Quantum Atomic Structure

Question 1

(a) Photon energy $E=h\nu = hc/\lambda$. $E = \dfrac{(6.626\times10^{-34})(3.00\times10^8)}{250\times10^{-9}} = 7.951\times10^{-19}\ \text{J} = 4.96\ \text{eV}$. (2) — correct $E=hc/\lambda$ (1), value (1). Einstein equation: $K_{\max}=h\nu-\phi = 4.96 - 2.28 = 2.68\ \text{eV}$. (3) — statement of photon model conservation (1), subtraction (1), answer (1).

(b) de Broglie: $\lambda = h/p \Rightarrow p = h/\lambda$. $p = \dfrac{6.626\times10^{-34}}{250\times10^{-9}} = 2.650\times10^{-27}\ \text{kg m/s}$. (2) $K = \dfrac{p^2}{2m_e} = \dfrac{(2.650\times10^{-27})^2}{2(9.109\times10^{-31})} = 3.856\times10^{-24}\ \text{J} = 2.41\times10^{-5}\ \text{eV}$. (2) Comment: the electron KE ($\sim10^{-5}$ eV) is ~$10^5$ times smaller than the photon energy at the same wavelength, because a photon (massless) carries energy $hc/\lambda \propto 1/\lambda$, whereas a massive electron of the same $\lambda$ carries $p^2/2m \propto 1/\lambda^2$ but is heavily suppressed by its mass. (2)

(c) Photon: $E_{ph}=hc/\lambda$. Particle: $E_p=\dfrac{p^2}{2m}=\dfrac{(h/\lambda)^2}{2m}=\dfrac{h^2}{2m\lambda^2}$. (2) Ratio: $\dfrac{E_{ph}}{E_p}=\dfrac{hc/\lambda}{h^2/(2m\lambda^2)}=\dfrac{hc}{\lambda}\cdot\dfrac{2m\lambda^2}{h^2}=\dfrac{2mc\lambda}{h}$. ∎ (2) Evaluate (electron): $\dfrac{2(9.109\times10^{-31})(3.00\times10^8)(250\times10^{-9})}{6.626\times10^{-34}} = 2.06\times10^{5}$. (1) Interpretation: at equal wavelength the photon carries ~$2\times10^5$ times more energy — consistent with part (b). (1)

(d) Any one (3): existence of a threshold frequency below which no emission occurs regardless of intensity; instantaneous emission; $K_{\max}$ depends on frequency not intensity (intensity only affects photocurrent). Classical waves predict energy accumulation → no threshold and a time-delay.

Question 2

(a) $\Delta p \ge \dfrac{\hbar}{2\Delta x}=\dfrac{1.055\times10^{-34}}{2(1.0\times10^{-10})}=5.28\times10^{-25}\ \text{kg m/s}$. (2) $\Delta v = \Delta p/m_e = \dfrac{5.28\times10^{-25}}{9.109\times10^{-31}} = 5.79\times10^{5}\ \text{m/s}$. (2) Energy scale $\sim \dfrac{(\Delta p)^2}{2m_e}=\dfrac{(5.28\times10^{-25})^2}{2(9.109\times10^{-31})}=1.53\times10^{-19}\ \text{J}=0.95\ \text{eV}$. (2) Comment: this is of order a few eV, consistent with actual atomic binding/ionisation energies (~eV–tens of eV), so quantum confinement naturally produces electron energies on the correct scale. (1)

(b) (i) $n=4,l=2$ ⇒ 4d subshell; $m_l=-2,-1,0,+1,+2$ (five values). (2) (ii) Each orbital holds 2 electrons (opposite spin); 5 orbitals ⇒ max 10 electrons. Pauli: no two electrons in an atom share all four quantum numbers, so within one orbital the two electrons must differ in $m_s$ ($\pm\tfrac12$). (2) (iii) d-orbital: cloverleaf (four lobes) shape (except $d_{z^2}$); angular nodes $=l=2$. (2)

(c) Rules: $n\ge1$ integer (OK, $n=3$); $l=0..n-1$ ⇒ for $n=3$, $l\le2$. Here $l=3$ violates $l\le n-1$. $m_l$ must lie in $-l..+l$; $m_s=\pm\tfrac12$ (OK). Set is NOT allowed — the azimuthal quantum number $l=3$ is impossible when $n=3$. (4) — identifying $l$ violation (2), correct statement of all rules (2).

(d) Heisenberg: $\Delta x\,\Delta p\ge\hbar/2$ means position and momentum cannot both be exactly known; a Bohr orbit specifies both a definite radius and momentum simultaneously, which is forbidden. Hence we replace fixed orbits with probability distributions (orbitals) giving the likelihood of finding the electron in a region. (3)

Question 3

(a) Madelung rule: subshells fill in order of increasing $n+l$; for equal $n+l$, lower $n$ first. (2) Naive: Cr → $[\text{Ar}]3d^4 4s^2$; Cu → $[\text{Ar}]3d^9 4s^2$. Actual: Cr $=[\text{Ar}]3d^5 4s^1$, Cu $=[\text{Ar}]3d^{10}4s^1$. (3) Explanation: a half-filled ($3d^5$) and fully-filled ($3d^{10}$) subshell gives extra stability (symmetric charge distribution + maximal exchange energy). Promoting one $4s$ electron to $3d$ achieves this favourable configuration; small $4s$–$3d$ energy gap makes the trade-off net-stabilising. (3)

(b) Fe ($Z=26$): $[\text{Ar}]3d^6 4s^2$. $3d$ boxes: ↑↓ ↑ ↑ ↑ ↑ ⇒ 4 unpaired electrons; $4s$ full. (3) Cr ($Z=24$): $[\text{Ar}]3d^5 4s^1$. $3d$: ↑ ↑ ↑ ↑ ↑ (all singly, Hund), $4s$: ↑ ⇒ 6 unpaired electrons. (3)

(c) Pseudocode / Python:

def fill_order(max_nl):
    subs = []
    for n in range(1, max_nl+1):
        for l in range(0, n):        # l = 0..n-1
            subs.append((n, l))
    subs.sort(key=lambda s: (s[0]+s[1], s[0]))  # n+l, then n
    return subs

Hand-trace (first 8, with $n+l$): $1s(1),2s(2),2p(3),3s(3),3p(4),4s(4),3d(5),4p(5)$ → matches required order. (6) — correct sort key (3), correct trace (3).

[
{"claim":"Kmax of Na photoelectrons at 250nm is 2.68 eV","code":"h=6.626e-34; c=3.00e8; lam=250e-9; eV=1.602e-19; phi=2.28; E=h*c/lam/eV; K=E-phi; result = abs(K-2.68)<0.02"},
{"claim":"Electron KE for lambda 250nm is ~2.41e-5 eV","code":"h=6.626e-34; lam=250e-9; me=9.109e-31; eV=1.602e-19; p=h/lam; K=p**2/(2*me)/eV; result = abs(K-2.41e-5)<1e-6"},
{"claim":"Photon/particle energy ratio for electron at 250nm ~2.06e5","code":"h=6.626e-34; c=3.00e8; me=9.109e-31; lam=250e-9; r=2*me*c*lam/h; result = abs(r-2.06e5)/2.06e5 < 0.01"},
{"claim":"Madelung fill_order reproduces standard first-8 order","code":"subs=[]; \nfor n in range(1,6):\n    for l in range(0,n):\n        subs.append((n,l))\nsubs.sort(key=lambda s:(s[0]+s[1],s[0])); expected=[(1,0),(2,0),(2,1),(3,0),(3,1),(4,0),(3,2),(4,1)]; result = subs[:8]==expected"}
]