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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 # a. Calculate and graph the hydrogen radial wave function ${R}_{2p}\left(r\right)$ over the interval $0\le r\le 8{a}_{BA}$.b. Determine the value of $r$ (in terms of ${a}_{B}$ ) for which ${R}_{2p}\left(r\right)$ is a maximum.c. Example $41.3$ and Figure $41.7$ showed that the radial probability density for the $2p$ state is a maximum at $r=4{a}_{B}$. Explain why this differs from your answer to part $b$.

The expression determined is ${R}_{2p}\left(r\right)=\frac{1}{\sqrt{24{\mathrm{\pi a}}_{\mathrm{B}}^{3}}}\left(\frac{r}{2{a}_{B}}\right){e}^{\frac{-r}{{}^{2{a}_{B}}}}$ and At $r=2{a}_{B}$, the ${R}_{2p}\left(r\right)$ is maximum

See the step by step solution

## Part (a)

We are given ${R}_{2p}\left(r\right)=\frac{1}{\sqrt{24{\mathrm{\pi a}}_{\mathrm{B}}^{3}}}\left(\frac{r}{2{a}_{B}}\right){e}^{\frac{-r}{{e}^{2{a}_{B}}}}$

Then we have the following plot ## Part (b)

We are given

${R}_{2p}\left(r\right)=\frac{1}{\sqrt{24{\mathrm{\pi a}}_{\mathrm{B}}^{3}}}\left(\frac{r}{2{a}_{B}}\right){e}^{\frac{-r}{{}^{2{a}_{B}}}}$

${R}_{2p}\left(r\right)$ is maximum when $\frac{d}{dr}\left({R}_{2p}\left(r\right)\right)=0$

$⇒\frac{d}{dr}\left[\frac{1}{24{\mathrm{\pi a}}_{\mathrm{B}}^{3}}\left(\frac{r}{2{a}_{B}}\right){e}^{\frac{-r}{2{a}_{B}}}\right]=0\phantom{\rule{0ex}{0ex}}⇒{e}^{\frac{r}{2{a}_{B}}}-\frac{r}{2{a}_{B}}{e}^{\frac{r}{2{a}_{B}}}=0$

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

At $r=2{a}_{B}$, the ${R}_{2p}\left(r\right)$ is maximum

## Part (c)

Probability density depends on the total wave function therefore we will not get the maximum value of the probability density at the same point where we get the maximum radial wave function. ### Want to see more solutions like these? 