 Solution

ID: past-exam-of-the-atomic-and-laser-physics-course-of-the-university-of-oxford/2021/4/e/i/solution

A nonrelativistic

2 p

wavefunction vanishes linearly at the origin. Its probability density is therefore proportional to

r^{2}

R^{4}

, rather than the

R^{2}

result for

1 s

. It is positive but much smaller.

More explicitly, the normalized hydrogen-like ion radial function obeys

R_{21}^{2} (r) ≃ Z^{5} r^{2} / (24 a_{0}^{5})

near the origin. Since the spherical harmonic has unit angular norm,

Δ E_{2 p} ≃ \frac{k _{Z} Z ^{5}}{24 a _{0}^{5}} \int_{0}^{R} (r^{3} - \frac{3 r ^{4}}{2 R} + \frac{r ^{6}}{2 R ^{3}}) d r = \frac{k _{Z} Z ^{5} R ^{4}}{1120 a _{0}^{5}} .

Thus

\frac{Δ E _{2 p}}{Δ E _{1 s}} ≃ \frac{1}{448} (\frac{ZR}{a _{0}})^{2} ≃ 2.71 \times 1 0^{- 7} .

This is about

2.4 μ eV

in the nonrelativistic model. At large

Z

, the relativistic

2 p_{1/2}

s

-type small component and a much stronger nuclear sensitivity than this nonrelativistic argument predicts;

2 p_{3/2}

remains much more suppressed. Neglecting spin-orbit interaction corrections would miss that distinction.

 New to topics? Read the docs here!