91Ó°ÊÓ

The radial probability distribution at a given radius is the probability density of an electron in an infinitely thin spherical shell at that radius and is a function of the radial distance from the nucleus.

The expression of radial probability density is given by,

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

Here, the Bohr radius is $\mathbf{a}\mathbf{=}\mathbf{52}\mathbf{.}\mathbf{292}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{12}}\mathbf{}\mathbf{m}$.

Substitute 0 for r in equation (1).

$$\begin{gathered} \mathrm{P}\left(0\right)=\frac{4}{{\mathrm{a}}^3}{\left(0\right)}^2{\mathrm{e}}^{-2\frac{0}{\mathrm{a}}} \\ =0{\mathrm{m}}^{-1} \end{gathered}$$

Therefore, the radial probability density is $0{\mathrm{m}}^{-1}$.

Substitute a for r in equation (1).

$$\begin{gathered} \mathrm{P}\left(0\right)=\frac{4}{{\mathrm{a}}^3}{\left(\mathrm{a}\right)}^2{\mathrm{e}}^{-2\frac{\mathrm{a}}{\mathrm{a}}} \\ =\frac{4}{\mathrm{a}}{\mathrm{e}}^{-2} \end{gathered}$$

Substitute known numerical values in the above equation, and you have,

$$\begin{gathered} \mathrm{P}\left(0\right)=\frac{4}{5.292×{10}^{-11}}{e}^{-2} \\ =1.02×{10}^{10}{\mathrm{m}}^{-1} \end{gathered}$$

Therefore, the radial probability density is $1.02×{10}^{10}{\mathrm{m}}^{-1}$.

Substitute 2a for r in equation (1).

$$\begin{gathered} \mathrm{P}\left(2\mathrm{a}\right)=\frac{4}{{\mathrm{a}}^3}{\left(2\mathrm{a}\right)}^2{\mathrm{e}}^{-2\frac{2\mathrm{a}}{\mathrm{a}}} \\ =\frac{16}{\mathrm{a}}{\mathrm{e}}^{-4} \\ =\frac{16}{5.292×{10}^{-11}}{\mathrm{e}}^{-4} \\ =5.53×{10}^9{\mathrm{m}}^{-1} \end{gathered}$$

Therefore, the radial probability density is $5.53×{10}^9{\mathrm{m}}^{-1}$.