91Ó°ÊÓ

The wave function for the orbital 1s for the hydrogen atom is given by

${\mathbf{Ψ}}_{\mathbf{1}\mathbf{s}}\mathbf{\left(}\mathbf{r}\mathbf{\right)}\mathbf{=}\frac{\mathbf{1}}{\sqrt{{\mathbf{Ï€²¹}}^{\mathbf{3}}}}{\mathbf{e}}^{\mathbf{-}\mathbf{r}\mathbf{/}\mathbf{a}}$

The probability distribution function over a region is given by

${\mathbf{p}}_{\mathbf{a}\mathbf{-}\mathbf{b}}\mathbf{=}{\mathbf{∫}}_{\mathbf{a}}^{\mathbf{b}}{\left|{\mathrm{Ψ}}_{1s}\left(r\right)\right|}^{\mathbf{2}}\mathbf{dV}$

For the probability that an electron in the 1s state of a hydrogen atom will be found at a distance less than a/2 from the nucleus,

Substituting we get

${\mathrm{p}}_{0\left(\mathrm{a}/\mathrm{b}\right)}={∫}_{\mathrm{a}}^{\mathrm{a}/\mathrm{b}}{\left|{\mathrm{Ψ}}_{1\mathrm{s}}\left(\mathrm{r}\right)\right|}^2\mathrm{dV}$

The 1s orbital is spherically symmetrical,

$$\begin{gathered} {\mathrm{p}}_{0-\left(\mathrm{a}/\mathrm{b}\right)}={∫}_{\mathrm{a}}^{\mathrm{a}/\mathrm{b}}\frac{1}{{\mathrm{Ï€²¹}}^3}{\mathrm{e}}^{-2\mathrm{r}/\mathrm{a}}4{\mathrm{Ï€°ù}}^2\mathrm{dr} \\ {\mathrm{p}}_{0-\left(\mathrm{a}/\mathrm{b}\right)}=\frac{4}{{\mathrm{a}}^3}{∫}_{\mathrm{a}}^{\mathrm{a}/\mathrm{b}}{\mathrm{e}}^{-2\mathrm{r}/\mathrm{a}}{\mathrm{r}}^2\mathrm{dr} \end{gathered}$$

From the integral table, we know

${∫}_{\mathrm{a}}^{\mathrm{a}/\mathrm{b}}{\mathrm{e}}^{-2\mathrm{r}/\mathrm{a}}{\mathrm{r}}^2\mathrm{dr}=\left(-\frac{{\mathrm{ar}}^2}{2}\frac{{\mathrm{a}}^2\mathrm{r}}{2}\frac{{\mathrm{a}}^3}{4}\right){\mathrm{e}}^{-2\mathrm{r}/\mathrm{a}}$

Substitute,

${\mathrm{p}}_{0-\left(\mathrm{a}/2\right)}=\frac{4}{{\mathrm{a}}^3}{\left[\left(-\frac{{\mathrm{ar}}^2}{2}\frac{{\mathrm{a}}^2\mathrm{r}}{2}\frac{{\mathrm{a}}^2}{4}\right){\mathrm{e}}^{-2\mathrm{r}/\mathrm{a}}\right]}_0^{\mathrm{a}/2}$

Solving we get

$$\begin{gathered} {\mathrm{p}}_{0-\left(\mathrm{a}/2\right)}=\frac{-5}{2\mathrm{e}}+1 \\ {\mathrm{p}}_{0-\left(\mathrm{a}/2\right)}=0.0803 \end{gathered}$$

The probability is given by ${\mathrm{p}}_{0-\left(\mathrm{a}/2\right)}=0.0803$.

The probability of finding the electron at a distance less than a is${\mathrm{p}}_{0-\left(\mathrm{a}/2\right)}=0.323$

So the probability of finding the electron between a/2 and a from the nucleus is

$$\begin{gathered} {\mathrm{p}}_{\left(\mathrm{a}/2\right)-\mathrm{a}}={\mathrm{p}}_{0-\mathrm{a}}{\mathrm{p}}_{0-\left(\mathrm{a}/2\right)} \\ {\mathrm{p}}_{\left(\mathrm{a}/2\right)-\mathrm{a}}=0.323-0.0803 \\ {\mathrm{p}}_{\left(\mathrm{a}/2\right)-\mathrm{a}}=0.243 \end{gathered}$$

Hence, Probability is given by${\mathrm{p}}_{\left(\mathrm{a}/2\right)-\mathrm{a}}=0.243$