91Ó°ÊÓ

The probability of finding an electron in some finite volume element around a point at a distance of r from the nucleus is given by the radial wave function R(r), which is simply the value of the wave function at some radius r.

(a)

We need to find$\left(r\right)$and$\left(r^2\right)$for an electron in the ground state of hydrogen, the wave function of the ground state is:

${\mathrm{ψ}}_{100}=\frac{1}{\sqrt{{\mathrm{Ï€²¹}}^3}}{e}^{-r/a}$

The expectation value of $r^n$is therefore:

$$\begin{gathered} ⟨r^nâŸ\copyright =\frac{1}{\mathrm{Ï€}{a}^3}{∫}_0^{∞}{∫}_0^{\mathrm{Ï€}/2}{∫}_0^2{r}^n{e}^{-2r/a}\left(r^{2}sin\right(\mathrm{θ}\left)drd\mathrm{θ}d\mathrm{Ï•}\right) \\ =4\mathrm{Ï€}/\left(\mathrm{Ï€}{a}^3\right){∫}_0^{∞}{r}^{n+2}{e}^{-2r/a}dr \end{gathered}$$

Letrole="math" localid="1658381554869" $x=2/a$, so $dx=\left(2/a\right)dr$, thus:

$⟨r^nâŸ\copyright =\frac{4\mathrm{Ï€}}{\mathrm{Ï€}{a}^3}{∫}_0^{∞}{\left(\frac{a}{2}\right)}^{n+3}{x}^{n+2}{e}^{-x}dx$

But,

${∫}_0^{∞}{x}^{n+2}{e}^{-x}dx=\mathrm{Γ}(n+3)=(n+2)!$

Thus,

$⟨r^nâŸ\copyright =\frac{4}{a^3}{\left(\frac{a}{2}\right)}^{n+3}(n+2)!$

For,$\left(r\right),n=1$thus:

$$\begin{gathered} \left(r\right)=\frac{4}{a^3}{\left(\frac{a}{2}\right)}^4\left(3\right)! \\ \left(r\right)=\frac{3}{2}a \end{gathered}$$

For,$\left(r^2\right),n=2$thus:

$$\begin{gathered} \left(r^2\right)=\frac{4}{3}\left(4\right)!{\left(\frac{a}{2}\right)}^5 \\ \left(r^2\right)=3a^2 \end{gathered}$$

(b)

Since the ground state is symmetric, we can work out the means for the rectangular coordinates separately without doing any more integrals (note that $r^2=x^2+y^2+z^2$):$\left(x\right)=0$

And

$$\begin{gathered} ⟨x^2âŸ\copyright =\frac{1}{3}\left(r^2\right)=a^2\left(x^2\right) \\ ⟨x^2âŸ\copyright =a^2 \end{gathered}$$

(c)

From problem 4.11 we have:

${\mathrm{ψ}}_{211}=R_{21}{Y}_1^1=-\frac{1}{\sqrt{\mathrm{Ï€²¹}}}\frac{1}{8a^2}r{e}^{-r/2a}sin\left(\mathrm{θ}\right)e^i\mathrm{Ï•}$

Thus,

$\left(x^2\right)=\frac{1}{\mathrm{Ï€²¹}}\frac{1}{{\left(8a^2\right)}^2}=∫\left(r^2{e}^{-r/a}{\sin }^2\left(\mathrm{θ}\right)\right)x^2{r}^2\sin \left(\mathrm{θ}\right)drd\mathrm{θ}d\mathrm{Ï•}$

Note that $x=r\sin (\mathrm{θ})\cos (\mathrm{ϕ})$, so$x^2=r^2{\sin }^2\left(\mathrm{θ}\right){\cos }^2\left(\mathrm{ϕ}\right)$..

So:

role="math" localid="1658384023991" $$\begin{gathered} \left(x^2\right)=\frac{1}{\mathrm{Ï€²¹}}\frac{1}{{\left(8a^2\right)}^2}∫\left(r^2{e}^{-r/a}{\sin }^2\left(\mathrm{θ}\right)\right)\left(r^2{\sin }^2\left(\mathrm{θ}\right){\cos }^2\left(\mathrm{Ï•}\right)\right)r^2\sin \left(\mathrm{θ}\right)drd\mathrm{θ}d\mathrm{Ï•} \\ \left(x^2\right)=\frac{1}{64{\mathrm{Ï€²¹}}^5}{∫}_0^{∞}{r}^6{e}^{-r/a}dr{∫}_0^{\mathrm{Ï€}}{\cos }^2\left(\mathrm{Ï•}\right)d\mathrm{Ï•} \\ \left(x^2\right)=\frac{1}{64{\mathrm{Ï€²¹}}^5}\left(6!a^7\right)\left(2\frac{2.4}{1.3.5}\right)\left(\frac{1}{2}.2\mathrm{Ï€}\right) \\ \left(x^2\right)=12a^2 \end{gathered}$$

Hence the state$\left(x^2\right)=12a^2$