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Level 4 — ApplicationQuantum Atomic Structure

Quantum Atomic Structure

⏱ 60 minutes50 marksprintable — key stays hidden on paper

Time limit: 60 minutes
Total marks: 50
Instructions: Answer all questions. Show full working. Use h=6.626×10−34 J sh = 6.626\times10^{-34}\ \text{J s}, c=3.00×108 m s−1c = 3.00\times10^8\ \text{m s}^{-1}, me=9.11×10−31 kgm_e = 9.11\times10^{-31}\ \text{kg}, ℏ=1.055×10−34 J s\hbar = 1.055\times10^{-34}\ \text{J s}, 1 eV=1.602×10−19 J1\ \text{eV} = 1.602\times10^{-19}\ \text{J}, NA=6.022×1023 mol−1N_A = 6.022\times10^{23}\ \text{mol}^{-1}.

Question 1 (Photoelectric effect + Planck) — (11 marks)

A clean sodium surface is irradiated. When light of wavelength 400 nm400\ \text{nm} is used, the ejected electrons have a maximum kinetic energy of 0.60 eV0.60\ \text{eV}.

(a) Calculate the work function of sodium in eV. (4)

(b) Determine the threshold wavelength (longest wavelength that will still eject electrons). (3)

(c) The surface is now illuminated with 300 nm300\ \text{nm} light at a power of 2.0 mW2.0\ \text{mW}, and 5.0% of incident photons eject an electron. Calculate the number of electrons ejected per second. (4)

Question 2 (de Broglie + Uncertainty) — (11 marks)

(a) An electron is accelerated from rest through a potential difference VV. Derive an expression for its de Broglie wavelength λ\lambda in terms of VV (and fundamental constants), then evaluate λ\lambda for V=100 VV = 100\ \text{V}. (6)

(b) An electron is confined within a region of size equal to a typical atomic diameter, 1.0×10−10 m1.0\times10^{-10}\ \text{m}. Using the Heisenberg uncertainty principle, estimate the minimum uncertainty in its speed. Comment on whether describing the electron with a fixed classical orbit is justified. (5)

Question 3 (Quantum numbers — reasoning) — (10 marks)

(a) State, with a one-line justification, whether each of the following quantum-number sets is allowed or forbidden: (6)

(i) n=3, l=3, ml=0, ms=+12n=3,\ l=3,\ m_l=0,\ m_s=+\tfrac12
(ii) n=4, l=2, ml=−2, ms=+12n=4,\ l=2,\ m_l=-2,\ m_s=+\tfrac12
(iii) n=2, l=1, ml=+2, ms=−12n=2,\ l=1,\ m_l=+2,\ m_s=-\tfrac12

(b) An electron in a certain atom has the set n=4, l=1n=4,\ l=1. Give the sub-shell name, the number of orbitals available, and the maximum number of electrons that can share these two values of nn and ll. (4)

Question 4 (Electronic configuration + stability) — (10 marks)

(a) Write the full ground-state electron configuration of chromium (Z=24Z=24) and explain, using the concept of subshell stability, why it deviates from the naïve Aufbau prediction. (4)

(b) A neutral atom of element X has exactly three unpaired electrons in its 3d3d subshell and no 4s4s vacancy issues (i.e. 4s24s^2 filled). Identify X and give its atomic number. (3)

(c) Using the Madelung (n+ln+l) rule, list the filling order of the subshells from 1s1s up to and including 4p4p, and state which rule you would invoke to explain the electron arrangement within a partly-filled 2p32p^3 configuration. (3)

Question 5 (Synthesis) — (8 marks)

A photon is emitted when an electron transition releases exactly the energy needed to just ionise a ground-state hydrogen-like assumption is not required here; instead:

The work function of a metal equals the energy of a photon whose momentum is p=1.10×10−27 kg m s−1p = 1.10\times10^{-27}\ \text{kg m s}^{-1}.

(a) Find the wavelength and energy (in eV) of this photon. (4)

(b) If this metal is used in a photocell and irradiated by 250 nm250\ \text{nm} light, find the maximum kinetic energy (in eV) and the stopping potential of the ejected electrons. (4)

Answer keyMark scheme & solutions

Question 1

(a) Photon energy at 400 nm: E=hcλ=(6.626×10−34)(3.00×108)400×10−9=4.97×10−19 J=3.10 eVE=\frac{hc}{\lambda}=\frac{(6.626\times10^{-34})(3.00\times10^8)}{400\times10^{-9}}=4.97\times10^{-19}\ \text{J}=3.10\ \text{eV} (2)

Einstein equation: KEmax=E−ϕ⇒ϕ=3.10−0.60=2.50 eVKE_{max}=E-\phi \Rightarrow \phi = 3.10-0.60=2.50\ \text{eV}. (2)

(b) Threshold: ϕ=hcλ0\phi=\dfrac{hc}{\lambda_0}, so λ0=hcϕ=(6.626×10−34)(3.00×108)2.50×1.602×10−19=4.96×10−7 m≈496 nm\lambda_0=\frac{hc}{\phi}=\frac{(6.626\times10^{-34})(3.00\times10^8)}{2.50\times1.602\times10^{-19}}=4.96\times10^{-7}\ \text{m}\approx 496\ \text{nm} (3)
(3 marks: rearrangement 1, substitution 1, answer 1.)

(c) Energy per 300 nm photon: E=hcλ=(6.626×10−34)(3.00×108)300×10−9=6.63×10−19 JE=\frac{hc}{\lambda}=\frac{(6.626\times10^{-34})(3.00\times10^8)}{300\times10^{-9}}=6.63\times10^{-19}\ \text{J} (1) Photons per second =PE=2.0×10−36.63×10−19=3.02×1015 s−1=\dfrac{P}{E}=\dfrac{2.0\times10^{-3}}{6.63\times10^{-19}}=3.02\times10^{15}\ \text{s}^{-1}. (2) Electrons/s =0.05×3.02×1015=1.5×1014 s−1=0.05\times3.02\times10^{15}=1.5\times10^{14}\ \text{s}^{-1}. (1)

Question 2

(a) Energy from acceleration: eV=12mv2=p22meV=\tfrac12 m v^2 = \dfrac{p^2}{2m}, so p=2meVp=\sqrt{2meV}. (2) λ=hp=h2meV\lambda=\frac{h}{p}=\frac{h}{\sqrt{2meV}} (2) For V=100V=100 V: λ=6.626×10−342(9.11×10−31)(1.602×10−19)(100)\lambda=\frac{6.626\times10^{-34}}{\sqrt{2(9.11\times10^{-31})(1.602\times10^{-19})(100)}} Denominator =2.919×10−47=5.40×10−24=\sqrt{2.919\times10^{-47}}=5.40\times10^{-24}. λ=1.23×10−10 m=123 pm\lambda=1.23\times10^{-10}\ \text{m}=123\ \text{pm} (2)

(b) Δx=1.0×10−10\Delta x=1.0\times10^{-10} m. Minimum: Δp=ℏ2Δx\Delta p=\dfrac{\hbar}{2\Delta x}. (1) Δp=1.055×10−342(1.0×10−10)=5.27×10−25 kg m s−1\Delta p=\frac{1.055\times10^{-34}}{2(1.0\times10^{-10})}=5.27\times10^{-25}\ \text{kg m s}^{-1} (1) Δv=Δpme=5.27×10−259.11×10−31=5.8×105 m s−1\Delta v=\frac{\Delta p}{m_e}=\frac{5.27\times10^{-25}}{9.11\times10^{-31}}=5.8\times10^{5}\ \text{m s}^{-1} (2) Comment: Δv∼106\Delta v\sim10^6 m/s is comparable to the electron's actual orbital speed, so position and velocity cannot both be well-defined; a fixed classical orbit is not justified — a probabilistic (orbital) description is required. (1)

Question 3

(a) (each: 1 for verdict, 1 for reason)

  • (i) Forbidden: ll must satisfy 0≤l≤n−10\le l\le n-1; with n=3n=3, lmax=2l_{max}=2, so l=3l=3 is impossible. (2)
  • (ii) Allowed: n=4n=4 permits l=2l=2 (d), and ml=−2m_l=-2 lies in [−2,+2][-2,+2], ms=±12m_s=\pm\tfrac12 valid. (2)
  • (iii) Forbidden: with l=1l=1, mlm_l ranges −1,0,+1-1,0,+1; ml=+2m_l=+2 is out of range. (2)

(b) n=4,l=1n=4,l=1 → sub-shell 4p. (1) Orbitals =2l+1=3=2l+1=3. (2) Max electrons =2(2l+1)=6=2(2l+1)=6. (1)

Question 4

(a) Cr configuration: 1s2 2s2 2p6 3s2 3p6 3d5 4s11s^2\,2s^2\,2p^6\,3s^2\,3p^6\,3d^5\,4s^1. (2) Naïve Aufbau predicts 3d44s23d^4 4s^2. The observed 3d54s13d^5 4s^1 arises because a half-filled 3d53d^5 subshell gives extra stability (symmetric charge distribution + maximum exchange energy from parallel-spin electrons), which outweighs the small 4s4s–3d3d promotion cost. (2)

(b) Three unpaired 3d electrons with 4s24s^2 full ⇒ 3d33d^3. Total = 1818 (Ar core) +2(4s)+3(3d)=23+2(4s)+3(3d)=23 → Vanadium, Z = 23. (3)

(c) Filling order (increasing n+ln+l, then nn): 1s,2s,2p,3s,3p,4s,3d,4p1s,2s,2p,3s,3p,4s,3d,4p (2) Within 2p32p^3, electrons occupy separate orbitals with parallel spins per Hund's rule of maximum multiplicity. (1)

Question 5

(a) λ=hp=6.626×10−341.10×10−27=6.02×10−7 m=602 nm\lambda=\dfrac{h}{p}=\dfrac{6.626\times10^{-34}}{1.10\times10^{-27}}=6.02\times10^{-7}\ \text{m}=602\ \text{nm}. (2) E=pc=(1.10×10−27)(3.00×108)=3.30×10−19 J=2.06 eVE=pc=(1.10\times10^{-27})(3.00\times10^8)=3.30\times10^{-19}\ \text{J}=2.06\ \text{eV} (2) (So ϕ=2.06\phi=2.06 eV.)

(b) 250 nm photon energy: E=hcλ=(6.626×10−34)(3.00×108)250×10−9=7.95×10−19 J=4.96 eVE=\frac{hc}{\lambda}=\frac{(6.626\times10^{-34})(3.00\times10^8)}{250\times10^{-9}}=7.95\times10^{-19}\ \text{J}=4.96\ \text{eV} (2) KEmax=4.96−2.06=2.90 eVKE_{max}=4.96-2.06=2.90\ \text{eV} (1) Stopping potential Vs=KEmax/e=2.90 VV_s=KE_{max}/e = 2.90\ \text{V}. (1)

[
  {"claim":"Q1a work function = 2.50 eV (photon 3.10 eV, KE 0.60 eV)",
   "code":"h=6.626e-34; c=3.00e8; lam=400e-9; eV=1.602e-19; Ephot=h*c/lam/eV; phi=Ephot-0.60; result = abs(phi-2.50)<0.03"},
  {"claim":"Q1b threshold wavelength approx 496 nm",
   "code":"h=6.626e-34; c=3.00e8; eV=1.602e-19; phi=2.50*eV; lam0=h*c/phi; result = abs(lam0-4.96e-7)<0.05e-7"},
  {"claim":"Q2a de Broglie wavelength at 100 V approx 1.23e-10 m",
   "code":"import sympy as sp; h=6.626e-34; me=9.11e-31; e=1.602e-19; V=100; lam=h/sp.sqrt(2*me*e*V); result = abs(float(lam)-1.23e-10)<0.03e-10"},
  {"claim":"Q5 stopping potential 2.90 V from 250 nm on phi=2.06 eV metal",
   "code":"h=6.626e-34; c=3.00e8; eV=1.602e-19; E=h*c/(250e-9)/eV; phi=1.10e-27*c/eV; Vs=E-phi; result = abs(Vs-2.90)<0.05"}
]