Level 2 — RecallQuantum Atomic Structure

Quantum Atomic Structure

30 minutes40 marksprintable — key stays hidden on paper

Level 2 — Recall & Standard Problems Time Limit: 30 minutes Total Marks: 40

Use the following constants where needed: h=6.626×1034 J sh = 6.626\times10^{-34}\ \text{J s}, c=3.0×108 m s1c = 3.0\times10^8\ \text{m s}^{-1}, me=9.11×1031 kgm_e = 9.11\times10^{-31}\ \text{kg}, 1 eV=1.602×1019 J1\ \text{eV} = 1.602\times10^{-19}\ \text{J}, =1.055×1034 J s\hbar = 1.055\times10^{-34}\ \text{J s}


Q1. State Planck's quantum hypothesis and write the mathematical relation between the energy of a quantum and the frequency of radiation. (3 marks)

Q2. Write Einstein's photoelectric equation and define each term. What is meant by the threshold frequency? (4 marks)

Q3. The work function of a metal is 2.3 eV2.3\ \text{eV}. Calculate: (a) the threshold frequency, and (b) the maximum kinetic energy (in joules) of the emitted electron when light of frequency 8.0×1014 Hz8.0\times10^{14}\ \text{Hz} falls on it. (5 marks)

Q4. Calculate the de Broglie wavelength of an electron moving with a velocity of 2.19×106 m s12.19\times10^{6}\ \text{m s}^{-1}. (4 marks)

Q5. State the Heisenberg uncertainty principle and write its mathematical form. If the uncertainty in the position of an electron is 1.0×1010 m1.0\times10^{-10}\ \text{m}, calculate the minimum uncertainty in its momentum. (5 marks)

Q6. List the four quantum numbers with the symbol and the physical property each describes. Give the allowed values of ll and mlm_l for n=3n = 3. (5 marks)

Q7. State the Aufbau principle and the Madelung (n+ln+l) rule. Using it, explain which orbital fills first: 3d3d or 4s4s. (4 marks)

Q8. State Hund's rule of maximum multiplicity and the Pauli exclusion principle. Illustrate Hund's rule with the orbital-box diagram for the 2p2p electrons of nitrogen (Z=7Z=7). (5 marks)

Q9. Write the ground-state electronic configuration of chromium (Z=24Z=24) and copper (Z=29Z=29). Explain briefly why these are exceptions to the expected filling order. (5 marks)


End of Paper

Answer keyMark scheme & solutions

Q1. (3 marks)

  • Planck's hypothesis: energy is emitted or absorbed by matter not continuously but in discrete packets called quanta. (2)
  • Relation: E=hνE = h\nu, where hh is Planck's constant and ν\nu the frequency. (1) Why: Explaining the black-body spectrum required quantised oscillator energies E=nhνE = nh\nu.

Q2. (4 marks)

  • Equation: hν=ϕ+12mvmax2h\nu = \phi + \tfrac{1}{2}m v^2_{\max}, or KEmax=hνhν0KE_{\max} = h\nu - h\nu_0. (2)
  • Terms: hνh\nu = photon energy; ϕ=hν0\phi = h\nu_0 = work function; 12mvmax2\tfrac12 m v^2_{\max} = max KE of ejected electron. (1)
  • Threshold frequency ν0\nu_0: minimum frequency of incident light below which no electrons are emitted. (1)

Q3. (5 marks) (a) ϕ=2.3 eV=2.3×1.602×1019=3.685×1019 J\phi = 2.3\ \text{eV} = 2.3\times1.602\times10^{-19} = 3.685\times10^{-19}\ \text{J}. (1) ν0=ϕ/h=3.685×10196.626×1034=5.56×1014 Hz\nu_0 = \phi/h = \dfrac{3.685\times10^{-19}}{6.626\times10^{-34}} = 5.56\times10^{14}\ \text{Hz}. (2) (b) KEmax=hνϕ=(6.626×1034)(8.0×1014)3.685×1019KE_{\max} = h\nu - \phi = (6.626\times10^{-34})(8.0\times10^{14}) - 3.685\times10^{-19} (1) =5.301×10193.685×1019=1.62×1019 J= 5.301\times10^{-19} - 3.685\times10^{-19} = 1.62\times10^{-19}\ \text{J}. (1)


Q4. (4 marks) λ=hmv\lambda = \dfrac{h}{m v} (1) =6.626×1034(9.11×1031)(2.19×106)= \dfrac{6.626\times10^{-34}}{(9.11\times10^{-31})(2.19\times10^{6})} (2) =6.626×10341.995×1024=3.32×1010 m (0.33 nm)= \dfrac{6.626\times10^{-34}}{1.995\times10^{-24}} = 3.32\times10^{-10}\ \text{m}\ (\approx0.33\ \text{nm}). (1)


Q5. (5 marks)

  • Statement: it is impossible to determine simultaneously and exactly both the position and momentum of a microscopic particle. (1)
  • Form: ΔxΔp2\Delta x\,\Delta p \ge \dfrac{\hbar}{2} (or h4π\ge \tfrac{h}{4\pi}). (1)
  • Δpmin=2Δx=1.055×10342×1.0×1010\Delta p_{\min} = \dfrac{\hbar}{2\Delta x} = \dfrac{1.055\times10^{-34}}{2\times1.0\times10^{-10}} (2) =5.28×1025 kg m s1= 5.28\times10^{-25}\ \text{kg m s}^{-1}. (1)

Q6. (5 marks)

  • nn (principal) — energy/size of shell. (1)
  • ll (azimuthal) — subshell shape. (1)
  • mlm_l (magnetic) — orbital orientation. (0.5)
  • msm_s (spin) — spin direction (+12,12+\tfrac12,-\tfrac12). (0.5)
  • For n=3n=3: l=0,1,2l = 0,1,2. (1)
  • mlm_l ranges l-l to +l+l: for l=2l=2, ml=2,1,0,+1,+2m_l = -2,-1,0,+1,+2 (and 1,0,+1-1,0,+1 for l=1l=1; 00 for l=0l=0). (1)

Q7. (4 marks)

  • Aufbau: orbitals fill in order of increasing energy, lowest first. (1)
  • Madelung rule: fill in order of increasing (n+l)(n+l); if equal, lower nn fills first. (1)
  • 4s4s: n+l=4+0=4n+l = 4+0 = 4; 3d3d: n+l=3+2=5n+l = 3+2 = 5. (1)
  • Since 4<54 < 5, 4s4s fills before 3d3d. (1)

Q8. (5 marks)

  • Pauli exclusion: no two electrons in an atom can have all four quantum numbers identical (max 2 per orbital, opposite spins). (1.5)
  • Hund's rule: electrons occupy degenerate orbitals singly with parallel spins before pairing, giving maximum multiplicity. (1.5)
  • Nitrogen 2p32p^3: ↑ | ↑ | ↑ (three orbitals each singly occupied, parallel spins). (2)

Q9. (5 marks)

  • Cr (Z=24Z=24): 1s22s22p63s23p63d54s11s^2\,2s^2\,2p^6\,3s^2\,3p^6\,3d^5\,4s^1 (1.5)
  • Cu (Z=29Z=29): 1s22s22p63s23p63d104s11s^2\,2s^2\,2p^6\,3s^2\,3p^6\,3d^{10}\,4s^1 (1.5)
  • Reason: a half-filled (3d53d^5) and fully-filled (3d103d^{10}) subshell gives extra stability (symmetrical distribution + exchange energy), so one 4s4s electron shifts to 3d3d. (2)

[
  {"claim":"Q3a threshold frequency ≈ 5.56e14 Hz","code":"phi=2.3*1.602e-19; h=6.626e-34; nu0=phi/h; result = abs(nu0-5.56e14) < 2e12"},
  {"claim":"Q3b max KE ≈ 1.62e-19 J","code":"phi=2.3*1.602e-19; h=6.626e-34; KE=h*8.0e14-phi; result = abs(KE-1.62e-19) < 5e-21"},
  {"claim":"Q4 de Broglie wavelength ≈ 3.32e-10 m","code":"h=6.626e-34; me=9.11e-31; v=2.19e6; lam=h/(me*v); result = abs(lam-3.32e-10) < 5e-12"},
  {"claim":"Q5 min uncertainty in momentum ≈ 5.28e-25 kg m/s","code":"hbar=1.055e-34; dx=1.0e-10; dp=hbar/(2*dx); result = abs(dp-5.28e-25) < 5e-27"}
]