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Answers without the blur. Sign up and see all textbooks for free! Q.35

Expert-verified Found in: Page 1296 ### Physics For Scientists & Engineers

Book edition 9th Edition
Author(s) Raymond A. Serway, John W. Jewett
Pages 1624 pages
ISBN 9781133947271 # Prove that the radial probability density peaks at $r={a}_{B}$ for the $1s$ state of hydrogen.

${p}_{r}\left(r\right)$ has its maximum peak at $r={a}_{B}$

See the step by step solution

## Given information:

Radial probability density ${p}_{r}\left(r\right)=4{\mathrm{\pi r}}^{2}{\left|{\mathrm{R}}_{\mathrm{nl}}\left(\mathrm{r}\right)\right|}^{2}$

For the $1s$ state where role="math" localid="1648542807295" $n=1,l$ is $s$

$\therefore {R}_{nl}\left(r\right)={R}_{1s}\left(r\right)=\frac{1}{\sqrt{{\mathrm{\pi a}}_{\mathrm{B}}}}{e}^{\frac{-r}{{e}^{{a}_{B}}}}\phantom{\rule{0ex}{0ex}}\therefore {p}_{r}\left(r\right)=4{\mathrm{\pi r}}^{2}{\left(\frac{1}{\sqrt{{\mathrm{\pi a}}_{\mathrm{B}}}}{\mathrm{e}}^{\frac{-\mathrm{r}}{{}^{{\mathrm{a}}_{\mathrm{B}}}}}\right)}^{2}\phantom{\rule{0ex}{0ex}}=\frac{4{\mathrm{\pi r}}^{2}}{{\mathrm{\pi a}}_{\mathrm{B}}}{\mathrm{e}}^{\frac{-2\mathrm{r}}{{}^{{\mathrm{a}}_{\mathrm{B}}}}}\phantom{\rule{0ex}{0ex}}=\frac{4}{{\mathrm{a}}_{\mathrm{B}}{\mathrm{r}}^{2}}{\mathrm{e}}^{\frac{-2\mathrm{r}}{{}^{{\mathrm{a}}_{\mathrm{B}}}}}$

## Calculation:

When ${p}_{r}\left(r\right)$ is maximum, then $\frac{d}{dr}\left[{p}_{r}\left(r\right)\right]=0$

$⇒\frac{d}{dr}\left[\frac{4}{{a}_{B}}{r}^{2}{e}^{\frac{-2r}{{a}_{B}}}\right]=0$

$⇒2r{e}^{\frac{-2r}{{a}_{B}}}\frac{-2{r}^{2}}{{a}_{B}}{e}^{\frac{-2r}{{a}_{B}}}=0\phantom{\rule{0ex}{0ex}}⇒2r{e}^{\frac{-2r}{{a}_{B}}}\left(1-\frac{r}{{a}_{B}}\right)-0\phantom{\rule{0ex}{0ex}}⇒1-\frac{r}{{a}_{B}}=0\phantom{\rule{0ex}{0ex}}⇒\frac{r}{{a}_{B}}=1\phantom{\rule{0ex}{0ex}}⇒r={a}_{B}$

At $r={a}_{B,}{p}_{r}\left(r\right)$ has either a maximum or a minimum. To determine whether it is a maximum or minimum. we find $\frac{{d}^{2}}{d{r}^{2}}\left[{p}_{r}\left(r\right)\right]$

$\frac{{d}^{2}}{d{r}^{2}}\left[{p}_{r}\left(r\right)\right]=\frac{d}{dr}\left[2r{e}^{\frac{-24}{{a}_{B}}}\left(1-\frac{r}{{a}_{B}}\right)\right]$

localid="1648543956396" $=\frac{d}{dr}\left(2r{e}^{\frac{-2r}{{a}_{B}}}\right)-\frac{d}{dr}\left(\frac{2{r}^{2}}{{a}_{B}}{e}^{\frac{-2r}{{a}_{B}}}\right)\phantom{\rule{0ex}{0ex}}=2{e}^{\frac{-2r}{{a}_{B}}}-\frac{4r}{{a}_{B}}{e}^{\frac{-2r}{{a}_{B}}}-\frac{4r}{{a}_{B}}{e}^{\frac{-2r}{{a}_{B}}}+\frac{4{r}^{2}}{{a}_{B}^{2}}{e}^{\frac{-2r}{{a}_{B}}}$

Now,at $r={a}_{B}$ $\frac{{d}^{2}}{d{r}^{2}}\left[{p}_{r}\left(r\right)\right]=2{e}^{\frac{-2{a}_{B}}{{a}_{B}}}-\frac{4{a}_{B}}{{a}_{B}}{e}^{\frac{-2{a}_{B}}{{a}_{B}}}-\frac{4{a}_{B}}{{a}_{B}}{e}^{\frac{-2{a}_{B}}{{a}_{B}}}+\frac{4{\left({a}_{B}\right)}^{2}}{{a}_{B}^{2}}{e}^{\frac{-2{a}_{B}}{{a}_{B}}}\phantom{\rule{0ex}{0ex}}=2{e}^{-2}-4{e}^{-2}-4{e}^{-2}+4{e}^{-2}\phantom{\rule{0ex}{0ex}}=-2{e}^{-2}\phantom{\rule{0ex}{0ex}}<0$

$\therefore {p}_{r}\left(r\right)$ has its maximum peak at $r={a}_{B}$. ### Want to see more solutions like these? 