91Ó°ÊÓ

To be considered an electron in the ground state of a hydrogen atom

The law of conservation of momentum states that the sum of kinetic energy and the potential energy for a particle always remains constant.

(a)

The expectation value of potential energy is,

$$\begin{gathered} PE={∫}_0^{∞}-\frac{e^2}{4\mathrm{Ï€}{\mathrm{Î\mu }}_{0}r}P\left(r\right)dr \\ =-\frac{e^2}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{4}{a_0^3}{∫}_0^{∞}r{e}^{-2rl{a}_0}dr \end{gathered}$$

Substitute x for$\frac{2r}{a_0}$in the integration.

$$\begin{gathered} PE=-\frac{e^2}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{1}{a}{∫}_0^{∞}x{e}^{x}dr \\ =-\frac{e^2}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{1}{a_0} \end{gathered}$$

Therefore, the expectation value of potential energy is$-\frac{e^2}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{1}{a_0}$ .

(b)

The total energy isrole="math" localid="1659471003767" $\left(-\frac{1}{2}\frac{e^2}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{1}{a_0}\right)$

The kinetic energy can be calculated by subtracting potential energy from the total energy

KE= E - PE

for $E=-\frac{1}{2}\frac{e^2}{4\mathrm{Ï€}{r}_0}\frac{1}{a_0}$and$PE=-\frac{e^2}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{1}{a_0}$, we have-
$$\begin{gathered} KE=\left(-\frac{1}{2}\frac{e^2}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{1}{a_0}\right)-\left(-\frac{e^2}{4\mathrm{Ï€}{r}_0}\frac{1}{a_0}\right) \\ =\frac{1}{2}\frac{e^2}{4{\mathrm{Ï€Î\mu }}_0}\frac{1}{a_0} \end{gathered}$$

Therefore, the expectation value of kinetic energy is$\frac{1}{2}\frac{e^2}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{1}{a_0}$.