Fine and Hyperfine Structure Questions and Answers

Answer: C

The correct matching is P-4, Q-2, R-1, S-3.

• The Bohr energy, \(\tfrac{1}{2}\alpha^{2}m_{e}c^{2} = 13.6\) eV, is of order \(\alpha^{2}m_{e}c^{2}\) (R-1).
• Fine structure, from relativistic and spin-orbit corrections, is smaller by a further \(\alpha^{2}\): order \(\alpha^{4}m_{e}c^{2}\) (Q-2).
• The Lamb shift, a quantum electrodynamics correction, is smaller again by about \(\alpha\): order \(\alpha^{5}m_{e}c^{2}\) (P-4).
• Hyperfine structure comes from the nuclear magnetic moment, which is smaller than the electron’s by \(m_{e}/m_{p}\): order \(\alpha^{4}m_{e}^{2}c^{2}/m_{p}\) (S-3).

The energy scales therefore fall in the order Bohr, fine structure, Lamb shift, hyperfine. Options giving the Bohr energy \(\alpha^{4}\) or putting fine structure at \(\alpha^{2}\) reverse the hierarchy.

Answer: D

The dimensionless constant is the fine structure constant.

\[ \begin{aligned} \alpha &= \frac{e^{2}}{4\pi\epsilon_{0}\hbar c} \\ &\approx \frac{1}{137} \end{aligned} \]

The units of \(e^{2}/4\pi\epsilon_{0}\) (energy times length) cancel exactly with those of \(\hbar c\), so \(\alpha\) is a pure number and has the same value in every system of units.

The permittivity of free space has units F/m, the permeability of free space H/m (or N/A²), and the Bohr magneton, \(e\hbar/2m_{e}\), J/T. All three carry dimensions.

Answer: B

The splitting is \(\dfrac{3}{2}k\hbar^{2}\).

With J = L + S, the operator L·S = (J² - L² - S²)/2, so for a p electron (l = 1, s = 1/2):

\[ \begin{aligned} \langle\vec{L}\cdot\vec{S}\rangle_{j=3/2} &= \tfrac{\hbar^{2}}{2}(\tfrac{15}{4} - 2 - \tfrac{3}{4}) = \tfrac{1}{2}\hbar^{2} \\ \langle\vec{L}\cdot\vec{S}\rangle_{j=1/2} &= \tfrac{\hbar^{2}}{2}(\tfrac{3}{4} - 2 - \tfrac{3}{4}) = -\hbar^{2} \end{aligned} \]

The two levels differ in L·S by 3ℏ²/2, so with H′ = -kL·S the splitting is (3/2)kℏ² in magnitude (with k > 0 the j = 1/2 level lies higher).

1/2 kℏ² is the shift of the j = 3/2 level alone. 3/4 and 2 come from slips in j(j + 1), l(l + 1) or s(s + 1).

Answer: C

The corrections are \(\frac{5\alpha^{2}E_{2}}{16}\) and \(\frac{\alpha^{2}E_{2}}{16}\).

For \(n = 2\), the orbital quantum number is \(l = 0\) or 1, so \(j = 1/2\) (from \(2s_{1/2}\) and \(2p_{1/2}\)) or \(j = 3/2\) (from \(2p_{3/2}\)). Substituting into the formula:

\[ \begin{aligned} j = \tfrac{1}{2}: \ \Delta E &= \frac{E_{2}\alpha^{2}}{4}\left(2 - \frac{3}{4}\right) \\ &= \frac{5\alpha^{2}E_{2}}{16} \\ j = \tfrac{3}{2}: \ \Delta E &= \frac{E_{2}\alpha^{2}}{4}\left(1 - \frac{3}{4}\right) \\ &= \frac{\alpha^{2}E_{2}}{16} \end{aligned} \]

The correction depends only on \(n\) and \(j\), so there are exactly two values, and neither is zero; any option containing 0 is wrong. (Since \(E_{2}\) is negative, both levels are pushed down, the \(j = 1/2\) pair more.)