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

Question

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?

Answer 1

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.

Answer 2

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

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

where a₀ is the Bohr radius:

\begin{equation} a_0=\frac{4\pi \epsilon_0 \hbar^2}{m_e e^2} \end{equation}

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

\begin{equation} \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 \end{equation}

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

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

now, using the identity

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

(which can easily be proven by repeatedly differentiating

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

w.r.t. alpha), we obtain

\begin{equation} \langle R \rangle = \frac{3}{2}a_0 \end{equation}

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

\begin{equation} \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 \end{equation}

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

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

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).