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Found in: Page 1217

### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718

# Calculate the probability that the electron in the hydrogen atom, in its ground state, will be found between spherical shells whose radii are a and 2a , where a is the Bohr radius?

The required probability is P=0.439.

See the step by step solution

## Step 1: Identification of the given data:

The given data is listed below.

• Radii of the spherical shells are given as a and 2a .

## Step 2: Formula for finding the probability of electron:

The ground state wave function of hydrogen atom is given by,

${{\mathbf{\psi }}}_{{\mathbf{100}}}{\mathbf{\left(}}{\mathbf{r}}{\mathbf{,}}{\mathbf{\theta }}{\mathbf{,}}{\mathbf{\varphi }}{\mathbf{\right)}}{\mathbf{=}}\frac{\mathbf{1}}{\sqrt{\mathbf{\pi }}}{\left(\frac{1}{a}\right)}^{\frac{\mathbf{3}}{\mathbf{2}}}{{\mathbf{e}}}^{\mathbf{-}\frac{\mathbf{r}}{\mathbf{a}}}$

Here, the Bohr radius is $\mathrm{a}=5.292×{10}^{-11}\mathrm{m}$.

## Step 3: Determine the probability of the electron in the hydrogen atom in its ground state:

The probability of finding the electron found between spherical shells is,

$\mathrm{P}={\int }_{\mathrm{a}}^{2\mathrm{a}}{\int }_{0}^{\mathrm{\pi }}{\int }_{0}^{2\mathrm{\pi }}{\left|{\mathrm{\psi }}_{100}\left(\mathrm{r},\mathrm{\theta },\mathrm{\varphi }\right)\right|}^{2}{\mathrm{r}}^{2}\mathrm{drsin\theta d\theta d\varphi }\phantom{\rule{0ex}{0ex}}={\int }_{\mathrm{a}}^{2\mathrm{a}}{\left|\frac{1}{\sqrt{\mathrm{\pi }}}{\left(\frac{1}{\mathrm{a}}\right)}^{3}{2}}{\mathrm{e}}^{-\mathrm{r}}{\mathrm{a}}}\right|}^{2}{\mathrm{r}}^{2}\mathrm{dr}{\int }_{0}^{\mathrm{\pi }}\mathrm{d\varphi }\phantom{\rule{0ex}{0ex}}=\frac{1}{\mathrm{\pi }}{\left(\frac{1}{\mathrm{a}}\right)}^{3}{\int }_{\mathrm{a}}^{2\mathrm{a}}{\left|{\mathrm{e}}^{-\mathrm{r}}{\mathrm{a}}}\right|}^{2}{\mathrm{r}}^{2}\mathrm{dr}×{\left(-\mathrm{cos\theta }\right)}_{0}^{\mathrm{\pi }}{\overline{)\left(\mathrm{\theta }\right)}}_{0}^{2\mathrm{\pi }}\phantom{\rule{0ex}{0ex}}=\frac{1}{\mathrm{\pi }}{\left(\frac{1}{\mathrm{a}}\right)}^{3}{\left[-\frac{1}{4}{\mathrm{ae}}^{-2\mathrm{r}}{\mathrm{a}}}\left({\mathrm{a}}^{2}+2\mathrm{ar}+2{\mathrm{r}}^{2}\right)\right]}_{\mathrm{a}}^{2\mathrm{a}}×\left(-\mathrm{cos\pi }+\mathrm{cos}0\right)\left(2\mathrm{\pi }-0\right)$

$\mathrm{P}=\frac{1}{\mathrm{\pi }}{\left(\frac{1}{\mathrm{a}}\right)}^{3}\left[-\frac{\mathrm{a}}{4}\left\{{\mathrm{e}}^{-2\left(2\mathrm{a}\right)}{\mathrm{a}}}\left({\mathrm{a}}^{2}+2\mathrm{a}\left(2\mathrm{a}\right)+2{\left(2\mathrm{a}\right)}^{2}\right)-{\mathrm{e}}^{-2\left(\mathrm{a}\right)}{\mathrm{a}}}\left({\mathrm{a}}^{2}+2{\mathrm{a}}^{2}+2{\mathrm{a}}^{2}\right)\right\}\right]×\left(-\left(-1\right)+1\right)\left(2\mathrm{\pi }\right)\phantom{\rule{0ex}{0ex}}=\frac{1}{\mathrm{\pi }}{\left(\frac{1}{\mathrm{a}}\right)}^{3}\left[-\frac{\mathrm{a}}{4}\left\{{\mathrm{e}}^{-4}\left(13{\mathrm{a}}^{2}\right)-{\mathrm{e}}^{-2}\left(5{\mathrm{a}}^{2}\right)\right\}\right]×\left(4\mathrm{\pi }\right)\phantom{\rule{0ex}{0ex}}=4{\left(\frac{1}{\mathrm{a}}\right)}^{3}\left[-\frac{\mathrm{a}}{4}\left\{0.0183\left(13{\mathrm{a}}^{2}\right)-0.1353\left(5{\mathrm{a}}^{2}\right)\right\}\right]\phantom{\rule{0ex}{0ex}}\mathrm{P}=4\left[-\frac{1}{4}\left\{0.0183\left(13\right)-0.1353\left(5\right)\right\}\right]\phantom{\rule{0ex}{0ex}}=-4\left[\frac{1}{4}\left\{0.2379-0.6765\right\}\right]\phantom{\rule{0ex}{0ex}}=0.439$

Hence, the probability of the electron in the hydrogen atom in its ground state is 0.439.

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