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Past exam of the quantum theory course of the University of Oxford / 2015 / 3 / v / Solution by

Codex  0 2026-10-31

The radial probability density includes the spherical-shell volume factor:

p (r) = 4 π r^{2} ∣ ψ_{0} (r) ∣^{2} = 4 α_{0}^{3} r^{2} e^{- 2 α_{0} r}, r \geq 0.

Consequently the radial expectation value is

E_{ψ_{0}} (r) = 4 α_{0}^{3} \int_{0}^{\infty} r^{3} e^{- 2 α_{0} r} d r = 4 α_{0}^{3} \frac{3 !}{( 2 α _{0} ) ^{4}} = \frac{3}{2 α _{0}} = \frac{3 ℏ ^{2}}{2 m K Z} .

Keeping

m

and

K

fixed, the hydrogenic ground-state radius scaling is therefore $⟨ r ⟩ \propto 1/ Z$ : the

Z = 2

state has half the mean radius of the

Z = 1

state. Equivalently, with

a_{*} = ℏ^{2} / (m K)

, the mean radii are

3 a_{*} /2

and

3 a_{*} /4

. For the physical electron-nucleus Coulomb potential,

a_{*}

is the Bohr radius of the fixed-nucleus hydrogen atom model.

Figure 1.
Radial probability densities for the one-electron $Z = 1$ and $Z = 2$ Coulomb bound states; dashed lines mark their mean radii
. The horizontal coordinate is $s = r / a_{*}$ . The normalized densities are $4 Z^{3} s^{2} e^{- 2 Z s}$ , so their means are $3/ (2 Z)$ . Details.

The exact

Z = 2

comparison in this one-electron model describes the hydrogen-like ion

He^{+}

. A neutral helium atom has two electrons; its exact size also depends on their mutual Coulomb interaction and is not determined by this one-electron calculation. The stated half-radius conclusion is the prediction of the model supplied in the paper.

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