Discussion
Average value of distance of an electron from nucleus in hydrogen atom in its lowest energy is expressed as :
Answer: C
The correct expression is (C), the product of the radial integral and the two angular integrals.
The hydrogen 1s function is \(\psi = (\pi a_{0}^{3})^{-1/2} e^{-r/a_{0}}\). The average distance is \(\langle r \rangle = \int \psi^{*} r \psi\, d\tau\), with the volume element \(d\tau = r^{2}\sin\theta\, dr\, d\theta\, d\phi\). The extra r2 turns r into r3, and \(\psi^{*}\psi\) gives \(e^{-2r/a_{0}}\), so the integral separates into a radial part times \(\int_{0}^{\pi}\sin\theta\, d\theta = 2\) times \(\int_{0}^{2\pi} d\phi = 2\pi\). That is (C).
\[ \begin{aligned} \langle r \rangle &= \frac{4\pi}{\pi a_{0}^{3}} \cdot \frac{3!}{(2/a_{0})^{4}} \\ &= \frac{3}{2}a_{0} \end{aligned} \]
(A) and (B) leave out the angular integrals (and (A) has a stray minus sign), so they are short by a factor of 4π. (D), 5/2 a0, is not the 1s value, which is 3/2 a0.
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