# Calculating most probable measurement result and expectation value of distance between nucleus and electron [closed]

Considering the ground state wavefunction of a hydrogen atom, how do you calculate the most probable measurement result for the distance r between the elctron and the nucleus? How do you calculate the expectation value of r?

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Hi VVV - this looks like exactly the kind of "do-my-homework"-type question that is prohibited in our FAQ. If you can expand the question to indicate what you've tried and why you weren't able to do it, and focus on the specific physical concept that is giving you trouble, I'll be happy to reopen it. – David Zaslavsky Dec 12 '11 at 0:22

## closed as too localized by David Zaslavsky♦Dec 12 '11 at 0:22

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The answer comes right from the definition of wavefunction and its relationship with the probability density. This question seems like a homework question, so I'll just say that you should read the definition of the space wavefunction and expectation value.

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You should start with the normalized wavefunction of the ground state, which is

$$\psi(r, \phi, \theta) = \sqrt{\frac{1}{\pi a_0^3}}exp(-\frac{r}{a_0})$$

where a0 is the Bohr radius:

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

now we can compute the expected value of the distance R of the electron from the nucleus as

$$\langle{R}\rangle = \langle\psi|R|\psi\rangle=\int_0^{2\pi} \int_{-\pi}^{\pi} \int_0^\infty \psi^* r \psi r^2 sin \theta drd\theta d\phi$$

substituting our wavefunction and noting that it's real, we get

$$\langle R \rangle = \frac{1}{\pi a_0^3}4\pi \int_0^\infty r^3exp(-\frac{2r}{a_0})dr$$

now, using the identity

$$\int_0^\infty r^n exp(-\alpha r)dr = \frac{n!}{\alpha^{n+1}}$$

(which can easily be proven by repeatedly differentiating

$$\int_0^\infty exp(-\alpha r)dr=\frac{1}{\alpha}$$

w.r.t. alpha), we obtain

$$\langle R \rangle = \frac{3}{2}a_0$$

As for the most likely value of R, taking it to mean the maximum of the probability density function (remember that we are dealing with continuous random variables here) we can find it by looking for zeros of pdf's derivative w.r.t. r.

The probability of finding the electron in volume V between the sphere with radius r and the sphere with radius r+dr is

$$\int_\Omega \psi^*(r, \theta, \phi) \psi(r, \theta, \phi) r^2 d\Omega dr = \frac{4}{a_0^3} exp(-\frac{2r}{a_0}) r^2 dr$$

Differentiating the expression w.r.t. r, we get

$$\frac{d}{dr}(r^2 exp(-\frac{2r}{a_0})) = 2r(1-\frac{r}{a_0})exp(-\frac{2r}{a_0})$$

This can only be zero when r=0 or r=a0. Probability of finding the electron at r=0 is zero. Thus a0 is the most probable measurement outcome for R (this can be verified by checking the sign of the second derivative).

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 Hi Adam - although this is a very good answer, our policy on homework-like questions is that complete answers will be (temporarily) deleted. Accordingly, I'll delete this now and come back and restore it after a few days. – David Zaslavsky♦ Dec 12 '11 at 0:23