Book edition 9th Edition

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

Pages 1624 pages

ISBN 9781133947271

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Short Answer

Prove that the radial probability density peaks at $r = a_{B}$ for the $1 s$ state of hydrogen.

$p_{r} (r)$ has its maximum peak at $r = a_{B}$

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Given information:

Radial probability density $p_{r} (r) = 4 {πr}^{2} {|R_{nl} (r)|}^{2}$

For the

1 s

state where role="math" localid="1648542807295"

n = 1, l

s

$∴ R_{n l} (r) = R_{1 s} (r) = \frac{1}{\sqrt{{πa}_{B}}} e^{\frac{- r}{e^{a_{B}}}} ∴ p_{r} (r) = 4 {πr}^{2} {(\frac{1}{\sqrt{{πa}_{B}}} e^{\frac{- r}{^{a_{B}}}})}^{2} = \frac{4 {πr}^{2}}{{πa}_{B}} e^{\frac{- 2 r}{^{a_{B}}}} = \frac{4}{a_{B} r^{2}} e^{\frac{- 2 r}{^{a_{B}}}}$

Calculation:

When $p_{r} (r)$ is maximum, then $\frac{d}{d r} [p_{r} (r)] = 0$

$\Rightarrow \frac{d}{d r} [\frac{4}{a_{B}} r^{2} e^{\frac{- 2 r}{a_{B}}}] = 0$

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

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

$\frac{d^{2}}{d r^{2}} [p_{r} (r)] = \frac{d}{d r} [2 r e^{\frac{- 24}{a_{B}}} (1 - \frac{r}{a_{B}})]$

localid="1648543956396" $= \frac{d}{d r} (2 r e^{\frac{- 2 r}{a_{B}}}) - \frac{d}{d r} (\frac{2 r^{2}}{a_{B}} e^{\frac{- 2 r}{a_{B}}}) = 2 e^{\frac{- 2 r}{a_{B}}} - \frac{4 r}{a_{B}} e^{\frac{- 2 r}{a_{B}}} - \frac{4 r}{a_{B}} e^{\frac{- 2 r}{a_{B}}} + \frac{4 r^{2}}{a_{B}^{2}} e^{\frac{- 2 r}{a_{B}}}$

Now,at $r = a_{B}$ $\frac{d^{2}}{d r^{2}} [p_{r} (r)] = 2 e^{\frac{- 2 a_{B}}{a_{B}}} - \frac{4 a_{B}}{a_{B}} e^{\frac{- 2 a_{B}}{a_{B}}} - \frac{4 a_{B}}{a_{B}} e^{\frac{- 2 a_{B}}{a_{B}}} + \frac{4 {(a_{B})}^{2}}{a_{B}^{2}} e^{\frac{- 2 a_{B}}{a_{B}}} = 2 e^{- 2} - 4 e^{- 2} - 4 e^{- 2} + 4 e^{- 2} = - 2 e^{- 2} < 0$

$∴ p_{r} (r)$ has its maximum peak at $r = a_{B}$ .

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Short Answer

Prove that the radial probability density peaks at $r = a_{B}$ for the $1 s$ state of hydrogen.

Step by Step Solution

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Physics For Scientists & Engineers

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Short Answer

Prove that the radial probability density peaks at r=aB for the 1s state of hydrogen.

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Calculation:

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