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Found in: Page 1217

### Fundamentals Of Physics

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

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# The wave function for the hydrogen-atom quantum state represented by the dot plot shown in Fig. 39-21, which has n = 2 and ${\mathsc{l}}{\mathbf{=}}{{\mathbit{m}}}_{{\mathbf{l}}}{\mathbf{=}}{\mathbf{0}}$, is ${{\mathbit{\Psi }}}_{{\mathbf{200}}}{\mathbf{\left(}}{\mathbit{r}}{\mathbf{\right)}}{\mathbf{=}}\frac{\mathbf{1}}{\mathbf{4}\sqrt{\mathbf{2}\mathbf{\pi }}}{{\mathbit{a}}}^{\mathbf{-}\mathbf{3}\mathbf{/}\mathbf{2}}\mathbf{\left(}\mathbf{2}\mathbf{-}\frac{\mathbf{r}}{\mathbf{a}}\mathbf{\right)}{{\mathbit{e}}}^{\mathbf{-}\mathbf{r}\mathbf{/}\mathbf{2}\mathbf{a}}$ in which a is the Bohr radius and the subscript on ${\mathbit{\Psi }}{\mathbf{\left(}}{\mathbit{r}}{\mathbf{\right)}}$ gives the values of the quantum numbers ${\mathbit{n}}{\mathbf{,}}{\mathsc{l}}{\mathbf{,}}{{\mathbit{m}}}_{{\mathcal{l}}}$. (a) Plot ${\mathbit{\Psi }}{}_{{\mathbf{200}}}{}^{{\mathbf{2}}}{\mathbf{\left(}}{\mathbit{r}}{\mathbf{\right)}}$ and show that your plot is consistent with the dot plot of Fig. 39-21. (b) Show analytically that ${\mathbit{\Psi }}{}_{{\mathbf{200}}}{}^{{\mathbf{2}}}{\mathbf{\left(}}{\mathbit{r}}{\mathbf{\right)}}$ has a maximum at ${\mathbit{r}}{\mathbf{=}}{\mathbf{4}}{\mathbit{a}}$. (c) Find the radial probability density ${{\mathbit{P}}}_{{\mathbf{200}}}\mathbf{\left(}\mathbf{r}\mathbf{\right)}$ for this state. (d) Show that ${{\mathbf{\int }}}_{{\mathbf{0}}}^{{\mathbf{\infty }}}{{\mathbit{P}}}_{{\mathbf{200}}}{\mathbf{\left(}}{\mathbit{r}}{\mathbf{\right)}}{\mathbit{d}}{\mathbit{r}}{\mathbf{=}}{\mathbf{1}}$and thus that the expression above for the wave function ${{\mathbit{\Psi }}}_{{\mathbf{200}}}\mathbf{\left(}\mathbf{r}\mathbf{\right)}$ has been properly normalized.

1. The plot is shown in the below figure.

b. It is proved that ${\Psi }_{200}^{2}\left(r\right)$ has a maximum at r = 4a.

c. The radial probability is ${P}_{200}\left(r\right)=\frac{{r}^{2}}{8{a}^{3}}\left(2-\frac{r}{a}\right){e}^{-r/a}$.

d. It is proved that ${\int }_{0}^{\infty }{P}_{200}\left(r\right)dr=1$.

See the step by step solution

## Step 1: Describe the given information

The wave function is given by,

${{\mathbf{\Psi }}}_{{\mathbf{200}}}\left(r\right){\mathbf{=}}\frac{\mathbf{1}}{\mathbf{4}\sqrt{\mathbf{2}}\mathbf{\pi }}{{\mathbf{a}}}^{\mathbf{-}\mathbf{3}\mathbf{/}\mathbf{2}}\left(2-\frac{r}{a}\right){{\mathbf{e}}}^{\mathbf{-}\mathbf{r}\mathbf{/}\mathbf{2}\mathbf{a}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{\left(}}{\mathbf{1}}{\mathbf{\right)}}$

## Step 2: Plot Ψ2002(r) and show that the plot is consistent with the dot plot of the given figure(a)

Take the values of r on the horizontal axis, but actually, the values of r/a must be taken on the horizontal axis. Here, a is the Bohr radius, and it is a constant value.

The plot is shown in the below figure.

From the above figure, it can be observed that there is a high central peak between r = 0 and r = 2A. At r = 2a, the value of the wave function ${\Psi }_{200}\left(r\right)$ must be equal to zero . Also, the low peak value is reached its maximum value at r = 4a. The graph is not consistent with the dot plot of the given figure.

## Step 3: Show that Ψ2002(r) has a maximum at r = 4a(b)

Differentiate the given wave function and equate to zero to find the maximum.

$\frac{\partial }{\partial r}\left[{\Psi }_{200}\left(r\right)\right]=0\phantom{\rule{0ex}{0ex}}\frac{\partial }{\partial r}\left[\frac{1}{4\sqrt{2\mathrm{\pi }}}{a}^{-3/2}\left(2-\frac{r}{a}\right){e}^{-r/2a}\right]=0\phantom{\rule{0ex}{0ex}}\frac{\partial }{\partial r}\left(2-\frac{r}{a}\right){e}^{-r/2a}=0\phantom{\rule{0ex}{0ex}}\left(2-\frac{r}{a}\right){e}^{-r/2a}\left(-\frac{1}{2a}\right)+\left(-\frac{r}{a}\right){e}^{-r/2a}=0$

Simplify further.

$\left(2-\frac{r}{a}\right)\left(-\frac{1}{2}\right)-1=0\phantom{\rule{0ex}{0ex}}-1+\frac{r}{2a}-1=0\phantom{\rule{0ex}{0ex}}\frac{r}{2a}=2\phantom{\rule{0ex}{0ex}}r=4a$

Therefore, it is proved that ${\Psi }_{200}^{2}\left(r\right)$ has a maximum at r = 4a.

## Step 4: Find the radial probability density P200(r) for this state(c)

The expression for the radial probability is as follows.

${P}_{200}\left(r\right)=4{\mathrm{\pi r}}^{2}{\mathrm{\Psi }}_{200}^{2}\left(\mathrm{r}\right)\phantom{\rule{0ex}{0ex}}=\frac{{\mathrm{r}}^{2}}{8{\mathrm{a}}^{3}}\left(2-\frac{\mathrm{r}}{\mathrm{a}}\right){\mathrm{e}}^{-\mathrm{r}/\mathrm{a}}$

Therefore, the radial probability for the given state is ${P}_{200}\left(r\right)=\frac{{\mathrm{r}}^{2}}{8{\mathrm{a}}^{3}}\left(2-\frac{\mathrm{r}}{\mathrm{a}}\right){\mathrm{e}}^{-\mathrm{r}/\mathrm{a}}$.

## Step 5: Show that ∫0∞P200(r)dr=1 (d)

Show that the value of ${\int }_{0}^{\infty }{P}_{200}\left(r\right)dr=1$ as follows.

${\int }_{0}^{\infty }{P}_{200}\left(r\right)dr={\int }_{0}^{\infty }\left[\frac{{r}^{2}}{8{a}^{3}}{\left(2-\frac{r}{a}\right)}^{2}{e}^{-r/a}\right]dr\phantom{\rule{0ex}{0ex}}=\frac{1}{8}{\int }_{0}^{\infty }{x}^{2}{\left(2-x\right)}^{2}{e}^{-x}dx\phantom{\rule{0ex}{0ex}}=\frac{1}{8}{\int }_{0}^{\infty }\left[{x}^{4}-4{x}^{3}+4{x}^{2}\right]{e}^{-x}dx\phantom{\rule{0ex}{0ex}}=\frac{1}{8}\left[4!-4\left(3!\right)+4\left(2!\right)\right]$

Simplify further.

${\int }_{0}^{\infty }{P}_{200}\left(r\right)dr=\frac{1}{8}\left[24-24+8\right]\phantom{\rule{0ex}{0ex}}=1$

Therefore, it is proved that ${\int }_{0}^{\infty }{P}_{200}\left(r\right)dr=1$.

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