Chapter 8, Problem 8.8P

(a)

Interpretation Introduction

Interpretation:

The radius at which hydrogen atom have thirty percent of radial distribution of the ground state has to be calculated.

Explanation of Solution

The radial distribution function varies as,

    $\begin{array}{l}\text{P}\text{=}\frac{\text{4}}{{\text{a}}_{\text{0}}^{\text{3}}}{\text{r}}^{\text{2}}{\text{e}}^{{\text{-2r/a}}_{\text{0}}}\text{At}\text{maximum}\text{P,}\text{r}\text{=}{\text{a}}_{\text{0}}\text{since}\\ \\ \frac{\text{dp}}{\text{dr}}\infty \left(\text{2r}\text{-}\frac{{\text{2r}}^{\text{2}}}{{\text{a}}_{\text{0}}}\right){\text{e}}^{{\text{-2r/a}}_{\text{0}}}\text{at}\text{r}\text{=}{\text{a}}_{\text{0}}\text{and}{\text{P}}_{\text{max}}\text{=}\frac{\text{4}}{{\text{a}}_{\text{0}}}{\text{e}}^{\text{-2}}\\ \\ \text{P}\text{falls}\text{to}\text{fraction}\text{f}\text{of}\text{its}\text{maximum}\text{given}\text{by}\\ \\ \text{f}\text{=}\frac{{\text{r}}^{\text{2}}}{{\text{a}}_{\text{0}}^{\text{2}}}{\text{e}}^{\text{2}}{\text{e}}^{{\text{-2r/a}}_{\text{0}}}\frac{{\text{f}}^{\text{1/2}}}{\text{e}}\text{=}\frac{\text{r}}{{\text{a}}_{\text{0}}}{\text{e}}^{{\text{-r/a}}_{\text{0}}}\\ \\ \text{f}\text{=}\text{0}\text{.30}\\ \\ \frac{{\text{(0}\text{.30)}}^{\text{1/2}}}{\text{e}}\text{=}\frac{\text{r}}{{\text{a}}_{\text{0}}}{\text{e}}^{{\text{-r/a}}_{\text{0}}}\\ \\ \text{r}\text{=}\frac{\text{0}{\text{.201a}}_{\text{0}}}{{\text{e}}^{{\text{-r/a}}_{\text{0}}}}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The radius at which hydrogen atom have five percent of radial distribution of the ground state has to be calculated.

Explanation of Solution

The radial distribution function varies as,

    $\begin{array}{l}\text{P}\text{=}\frac{\text{4}}{{\text{a}}_{\text{0}}^{\text{3}}}{\text{r}}^{\text{2}}{\text{e}}^{{\text{-2r/a}}_{\text{0}}}\text{At}\text{maximum}\text{P,}\text{r}\text{=}{\text{a}}_{\text{0}}\text{since}\\ \\ \frac{\text{dp}}{\text{dr}}\infty \left(\text{2r}\text{-}\frac{{\text{2r}}^{\text{2}}}{{\text{a}}_{\text{0}}}\right){\text{e}}^{{\text{-2r/a}}_{\text{0}}}\text{at}\text{r}\text{=}{\text{a}}_{\text{0}}\text{and}{\text{P}}_{\text{max}}\text{=}\frac{\text{4}}{{\text{a}}_{\text{0}}}{\text{e}}^{\text{-2}}\\ \\ \text{P}\text{falls}\text{to}\text{fraction}\text{f}\text{of}\text{its}\text{maximum}\text{given}\text{by}\\ \\ \text{f}\text{=}\frac{{\text{r}}^{\text{2}}}{{\text{a}}_{\text{0}}^{\text{2}}}{\text{e}}^{\text{2}}{\text{e}}^{{\text{-2r/a}}_{\text{0}}}\frac{{\text{f}}^{\text{1/2}}}{\text{e}}\text{=}\frac{\text{r}}{{\text{a}}_{\text{0}}}{\text{e}}^{{\text{-r/a}}_{\text{0}}}\\ \\ \text{f}\text{=}\text{0}\text{.05}\\ \\ \frac{{\text{(0}\text{.30)}}^{\text{1/2}}}{\text{e}}\text{=}\frac{\text{r}}{{\text{a}}_{\text{0}}}{\text{e}}^{{\text{-r/a}}_{\text{0}}}\\ \\ \text{r}\text{=}\frac{\text{0}{\text{.082a}}_{\text{0}}}{{\text{e}}^{{\text{-r/a}}_{\text{0}}}}\end{array}$