Solution

Take the spherical harmonic

Y_{00}

to have unit angular norm. The radial normalization condition is then

1 = \int_{0}^{\infty} ∣ R (r) ∣^{2} r^{2} d r = C^{2} \int_{0}^{\infty} r^{2} e^{- 2 γ r} d r = \frac{C ^{2}}{4 γ ^{3}} .

Therefore

C = 2 γ^{3/2} .

Equivalently, if

R

denotes the entire spherically symmetric wavefunction rather than the radial factor multiplying normalized

Y_{00}

, then

C = (γ^{3} / π)^{1/2}

The expected radius is

⟨ r ⟩_{R} = C^{2} \int_{0}^{\infty} r^{3} e^{- 2 γ r} d r = 4 γ^{3} \frac{3 !}{( 2 γ ) ^{4}} = \frac{3}{2 γ} .

For the ground state,

γ = m q^{2} / ℏ^{2}

a_{0} = ℏ^{2} / (m q^{2}) = 1/ γ

, so

⟨ r ⟩_{R} = \frac{3}{2} a_{0} .

The mean radius is therefore of the Bohr-radius scale and is one and a half times

a_{0}

Solved by gpt-5.6-sol high.

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