$, factor $= 1\left(\frac{1}{1}-\frac{1}{4}\right)=\frac{3}{4}$. (1) For Li²⁺: $Z=3$, need $9\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)=\frac{3}{4}$ $\Rightarrow \frac{1}{n_1^2}-\frac{1}{n_2^2}=\frac{1}{12}$. (2) Try $n_1=2,\ n_2=6$: $\frac14-\frac1{36}=\frac{9-1}{36}=\frac{8}{36}=\frac29$ — no. Try $n_1=2,\ n_2=4$: $\frac14-\frac1{16}=\frac{3}{16}$ — no. Try $n_1=3,\ n_2=6$: $\frac19-\frac1{36}=\frac{4-1}{36}=\frac{3}{36}=\frac1{12}$ ✓ (2) Transition: $n=6 \to n=3$ in Li²⁺.

(b) $\dfrac{1}{\lambda}=R\cdot\frac34 = 1.097\times10^7\times0.75 = 8.228\times10^6\ \text{m}^{-1}$. $\lambda = 1.215\times10^{-7}\ \text{m} = 121.5$ nm. (3)

(c) Both energies scale as $Z^2/n^2$; by choosing $n$ so that $Z^2(1/n_1^2-1/n_2^2)$ takes the same value, the emitted photon energies (hence wavelengths) coincide — the $Z^2$ enlargement in Li²⁺ is compensated by using higher $n$ levels. (2)

Q3 (10 marks)

(a) Any two: (2 each)

• Lighter metal has smaller nuclear charge Z, so the Coulomb repulsion on the alpha particle is weaker → less deflection.
• Lighter metal has smaller nuclear mass; a light nucleus recoils when struck, so it cannot turn the alpha particle straight back (no near-180° backscatter) — momentum is shared.

(b) At closest approach all KE converts to electrostatic PE: $KE = \dfrac{1}{4\pi\varepsilon_0}\dfrac{(2e)(Ze)}{d}$ (2) $\Rightarrow d = \dfrac{1}{4\pi\varepsilon_0}\dfrac{2Ze^2}{KE}$. (1)

(c) $KE = 5.0\times10^6\times1.602\times10^{-19}=8.01\times10^{-13}$ J. (1) $d = \dfrac{8.99\times10^9\times2\times13\times(1.602\times10^{-19})^2}{8.01\times10^{-13}}$ Numerator $=8.99\times10^9\times26\times2.566\times10^{-38}=5.999\times10^{-27}$. $d = 7.49\times10^{-15}$ m $\approx 7.5\times10^{-15}$ m. (2)

Q4 (10 marks)

(a) $r_n \propto 1/m$; muon mass $=207 m_e$: $r_1 = \dfrac{0.529}{207}\ \text{Å} = 2.556\times10^{-3}$ Å $\approx 2.56\times10^{-3}$ Å. (2.5) $E_n \propto m$: $E_1 = -13.6\times207 = -2815$ eV $\approx -2.82\times10^3$ eV. (2.5) (Reduced-mass corrections neglected as instructed via simple proportionality.)

(b) $\Delta E = 2815\left(1-\frac14\right)=2815\times0.75 = 2111$ eV. (1) $\lambda = \dfrac{hc}{\Delta E}=\dfrac{(6.626\times10^{-34})(3\times10^8)}{2111\times1.602\times10^{-19}}$ $=\dfrac{1.988\times10^{-25}}{3.382\times10^{-16}} = 5.88\times10^{-10}$ m $= 0.588$ nm. (2)

(c) The muon orbits ~207× closer to the nucleus, so the muonic atom is roughly 207 times smaller than ordinary hydrogen — an extremely compact atom (muon essentially "buried" near the proton). (2)

Q5 (10 marks)

(a) Bohr prediction: $I = 13.6\times Z^2 = 13.6\times4 = 54.4$ eV. (2) Percentage error $=\dfrac{54.4-24.6}{24.6}\times100 = 121\%$. (2) (Grossly overestimates — prediction more than double experimental.)

(b) Any two: (2 each)

• The Bohr model ignores electron–electron repulsion; in neutral He the second electron screens/repels, lowering the binding energy well below $13.6Z^2$.
• It uses the full nuclear charge Z=2, but each electron feels a shielded effective charge $Z_{eff}<2$ due to the other electron.
• (Also acceptable: Bohr treats only one-electron systems; no correct multi-electron quantisation / no accounting for electron correlation.)

(c) Any one: (2)

• Fine structure (splitting of spectral lines) — Bohr predicts single lines.
• (Also acceptable: relative line intensities, Zeeman/Stark splitting in fields, spin–orbit coupling.)

[
  {"claim":"Mg average atomic mass ~24.32 from abundances", "code":"x=Rational(1,9951)*1000; a24=7.85*x; a26=1.101*x; a25=x; tot=a24+a25+a26; m=24*a24+25*a25+26*a26; result = abs(float(m)-24.32)<0.05"},
  {"claim":"Li2+ 6->3 gives factor 1/12 matching H Lyman-alpha 3/4 with Z^2=9", "code":"factor_li=9*(Rational(1,9)-Rational(1,36)); factor_H=1*(Rational(1,1)-Rational(1,4)); result = (factor_li==factor_H)"},
  {"claim":"Lyman-alpha wavelength ~121.5 nm", "code":"R=1.097e7; inv=R*0.75; lam=1/inv; result = abs(lam*1e9-121.5)<1.0"},
  {"claim":"Distance of closest approach ~7.5e-15 m", "code":"ke=5.0e6*1.602e-19; d=8.99e9*2*13*(1.602e-19)**2/ke; result = abs(d-7.5e-15)<0.2e-15"},
  {"claim":"Muonic H ground energy -2815 eV and He Bohr IE error 121%", "code":"E=13.6*207; err=(54.4-24.6)/24.6*100; result = abs(E-2815.2)<1 and abs(err-121)<1"}
]