(a) Find〈r〉and〈r²〉for an electron in the ground state of hydrogen. Express your answers in terms of the Bohr radius.

(b) Find〈x〉and $\left(x^2\right)$for an electron in the ground state of hydrogen.

Hint: This requires no new integration—note that ${\mathit{r}}^{\mathbf{2}}\mathbf{=}{\mathit{x}}^{\mathbf{2}}\mathbf{+}{\mathit{y}}^{\mathbf{2}}\mathbf{+}{\mathit{z}}^{\mathbf{2}}$,and exploit the symmetry of the ground state.

(c) Find〈x²〉in the state $\mathit{n}\mathbf{=}\mathbf{2}\mathbf{,}\mathit{l}\mathbf{=}\mathbf{1}\mathbf{,}\mathit{m}\mathbf{=}\mathbf{1}$. Hint: this state is not symmetrical in x, y, z. Use$\mathit{x}\mathbf{=}\mathit{r}\mathbf{}\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }\mathbf{}\mathbf{}\mathit{c}\mathit{o}\mathit{s}$$\mathit{\pi }\mathit{x}\mathbf{=}\mathit{r}\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }\mathit{c}\mathit{o}\mathit{s}\mathit{\varphi }$

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:

${\psi }_{100}=\frac{1}{\sqrt{{\mathrm{\pi a}}^3}}{e}^{-r/a}$

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

$$\begin{gathered} ⟨r^n⟩=\frac{1}{\pi a^3}{\int }_0^{\infty }{\int }_0^{\mathrm{\pi }/2}{\int }_0^2{r}^n{e}^{-2r/a}\left(r^{2}sin\right(\theta \left)drd\theta d\varphi \right) \\ =4\pi /\left(\pi a^3\right){\int }_0^{\infty }{r}^{n+2}{e}^{-2r/a}dr \end{gathered}$$

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

$⟨r^n⟩=\frac{4\pi }{\pi a^3}{\int }_0^{\infty }{\left(\frac{a}{2}\right)}^{n+3}{x}^{n+2}{e}^{-x}dx$

But,

${\int }_0^{\infty }{x}^{n+2}{e}^{-x}dx=\Gamma (n+3)=(n+2)!$

Thus,

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

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

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

(c)

From problem 4.11 we have:

${\psi }_{211}=R_{21}{Y}_1^1=-\frac{1}{\sqrt{\mathrm{\pi a}}}\frac{1}{8a^2}r{e}^{-r/2a}sin\left(\theta \right)e^i\varphi$

Thus,

$\left(x^2\right)=\frac{1}{\mathrm{\pi a}}\frac{1}{{\left(8a^2\right)}^2}=\int \left(r^2{e}^{-r/a}{\sin }^2\left(\theta \right)\right)x^2{r}^2\sin \left(\theta \right)drd\theta d\varphi$

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

So:

role="math" localid="1658384023991" $$\begin{gathered} \left(x^2\right)=\frac{1}{\mathrm{\pi a}}\frac{1}{{\left(8a^2\right)}^2}\int \left(r^2{e}^{-r/a}{\sin }^2\left(\theta \right)\right)\left(r^2{\sin }^2\left(\theta \right){\cos }^2\left(\varphi \right)\right)r^2\sin \left(\theta \right)drd\theta d\varphi \\ \left(x^2\right)=\frac{1}{64{\mathrm{\pi a}}^5}{\int }_0^{\infty }{r}^6{e}^{-r/a}dr{\int }_0^{\mathrm{\pi }}{\cos }^2\left(\varphi \right)d\varphi \\ \left(x^2\right)=\frac{1}{64{\mathrm{\pi a}}^5}\left(6!a^7\right)\left(2\frac{2.4}{1.3.5}\right)\left(\frac{1}{2}.2\mathrm{\pi }\right) \\ \left(x^2\right)=12a^2 \end{gathered}$$

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