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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.