Question

For states where l = n - t the radial probability assumes the general form given in Exercise 54. The proportionality constant that normalizes this radial probability is given in Exercise 64.

(a) Show that the expectation value of the hydrogen atom potential energy is exactly twice the total energy. (It turns out that this holds no matter what l may be)

(b) Argue that the expectation value of the kinetic energy must be the negative of the total energy.

Short answer

a) The expectation value of the hydrogen atom potential energy is exactly twice the total energy

b) The expectation value of the kinetic energy must be the negative of the total energy.

Step-by-step solution

 Given data

The probability density is given by

$$\begin{gathered} \mathrm{P}\left(\mathrm{r}\right)={\left(\frac{2}{{\mathrm{na}}_0}\right)}^{2\mathrm{n}-1}\frac{1}{\left(2\mathrm{n}\right)!}{\mathrm{r}}^{2{\mathrm{ne}}^{-2\mathrm{r}/{\mathrm{na}}_0}} \\ \mathrm{K}.\mathrm{E}={\mathrm{E}}_{\mathrm{total}}-{\mathrm{E}}_{\mathrm{potential}} \end{gathered}$$

 Concept

The potential energy of an atom is generally negative. The negative sign represents the attractive nature of the Colombian forces that act between the nucleus and the surrounding electrons.

 Calculation

The expectation value of potential energy is given as follows to find the potential energy

$$\begin{gathered} PE={\int }_0^{\infty }-\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{e^2}{r}{\left(\frac{2}{n{a}_0}\right)}^{2n+1}\frac{1}{\left(2n\right)!}{r}^{2n}{e}^{-2r/n{a}_0}dr \\ =\left(\frac{e^2}{4{\mathrm{\pi \epsilon }}_0}\right){\left(\frac{2}{n{a}_0}\right)}^{2n+1}\frac{1}{\left(2n\right)!}{\int }_0^{\infty }{r}^{2n}{e}^{-2r/n{a}_0}dr \\ =\left(\frac{e^2}{4{\mathrm{\pi \epsilon }}_0}\right)\left(\frac{2}{n{a}_0}\right)\frac{1}{\left(2n\right)!}{\int }_0^{\infty }{x}^{2n-1}{e}^{-x}dx \\ =\frac{1}{2n}\frac{2}{n{a}_0}\left(-\frac{e^2}{4{\mathrm{\pi \epsilon }}_0}\right) \\ PE=\left(-\frac{e^2}{4{\mathrm{\pi \epsilon }}_0}\right)\frac{1}{n^2{a}_0} \end{gathered}$$

Using$a_0=\frac{4{\mathrm{\pi \epsilon }}_0{\mathrm{h}}^2}{m{e}^2}$

$$\begin{gathered} PE=\frac{\left(-{\displaystyle \frac{e^2}{4{\mathrm{\pi \epsilon }}_0}}\right)}{{\displaystyle \frac{4{\mathrm{\pi \epsilon }}_0{h}^2}{m{e}^2}}} \\ =-\frac{m{e}^4}{{\left(4{\mathrm{\pi \epsilon }}_0\right)}^2{\mathrm{h}}^2{\mathrm{n}}^2} \end{gathered}$$

On the other hand, potential energy is given by

$$\begin{gathered} E_n=-\frac{m{e}^4}{2{\left(4{\mathrm{\pi \epsilon }}_0\right)}^2{\mathrm{h}}^2{\mathrm{n}}^2} \\ =-\frac{PE}{2} \end{gathered}$$

Thus, the expectation value of the hydrogen atom potential energy is exactly twice the total energy.

(b)

The kinetic energy is given as-

$$\begin{gathered} K.E=E_n-PE \\ =E_n-2E_n \\ =-E_n \end{gathered}$$

Thus, the expectation value of the kinetic energy must be the negative of the total energy.