(a) Calculate the average orbital radius of a 3d electron in the hydrogen atom. Compare with the Bohr radius for a n = 3 electron. (b) What is the probability of a 3d electron in the hydrogen atom being at a greater radius than the n = 3 Bohr electron? Strategy We used a similar strategy in Example

Question

(a) Calculate the average orbital radius of a 3d electron in the hydrogen atom. Compare with the Bohr radius for a n = 3 electron. (b) What is the probability of a 3d electron in the hydrogen atom being at a greater radius than the n = 3 Bohr electron?

Strategy We used a similar strategy in Example 7.12 to find the expectation (or average) value of r. We determine the probability of a 3d electron being in a certain radial position (greater than [latex]r_{3}=n^{2} a_{0}=3^{2} a_{0}=9 a_{0}[/latex]) by integrating over the probability density from [latex]9 a_{0} ext { to } \infty[/latex].

Accepted Answer

(a) The expectation value of the radial position of a 3d electron is

[latex]\left\langle r_{3 d} ight angle=\int_{0}^{\infty} r P(r) d r=\int_{0}^{\infty} r^{3}\left|R_{3 d}(r) ight|^{2} d r[/latex]

where we have used Equations (6.20) and (7.39). We look up [latex]R_{3 d}(r)[/latex] in Table 7.1 and obtain

[latex]\langle g(x) angle=\int_{-\infty}^{\infty} \Psi *(x, t) g(x) \Psi(x, t) d x[/latex] (6.20)

[latex]P_{n \ell}(r) d r=r^{2}\left|R_{n \ell}(r) ight|^{2} d r[/latex] (7.39)

[latex]\left\langle r_{3 d} ight angle=\frac{1}{a_{0}^{7}}\left(\frac{4}{81 \sqrt{30}} ight)^{2} \int_{0}^{\infty} r^{7} e^{-2 r / 3 a_{0}} d r[/latex]

We use the integral [latex]\int_{0}^{\infty} x^{n} e^{-x / \alpha} d x=n ! \alpha^{n+1}[/latex] from Appendix 3 to determine

[latex]\left\langle r_{3 d} ight angle=\frac{1}{a_{0}{ }^{7}}\left(\frac{4}{81 \sqrt{30}} ight)^{2} 7 !\left(\frac{3 a_{0}}{2} ight)^{8}=10.5 a_{0}[/latex]

The average value of [latex]r_{3 d}[/latex] is more than 15% larger than the Bohr value of [latex]9 a_{0}[/latex]. We see in Figure 7.12b that the most probable value of [latex]P_{32} ext { is } 9 a_{0}[/latex], but because of the tail of the [latex]P_{32}[/latex] distribution for increasing radius, it is likely that the average value of [latex]r_{32}\left(r_{3 d} ight)[/latex] is somewhat larger, consistent with a value of [latex]10.5 a_{0}[/latex].

(b) The probability of the electron in the 3d state of the hydrogen atom being at a radius greater the Bohr radius [latex]9 a_{0}[/latex] is (see Example 7.13)

[latex]P=\int_{9 a_{0}}^{\infty} P(r) d r=\int_{9 a_{0}}^{\infty} r^{2}\left|R_{3 d}(r) ight|^{2} d r[/latex]

We again look up [latex]R_{3 d}(r)[/latex] in Table 7.1 and obtain

[latex]P=\frac{1}{a_{0}^{7}}\left(\frac{4}{81 \sqrt{30}} ight)^{2} \int_{9 a_{0}}^{\infty} r^{6} e^{-2 r / 3 a_{0}} d r[/latex]

This integral is somewhat more difficult (that is, tedious) than the one we did in (a). We must use an indefinite integral from Appendix 3, because the lower integration limit is [latex]9 a_{0}[/latex], not zero. If we use the integral formula [latex]\int x^{m} e^{b x} d x=e^{b x} \sum_{k=0}^{m}(-1)^{k} \frac{m ! x^{m-k}}{(m-k) ! b^{k+1}}[/latex] we end up with seven
terms and a high probability of making an error. This integral is a good candidate for computer integration, which gives a value of 0.606. The result for the probability P is 61%, which seems reasonable given the shape of the distribution.

Table 7.1 Hydrogen Atom Radial Wave Functions
n	[latex]\ell[/latex]	[latex]R_{n \ell}(r)[/latex]
1	0	[latex]\frac{2}{\left(a_{0} ight)^{3 / 2}} e^{-r / a_{0}}[/latex]
2	0	[latex]\left(2-\frac{r}{a_{0}} ight) \frac{e^{-r / 2 a_{0}}}{\left(2 a_{0} ight)^{3 / 2}}[/latex]
2	1	[latex]\frac{r}{a_{0}} \frac{e^{-r / 2 a_{0}}}{\sqrt{3}\left(2 a_{0} ight)^{3 / 2}}[/latex]
3	0	[latex]\frac{1}{\left(a_{0} ight)^{3 / 2}} \frac{2}{81 \sqrt{3}}\left(27-18 \frac{r}{a_{0}}+2 \frac{r^{2}}{a_{0}^{2}} ight) e^{-r / 3 a_{0}}[/latex]
3	1	[latex]\frac{1}{\left(a_{0} ight)^{3 / 2}} \frac{4}{81 \sqrt{6}}\left(6-\frac{r}{a_{0}} ight) \frac{r}{a_{0}} e^{-r / 3 a_{0}}[/latex]
3	2	[latex]\frac{1}{\left(a_{0} ight)^{3 / 2}} \frac{4}{81 \sqrt{30}} \frac{r^{2}}{a_{0}^{2}} e^{-r / 3 a_{0}}[/latex]

Question 7.14: (a) Calculate the average orbital radius of a 3d electron in...

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