Quantum Atomic Structure

Q1 (10 marks)

Derivation (6):

• Photon energy $E = h\nu$ (Planck/Einstein quantisation assumption). (1)
• Assumptions: one photon ejects one electron; energy $\phi$ needed to escape (work function); no energy losses; surplus becomes kinetic energy. (1)
• Energy conservation: $h\nu = \phi + KE_{max}$. (1)
• Stopping potential halts the fastest electron: $KE_{max} = eV_0$. (1)
• Therefore $eV_0 = h\nu - \phi$, i.e. (1) $V_0 = \frac{h}{e}\nu - \frac{\phi}{e}$
• A straight line vs $\nu$ with slope $h/e$ and intercept determined by $\phi$; threshold at $V_0=0 \Rightarrow \nu_0=\phi/h$. (1)

Explain-out-loud (4):

• (a) Classical waves deliver energy continuously; any frequency should eventually eject electrons if intensity is high enough — but experiment shows no emission below $\nu_0$ regardless of intensity. Photon model: energy per quantum $h\nu$ must exceed $\phi$. (2)
• (b) Classical theory predicts a time lag while the electron "soaks up" enough wave energy; experiment shows emission is essentially instantaneous. Photon model: single photon-electron collision transfers energy at once. (2)

Q2 (12 marks)

(a) (7): Photon energy: $E = \dfrac{hc}{\lambda} = \dfrac{(6.626\times10^{-34})(3.00\times10^8)}{400\times10^{-9}} = 4.97\times10^{-19}\,\text{J} = 3.10\,\text{eV}$. (2) $KE_{max} = E - \phi = 3.10 - 2.28 = 0.82\,\text{eV} = 1.31\times10^{-19}\,\text{J}$. (2) Momentum: $p = \sqrt{2 m_e KE} = \sqrt{2(9.11\times10^{-31})(1.31\times10^{-19})} = 4.89\times10^{-25}\,\text{kg m/s}$. (2) $\lambda_{dB} = \dfrac{h}{p} = \dfrac{6.626\times10^{-34}}{4.89\times10^{-25}} = 1.35\times10^{-9}\,\text{m} \approx 1.35\,\text{nm}$. (1)

(b) (5):

• For a photon $E = h\nu = \dfrac{hc}{\lambda}$. (1)
• Also $E = mc^2 \Rightarrow$ effective momentum $p = mc = \dfrac{E}{c}$. (1)
• Combine: $\dfrac{hc}{\lambda} = pc \Rightarrow \lambda = \dfrac{h}{p}$. (1)
• de Broglie's insight: postulate this relation holds for all matter, not just light — wave-particle duality is universal. (1)
• Hence a particle of momentum $p=mv$ has wavelength $\lambda = h/mv$. (1)

Q3 (10 marks)

• Uncertainty principle: $\Delta x\,\Delta p \ge \dfrac{\hbar}{2}$. (1)
• Minimum: $\Delta p = \dfrac{\hbar}{2\Delta x} = \dfrac{1.055\times10^{-34}}{2(1.0\times10^{-10})} = 5.28\times10^{-25}\,\text{kg m/s}$. (3)
• $\Delta v = \dfrac{\Delta p}{m_e} = \dfrac{5.28\times10^{-25}}{9.11\times10^{-31}} = 5.8\times10^{5}\,\text{m/s}$. (3)
• Comment: $\Delta v \sim 10^6\,\text{m/s}$ is a large fraction of typical atomic electron speeds, so position and velocity cannot be simultaneously well defined. A fixed Bohr orbit (definite radius and velocity) violates this. Hence the probabilistic orbital picture replaces sharp trajectories. (3)

Q4 (10 marks)

• Madelung rule: subshells fill in order of increasing $(n+l)$; for equal $(n+l)$, the one with lower $n$ fills first. (2)
• Justification: $n+l$ approximates the effective energy ordering including penetration/shielding — higher-$l$ orbitals are less penetrating and destabilised. (1)
• Filling order to $4p$: $1s,\,2s,\,2p,\,3s,\,3p,\,4s,\,3d,\,4p$. (2)
• Cr ($Z=24$): naïve $= [\text{Ar}]4s^2 3d^4$; actual $=[\text{Ar}]4s^1 3d^5$. Promotion of one $4s$ electron gives a half-filled $3d^5$ plus half-filled $4s^1$ — extra exchange-energy stability. (2.5)
• Cu ($Z=29$): naïve $=[\text{Ar}]4s^2 3d^9$; actual $=[\text{Ar}]4s^1 3d^{10}$. Gives a fully-filled $3d^{10}$, more stable. (2.5)

Q5 (10 marks)

• Pauli: no two electrons in an atom can have all four quantum numbers identical (equivalently, an orbital holds ≤2 electrons of opposite spin). (1.5)
• Hund: for degenerate orbitals, electrons occupy them singly with parallel spins first, maximising total spin multiplicity, before pairing. (1.5)
• $p$ holds 6 proof: for a $p$ subshell $l=1 \Rightarrow m_l = -1,0,+1$ (3 orbitals). Each $m_l$ pairs with $m_s = +\tfrac12$ or $-\tfrac12$, giving $3\times2 = 6$ distinct $(n,l,m_l,m_s)$ sets; Pauli forbids a 7th. (3)
• Nitrogen $Z=7$: $1s^2\,2s^2\,2p^3$. Box diagram: $1s\,[\uparrow\downarrow]\;\;2s\,[\uparrow\downarrow]\;\;2p\,[\uparrow][\uparrow][\uparrow]$ The three $2p$ electrons occupy separate orbitals with parallel spins (Hund) — no pairing yet; each obeys Pauli. (4)

Q6 (8 marks)

• Cause of stability: exactly half-filled ($d^5$, $p^3$, $f^7$) and fully-filled ($d^{10}$, $p^6$, $f^{14}$) subshells have (i) symmetric charge distribution and (ii) maximum exchange energy (many parallel-spin exchange pairs), lowering energy. (3)
• $\text{Cu}^+$: Cu is $[\text{Ar}]4s^1 3d^{10}$; removing the $4s$ electron gives $[\text{Ar}]3d^{10}$ — a fully-filled $3d$ subshell, stable. (2.5)
• Mn ($Z=25$): $[\text{Ar}]4s^2 3d^5$ — a half-filled $3d^5$ subshell (accounts for Mn's stability and common $+2$ state). (2.5)

[
  {"claim":"Photon energy of 400 nm light in eV ≈ 3.10",
   "code":"h=6.626e-34; c=3.00e8; lam=400e-9; E=h*c/lam/1.602e-19; result = abs(E-3.10)<0.03"},
  {"claim":"Max KE for phi=2.28 eV at 400 nm ≈ 0.82 eV",
   "code":"h=6.626e-34; c=3.00e8; lam=400e-9; E=h*c/lam/1.602e-19; KE=E-2.28; result = abs(KE-0.82)<0.03"},
  {"claim":"de Broglie wavelength of electron with KE=0.82 eV ≈ 1.35e-9 m",
   "code":"import sympy as sp; me=9.11e-31; KE=0.82*1.602e-19; p=sp.sqrt(2*me*KE); lam=6.626e-34/p; result = abs(float(lam)-1.35e-9)<0.1e-9"},
  {"claim":"Uncertainty velocity for dx=1e-10 m ≈ 5.8e5 m/s",
   "code":"hbar=1.055e-34; dx=1.0e-10; me=9.11e-31; dp=hbar/(2*dx); dv=dp/me; result = abs(dv-5.8e5)<0.2e5"}
]