91Ó°ÊÓ

The radial probability density for the hydrogen ground state is obtained by multiplying the square of the wave function by the spherical element of the shell volume.

The radial probability density for the ground state of hydrogen is given by,

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

Find the probability of finding the electron between two spherical shells whose radii are r and $\mathbf{∆}\mathbf{r}$. If $\mathbf{∆}\mathbf{r}$ is small enough then write the probability as follows:

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

….. (1)

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

Substitute 0.500a for r, 0.010a for $∆\mathrm{r}$, and $52.292×{10}^{-12}$m for a in equation (1).

$$\begin{gathered} \mathrm{p}\left(0.500\mathrm{a}\right)={\left(\frac{4{\left(0.500\mathrm{a}\right)}^2\left(0.010\mathrm{a}\right)}{{\mathrm{a}}^3}\right)}^{{\mathrm{e}}^{-\frac{2(0.500\mathrm{a})}{\mathrm{a}}}} \\ ={\left(4{\left(0.500\mathrm{a}\right)}^2\left(0.010\right)\right)}^{\mathrm{e}-1} \\ =\left(0.010\right)×0.3678 \\ =0.0037 \end{gathered}$$

Therefore, the required probability is 0.0037.

Substitute 1.00a for r, and 0.010a for $∆\mathrm{r}$in equation (1).

$$\begin{gathered} \mathrm{p}\left(0.500\mathrm{a}\right)=\left(\frac{4{\left(1.00\mathrm{a}\right)}^2(0.010\mathrm{a})}{{\mathrm{a}}^3}\right){\mathrm{e}}^{-\frac{2(1.00\mathrm{a})}{\mathrm{a}}} \\ =4{\left(1.00\mathrm{a}\right)}^2(0.010\mathrm{a}){\mathrm{e}}^{-2\left(1.00\right)} \\ =0.04×0.1353 \\ =0.0054 \end{gathered}$$

Therefore, the required probability is 0.0054.