# Expectation Value of 1/r for Hydrogen using Virial Theorem -Quantum Mechanics

In this post we will be evaluating the expectation value of 1/r : $\left<\frac{1}{r}\right>$ using the Virial Theorem, that we proved and talked about in the last post.

We will be using the expression for Energy of the nth energy level for Hydrogen Atom:

$E_n=-\frac{m}{2\hbar^2}\left(\frac{e^2}{4\pi \epsilon_0 }\right)^2\frac{1}{n^2}$

Virial Theorem states that for stationary states,
$2\left=\left$

For Hydrogen, $V=-\frac{e^2}{4\pi \epsilon_0 r}$

For 3-dimensions Virial Theorem can be written as:
$2\left=\left$

Plugging the value of $V$ in the above equation, we get

$\implies 2\left= \frac{e^2}{4\pi \epsilon_0 r}$

$\implies 2\left= -\left$

We know, $E_n=\left+\left$

$\implies E_n=-\frac{\left}{2} + \left$

$\implies E_n=\frac{\left}{2}$

Now plugging the expression of Energy for Hydrogen in the above equation, we get

$\implies -\frac{m}{2\hbar^2}\left(\frac{e^2}{4\pi \epsilon_0 }\right)^2\frac{1}{n^2}=\frac{\left}{2}$

$\implies \left = -\frac{m}{\hbar^2}\left(\frac{e^2}{4\pi \epsilon_0 }\right)^2\frac{1}{n^2}$

$\implies -\frac{e^2}{4\pi \epsilon_0} \left<\frac{1}{r}\right>= -\frac{m}{\hbar^2}\left(\frac{e^2}{4\pi \epsilon_0 }\right)^2\frac{1}{n^2}$

$\implies \left<\frac{1}{r}\right>= \frac{m}{\hbar^2}\frac{e^2}{4\pi \epsilon_0 }\frac{1}{n^2}$

This is the expectation value of 1/r for Hydrogen.

You can simplify it further by writing in terms of the Bohr’s radius $a_0$,

$a_0= \frac{4\pi \epsilon_0 \hbar^2}{m e^2}$

Using the above expression, one can write $\left<\frac{1}{r}\right>$, as

$\left<\frac{1}{r}\right> = \frac{1}{n^2 a_0}$

If you have any questions or doubts regarding the above proof, feel free to post a comment down below.