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Expert-verified Found in: Page 1217 ### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718 # The wave functions for the three states with the dot plots shown in Fig. 39-23, which have n = 2 , l = 1 , and 0, and ${{\mathbf{m}}}_{{\mathbf{l}}}{\mathbf{=}}{\mathbf{0}}{\mathbf{,}}{\mathbf{+}}{\mathbf{1}}{\mathbf{,}}{\mathbf{-}}{\mathbf{1}}$ , are${\mathbf{}}{\mathbf{}}{{\mathbf{\Psi }}}_{{\mathbf{210}}}\left(r,\theta \right){\mathbf{=}}\left(1/4\sqrt{2\pi }\right)\left({a}^{-3/2}\right)\left(r/a\right){{\mathbf{r}}}^{\mathbf{-}\mathbf{r}\mathbf{/}\mathbf{2}\mathbf{a}}{\mathbf{}}{\mathbf{cos}}{\mathbf{}}{\mathbf{\theta }}\phantom{\rule{0ex}{0ex}}{{\mathbf{\Psi }}}_{\mathbf{21}\mathbf{+}\mathbf{1}}\left(r,\theta \right){\mathbf{=}}\left(1/8\sqrt{\pi }\right)\left({a}^{-3/2}\right)\left(r/a\right){{\mathbf{r}}}^{\mathbf{-}\mathbf{r}\mathbf{/}\mathbf{2}\mathbf{a}}{\mathbf{}}\left(\mathrm{sin\theta }\right){\mathbf{}}{{\mathbf{e}}}^{\mathbf{+}\mathbf{i\varphi }}\phantom{\rule{0ex}{0ex}}{{\mathbf{\Psi }}}_{\mathbf{21}\mathbf{-}\mathbf{1}}\left(r,\theta \right){\mathbf{=}}\left(1/8\sqrt{\pi }\right)\left({a}^{-3/2}\right)\left(r/a\right){{\mathbf{r}}}^{\mathbf{-}\mathbf{r}\mathbf{/}\mathbf{2}\mathbf{a}}{\mathbf{}}\left(\mathrm{sin\theta }\right){\mathbf{}}{{\mathbf{e}}}^{\mathbf{-}\mathbf{i\varphi }}$ in which the subscripts on ${\mathbit{\Psi }}\left(r,\theta \right)$ give the values of the quantum numbers n , l , and ${{\mathbf{m}}}_{{\mathbf{l}}}$ the angles ${\mathbit{\theta }}$ and ${\mathbit{\varphi }}$ are defined in Fig. 39-22. Note that the first wave function is real but the others, which involve the imaginary number i, are complex. Find the radial probability density P(r) for (a) ${{\mathbf{\Psi }}}_{{\mathbf{210}}}$ and (b) ${{\mathbf{\Psi }}}_{\mathbf{21}\mathbf{+}\mathbf{1}}$ (same as for ${{\mathbf{\Psi }}}_{\mathbf{21}\mathbf{-}\mathbf{1}}$ ). (c) Show that each P(r) is consistent with the corresponding dot plot in Fig. 39-23. (d) Add the radial probability densities for ${{\mathbf{\Psi }}}_{{\mathbf{210}}}$ , ${{\mathbf{\Psi }}}_{\mathbf{21}\mathbf{+}\mathbf{1}}$ , and ${{\mathbf{\Psi }}}_{\mathbf{21}\mathbf{-}\mathbf{1}}$ and then show that the sum is spherically symmetric, depending only on r .

1. The probability density of the function ${\mathrm{\Psi }}_{210}$ is ${P}_{210}\left(r\right)=\left(\frac{{r}^{4}}{8{a}^{5}}\right){r}^{-r/a}{\mathrm{cos}}^{2}\theta$ .
2. The probability density of the function ${\mathrm{\Psi }}_{21+1}$ is ${P}_{21+1}\left(r\right)=\left(\frac{{r}^{4}}{16{a}^{5}}\right){r}^{-r/a}\left({\mathrm{sin}}^{2}\theta \right)$ .
3. The probability P(r) is consistent with the corresponding dot plot.
4. It is proved that the total probability density depends only on r , and it is spherically symmetric.
See the step by step solution

## Step 1: Give the expression for probability density function:

The probability density function (PDF) is used to define the probability of a random variable falling into a distinct range of values, as opposed to assuming a single value. The function explains the probability density function of the normal distribution and how the mean and variance exist.

The expression for the probability density of the function ${\mathbf{\Psi }}$ is given by,

$\mathbf{P}\mathbf{\left(}\mathbf{r}{\mathbf{\right)}}{\mathbf{=}}{|\Psi |}^{{\mathbf{2}}}{\mathbf{\left(}}{\mathbf{4}}{{\mathbf{\tau \tau r}}}^{{\mathbf{2}}}{\mathbf{\right)}}$

## Step 2: (a) Define the radial probability density P(r) for Ψ210 :

The probability density of the function ${\mathrm{\Psi }}_{210}$ is calculated as follows.

${P}_{210}\left(r\right)={\left|{\mathrm{\Psi }}_{210}\right|}^{2}\left(4{\mathrm{\pi r}}^{2}\right)\phantom{\rule{0ex}{0ex}}={\left|\left(1/4\sqrt{2\mathrm{\pi }}\right)\left({a}^{-3/2}\right)\left(r/a\right){r}^{-r/2a}\mathrm{cos}\theta \right|}^{2}\left(4{\mathrm{\pi r}}^{2}\right)\phantom{\rule{0ex}{0ex}}={\left(\frac{1}{4\sqrt{2\mathrm{\pi }}}\right)}^{2}{\left(\frac{r}{{a}^{5/2}}\right)}^{2}{\left({r}^{-r/2a}\mathrm{cos}\theta \right)}^{2}{\left(4\mathrm{\pi }r\right)}^{2}\phantom{\rule{0ex}{0ex}}=\left(\frac{1}{32\mathrm{\pi }}\right)\left(\frac{{r}^{2}}{{a}^{5}}\right)\left({r}^{-2r/2a}{\mathrm{cos}}^{2}\theta \right)\left(4{\mathrm{\pi r}}^{2}\right)$

Simplify further.

${P}_{210}\left(r\right)=\left(\frac{{r}^{4}}{8{a}^{5}}\right){r}^{-r/a}{\mathrm{cos}}^{2}\theta$

Therefore, the probability density of the function ${\Psi }_{210}$ is ${P}_{210}\left(r\right)=\left(\frac{{r}^{4}}{8{a}^{5}}\right){r}^{-r/a}{\mathrm{cos}}^{2}\theta$ .

## Step 3: (b) Find the radial probability density P(r) for Ψ21+1 :

Find the square of the function ${\mathrm{\Psi }}_{21+1}$ .

${\left|{\mathrm{\Psi }}_{21+1}\right|}^{2}={\left(\left(1/8\sqrt{\mathrm{\pi }}\right)\left({a}^{-3/2}\right)\left(r/a\right){r}^{-r/2a}\left(\mathrm{sin}\theta \right){e}^{+i\varphi }\right)}^{2}\phantom{\rule{0ex}{0ex}}={\left(\left(1/8\sqrt{\mathrm{\pi }}\right)\left(\frac{r}{{a}^{5/2}}\right){r}^{-r/2a}\left({\mathrm{sin}}^{2}\theta \right){e}^{+i\varphi }\right)}^{2}\phantom{\rule{0ex}{0ex}}=\left(\frac{1}{64\mathrm{\pi }}\right)\left(\frac{{r}^{2}}{{a}^{5}}\right){r}^{-2r/2a}\left({\mathrm{sin}}^{2}\theta \right){e}^{+2i\varphi }\phantom{\rule{0ex}{0ex}}=\left(\frac{{r}^{2}}{64{\mathrm{\pi a}}^{5}}\right){r}^{-r/a}\left({\mathrm{sin}}^{2}\theta \right)$

Find the square of the function ${\Psi }_{21-1}$ .

${\left|{\mathrm{\Psi }}_{21+1}\right|}^{2}={\left(\left(1/8\sqrt{\mathrm{\pi }}\right)\left({a}^{-3/2}\right)\left(r/a\right){r}^{-r/2a}\left(\mathrm{sin}\theta \right){e}^{-i\varphi }\right)}^{2}\phantom{\rule{0ex}{0ex}}={\left(\left(1/8\sqrt{\mathrm{\pi }}\right)\left(\frac{r}{{a}^{5/2}}\right){r}^{-r/2a}\left({\mathrm{sin}}^{2}\theta \right){e}^{-i\varphi }\right)}^{2}\phantom{\rule{0ex}{0ex}}=\left(\frac{1}{64\mathrm{\pi }}\right)\left(\frac{{r}^{2}}{{a}^{5}}\right){r}^{-2r/2a}\left({\mathrm{sin}}^{2}\theta \right){e}^{-2i\varphi }\phantom{\rule{0ex}{0ex}}=\left(\frac{{r}^{2}}{64{\mathrm{\pi a}}^{5}}\right){r}^{-r/a}\left({\mathrm{sin}}^{2}\theta \right)$

Find the probability density of the above two functions.

${P}_{21±1}\left(r\right)={\left|{\Psi }_{21±1}\right|}^{2}\left(4{\mathrm{\pi r}}^{2}\right)\phantom{\rule{0ex}{0ex}}=\left(\frac{{r}^{2}}{64{\mathrm{\pi a}}^{5}}\right){r}^{-r/a}\left({\mathrm{sin}}^{2}\theta \right)\left(4{\mathrm{\pi r}}^{2}\right)\phantom{\rule{0ex}{0ex}}=\left(\frac{{r}^{4}}{16{a}^{5}}\right){r}^{-r/a}\left({\mathrm{sin}}^{2}\theta \right)$

Hence, the probability density of the function ${\mathrm{\Psi }}_{21+1}$ is ${P}_{21+1}\left(r\right)=\left(\frac{{r}^{4}}{16{a}^{5}}\right){r}^{-r/a}\left({\mathrm{sin}}^{2}\theta \right)$ .

## Step 4: (c) Show that each P(r)  is consistent with the corresponding dot plot:

For the probability density in the first case where ${\mathrm{m}}_{\mathrm{l}}=0$ , the probability decreases with radial distance from the nucleus, also the probability is proportional to the factor of ${\mathrm{cos}}^{2}\theta$ which is the maximum along the z-axis of which the angle is $\theta =0$ . This is consistent with the dot plot.

For the probability density in the second and the third cases where ${\mathrm{m}}_{\mathrm{l}}=±1$ , the probability decreases with radial distance from the nucleus, also the probability is proportional to the factor of ${\mathrm{sin}}^{2}\theta$ which is the maximum in the xy -plane of which the angle is $\theta ={90}^{\circ }$ . This is also consistent with the dot plot.

Therefore, the probability P(r) is consistent with the corresponding dot plot.

## Step 5: (d) Show that the sum of Ψ210 , Ψ21+1 , and Ψ21-1  is spherically symmetric, and depending only on r :

Add the three probabilities as follows.

$P\left(r\right)={P}_{210}\left(r\right)+{P}_{21+1}\left(r\right)+{P}_{21-1}\left(r\right)\phantom{\rule{0ex}{0ex}}=\left(\frac{{r}^{4}}{8{a}^{5}}\right){r}^{-r/a}{\mathrm{cos}}^{2}\theta +\left(\frac{{r}^{4}}{16{a}^{5}}\right){r}^{-r/a}\left({\mathrm{sin}}^{2}\theta \right)+\left(\frac{{r}^{4}}{16{a}^{5}}\right){r}^{-r/a}\left({\mathrm{sin}}^{2}\theta \right)\phantom{\rule{0ex}{0ex}}=\left(\frac{{r}^{4}}{8{a}^{5}}\right){r}^{-r/a}\left({\mathrm{sin}}^{2}\theta +{\mathrm{cos}}^{2}\theta \right)\phantom{\rule{0ex}{0ex}}=\left(\frac{{r}^{4}}{8{a}^{5}}\right){r}^{-r/a}$

From the above equation it is clear that the total probability density depends only on r , and it is spherically symmetric. ### Want to see more solutions like these? 