Answer to Question #118515 in Quantum Mechanics for Moel Tariburu

Question

Consider a hydrogen atom in the 2p state. Determine at what radius will the electron probability be a maximum.

Answer 1

Radial wave function of hydrogen atom in 2p state is $R_{21}=\frac{1}{\sqrt{3}}(\frac{1}{2a})^{\frac{3}{2}}(\frac{r}{a})e^{-(\frac{r}{2a})}$

The radial probability is

$|R_{21}|^2r^2=\frac{1}{{3}}(\frac{1}{2a})^{3}(\frac{r}{a})^2e^{-(\frac{r}{a})}r^2$

To find the maximum value of above equation, Differentiating above equation with respect to r and equating to zero, we get

$\frac{d}{dr}|R_{21}|^2r^2=0$

$\frac{d}{dr}[\frac{1}{{3}}(\frac{1}{2a})^{3}(\frac{r}{a})^2e^{-(\frac{r}{a})}r^2]=0$

$\frac{d}{dr}[r^4e^{-(\frac{r}{a})}]=0$

$[4r^3e^{-(\frac{r}{a})}+r^4(-\frac{1}{a})e^{-(\frac{r}{a})}]=0$

$4r^3+r^4(-\frac{1}{a})=0 \implies r-4a=0$

$r=4a$ since $a$ is the Bohr radius.

Hence, An electron has maximum probability at radius is $r=4a$ .