Question

The wave function for the hydrogen-atom quantum state represented by the dot plot shown in Fig. 39-21, which has n = 2 and $\mathcal{l}\mathbf{=}{\mathit{m}}_{\mathbf{l}}\mathbf{=}\mathbf{0}$, is

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

in which a is the Bohr radius and the subscript on$\mathit{\Psi }\mathbf{\left(}\mathit{r}\mathbf{\right)}$gives the values of the quantum numbers $\mathit{n}\mathbf{,}\mathcal{l}\mathbf{,}{\mathit{m}}_{\mathcal{l}}$. (a) Plot$\mathit{\Psi }{}_{\mathbf{200}}{}^{\mathbf{2}}\mathbf{(}\mathit{r}\mathbf{)}$and show that your plot is consistent with the dot plot of Fig. 39-21. (b) Show analytically that$\mathit{\Psi }{}_{\mathbf{200}}{}^{\mathbf{2}}\mathbf{(}\mathit{r}\mathbf{)}$has a maximum at $\mathit{r}\mathbf{=}\mathbf{4}\mathit{a}$. (c) Find the radial probability density${\mathit{P}}_{\mathbf{200}}\mathbf{\left(}\mathbf{r}\mathbf{\right)}$for this state. (d) Show that

${\mathbf{\int }}_{\mathbf{0}}^{\mathbf{\infty }}{\mathit{P}}_{\mathbf{200}}\mathbf{\left(}\mathit{r}\mathbf{\right)}\mathit{d}\mathit{r}\mathbf{=}\mathbf{1}$

and thus that the expression above for the wave function ${\mathit{\Psi }}_{\mathbf{200}}\mathbf{\left(}\mathbf{r}\mathbf{\right)}$has been properly normalized.

Short answer

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 $P_{200}\left(r\right)=\frac{r^2}{8a^3}\left(2-\frac{r}{a}\right)e^{-r/a}$.

d. It is proved that ${\int }_0^{\infty }{P}_{200}\left(r\right)dr=1$.

Step-by-step solution

Describe the given information

The wave function is given by,

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

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.

Show that Ψ2002(r) has a maximum at r = 4a(b)

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

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

Simplify further.

$$\begin{gathered} \left(2-\frac{r}{a}\right)\left(-\frac{1}{2}\right)-1=0 \\ -1+\frac{r}{2a}-1=0 \\ \frac{r}{2a}=2 \\ r=4a \end{gathered}$$

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

Find the radial probability density P200(r) for this state(c)

The expression for the radial probability is as follows.

$$\begin{gathered} P_{200}\left(r\right)=4{\mathrm{\pi r}}^2{\mathrm{\Psi }}_{200}^2\left(\mathrm{r}\right) \\ =\frac{{\mathrm{r}}^2}{8{\mathrm{a}}^3}\left(2-\frac{\mathrm{r}}{\mathrm{a}}\right){\mathrm{e}}^{-\mathrm{r}/\mathrm{a}} \end{gathered}$$

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

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.

$$\begin{gathered} {\int }_0^{\infty }{P}_{200}\left(r\right)dr={\int }_0^{\infty }\left[\frac{r^2}{8a^3}{\left(2-\frac{r}{a}\right)}^2{e}^{-r/a}\right]dr \\ =\frac{1}{8}{\int }_0^{\infty }{x}^2{\left(2-x\right)}^2{e}^{-x}dx \\ =\frac{1}{8}{\int }_0^{\infty }\left[x^4-4x^3+4x^2\right]e^{-x}dx \\ =\frac{1}{8}\left[4!-4(3!)+4(2!)\right] \end{gathered}$$

Simplify further.

$$\begin{gathered} {\int }_0^{\infty }{P}_{200}\left(r\right)dr=\frac{1}{8}\left[24-24+8\right] \\ =1 \end{gathered}$$

Therefore, it is proved that ${\int }_0^{\infty }{P}_{200}\left(r\right)dr=1$.