91影视

The probability density is given by

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

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.

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

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

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

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

On the other hand, potential energy is given by

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

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.