Book edition 9th Edition

Author(s) Raymond A. Serway, John W. Jewett

Pages 1624 pages

ISBN 9781133947271

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Short Answer

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

The expression determined is $R_{2 p} (r) = \frac{1}{\sqrt{24 {πa}_{B}^{3}}} (\frac{r}{2 a_{B}}) e^{\frac{- r}{^{2 a_{B}}}}$ and At $r = 2 a_{B}$ , the $R_{2 p} (r)$ is maximum

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Part (a)

We are given $R_{2 p} (r) = \frac{1}{\sqrt{24 {πa}_{B}^{3}}} (\frac{r}{2 a_{B}}) e^{\frac{- r}{e^{2 a_{B}}}}$

Then we have the following plot

Part (b)

We are given

$R_{2 p} (r) = \frac{1}{\sqrt{24 {πa}_{B}^{3}}} (\frac{r}{2 a_{B}}) e^{\frac{- r}{^{2 a_{B}}}}$

$R_{2 p} (r)$ is maximum when $\frac{d}{d r} (R_{2 p} (r)) = 0$

$\Rightarrow \frac{d}{d r} [\frac{1}{24 {πa}_{B}^{3}} (\frac{r}{2 a_{B}}) e^{\frac{- r}{2 a_{B}}}] = 0 \Rightarrow e^{\frac{r}{2 a_{B}}} - \frac{r}{2 a_{B}} e^{\frac{r}{2 a_{B}}} = 0$

$\Rightarrow (1 - \frac{r}{2 a_{B}}) e^{\frac{- r}{2 a_{B}}} = 0 \Rightarrow 1 - \frac{r}{2 a_{B}} = 0 \Rightarrow \frac{r}{2 a_{B}} = 0 \Rightarrow r = 2 a_{B}$

At $r = 2 a_{B}$ , the $R_{2 p} (r)$ 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.

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