“Know how to solve every problem that has been solved.” “What I cannot create, I do not understand.” — Richard Feynman

Basis Sets by Hand — Exponent as inverse size

Exercises

Quantum Chemistry I › Unit 3 · Basis sets › Basis Sets by Hand › all problems

Practice Problem 4 of 8

Exponent as inverse size

Problem

For a normalized 1s Gaussian

g = (2 α / π)^{3/4} e^{- α r^{2}}

, compute the mean-square radius

⟨ r^{2} ⟩ = \int ∣ g ∣^{2} r^{2} d^{3} r

, and evaluate it for

α = 0.5

Solution

This is a Gaussian radial moment, $\int_{0}^{\infty} r^{4} e^{- 2 α r^{2}} d r$ against the $4 π r^{2}$ volume element. Carrying it out with the normalized prefactor gives the clean result:

⟨ r^{2} ⟩ = \int ∣ g ∣^{2} r^{2} d^{3} r = \frac{3}{4 α}

For $α = 0.5$ : $⟨ r^{2} ⟩ = 3/2 = 1.5 a_{0}^{2}$ , so the RMS size is $1.5 \approx 1.22 a_{0}$ . Larger $α$ means a smaller orbital — the exponent is an inverse-size-squared scale.