91影视

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:

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

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

$$\begin{gathered} 鉄�r^n鉄�=\frac{1}{蟺a^3}{鈭�}_0^{鈭�}{鈭�}_0^{\mathrm{蟺}/2}{鈭�}_0^2{r}^n{e}^{-2r/a}\left(r^{2}sin\right(胃\left)drd胃d蠒\right) \\ =4蟺/\left(蟺a^3\right){鈭�}_0^{鈭�}{r}^{n+2}{e}^{-2r/a}dr \end{gathered}$$

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

$鉄�r^n鉄�=\frac{4蟺}{蟺a^3}{鈭�}_0^{鈭�}{\left(\frac{a}{2}\right)}^{n+3}{x}^{n+2}{e}^{-x}dx$

But,

${鈭�}_0^{鈭�}{x}^{n+2}{e}^{-x}dx=螕(n+3)=(n+2)!$

Thus,

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

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

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

(c)

From problem 4.11 we have:

${蠄}_{211}=R_{21}{Y}_1^1=-\frac{1}{\sqrt{\mathrm{蟺补}}}\frac{1}{8a^2}r{e}^{-r/2a}sin\left(胃\right)e^i蠒$

Thus,

$\left(x^2\right)=\frac{1}{\mathrm{蟺补}}\frac{1}{{\left(8a^2\right)}^2}=鈭�\left(r^2{e}^{-r/a}{\sin }^2\left(胃\right)\right)x^2{r}^2\sin \left(胃\right)drd胃d蠒$

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

So:

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

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