For the hydrogen atom in its ground state, calculate (a) the probability density ${\mathbf{\psi }}^{\mathbf{2}}\left(r\right)$and (b) the radial probability density P(r) for r = a, where a is the Bohr radius.

The expression of the ground state wave function is given by,

$\mathbf{\psi }\mathbf{\left(}\mathbf{r}\mathbf{\right)}\mathbf{=}\frac{\mathbf{1}}{\sqrt{\mathbf{\pi }}{\mathbf{r}}^{{\displaystyle \frac{\mathbf{3}}{\mathbf{2}}}}}{\mathbf{e}}^{\mathbf{-}\frac{\mathbf{r}}{\mathbf{a}}}$

The probability density function is given by,

${\mathbf{\psi }}^{\mathbf{2}}\mathbf{\left(}\mathbf{r}\mathbf{\right)}\mathbf{=}\frac{\mathbf{1}}{{\mathbf{\pi a}}^{\mathbf{3}}}{\mathbf{e}}^{\mathbf{-}\frac{\mathbf{2}\mathbf{r}}{\mathbf{a}}}$ ….. (1)

The expression of radial probability density is given by,

$\mathbf{P}\left(r\right)\mathbf{=}\frac{\mathbf{4}}{{\mathbf{a}}^{\mathbf{3}}}{\mathbf{r}}^{\mathbf{2}}{\mathbf{e}}^{\mathbf{-}\mathbf{2}\frac{\mathbf{r}}{\mathbf{a}}}$ ….. (2)

Consider the known numerical value as below.

Bohr constant, $\mathrm{a}=5.292\times {10}^{-11}$

Substitute a for r in equation (1).

$$\begin{gathered} {\psi }^2\left(a\right)=\frac{1}{{\mathrm{\pi a}}^3}{e}^{-\frac{2a}{a}} \\ =\frac{1}{3.14{\left(5.292\times {10}^{-11}\right)}^3}{\mathrm{e}}^{-2} \\ =2.19\times {10}^{29}{\mathrm{m}}^{-3} \\ =291n{m}^{-3} \end{gathered}$$

Therefore, the probability density is $2.19\times {10}^{29}{\mathrm{m}}^{-3}$.

Substitute a for r in equation (2).

$$\begin{gathered} \mathrm{P}\left(\mathrm{a}\right)=\frac{4}{{\mathrm{a}}^3}{\left(\mathrm{a}\right)}^2{\mathrm{e}}^{-2\frac{\mathrm{a}}{\mathrm{a}}} \\ =\frac{4}{\mathrm{a}}{\mathrm{e}}^{-2} \end{gathered}$$

Substitute known values in the above equation.

$$\begin{gathered} \mathrm{P}\left(\mathrm{a}\right)=\frac{4}{5.292\times {10}^{-11}}{\mathrm{e}}^{-2} \\ =1.02\times {10}^{10}{\mathrm{m}}^{-1} \\ =10.2{\mathrm{nm}}^{-1} \end{gathered}$$

Therefore, the radial probability density is $10.2{\mathrm{nm}}^{-1}$.