Question

Question: Consider an electron in the ground state of a hydrogen atom. (a) Calculate the expectation value of its potential energy. (b) What is the expectation value of its kinetic energy? (Hint: What is the expectation value of the total energy?)

Short answer

a) The expectation value of potential energy in the ground state of the hydrogen atom is$-\frac{e^2}{4{\mathrm{\pi \epsilon }}_{\circ }}\frac{1}{a_{\circ }}$.

b) The expectation value of kinetic energy in the ground state of the hydrogen atom is $\frac{1}{2}\frac{e^2}{4{\mathrm{\pi \epsilon }}_{\circ }}\frac{1}{a_{\circ }}$.

Step-by-step solution

 Given data

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

 Concept

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

Solution

(a)

The expectation value of potential energy is,

$$\begin{gathered} PE={\int }_0^{\infty }-\frac{e^2}{4{\mathrm{\pi \epsilon }}_{\circ }\mathrm{r}}P\left(r\right)dr \\ =-\frac{e^2}{4{\mathrm{\pi \epsilon }}_{\circ }\mathrm{r}}\frac{4}{{a_{\circ }}^3}{\int }_0^{\infty }r{e}^{-2r/a_{\circ }}dr \end{gathered}$$

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

role="math" localid="1659615862149" $$\begin{gathered} PE=-\frac{e^2}{4{\mathrm{\pi \epsilon }}_{\circ }}\frac{1}{a_{\circ }}{\int }_0^{\infty }x{e}^{x}dr \\ =-\frac{e^2}{4{\mathrm{\pi \epsilon }}_{\circ }}\frac{1}{a_{\circ }} \end{gathered}$$

Therefore, the expectation value of potential energy is $-\frac{e^2}{4{\mathrm{\pi \epsilon }}_{\circ }}\frac{1}{a_{\circ }}$.

(b)

The total energy is $\left(-\frac{1}{2}\frac{e^2}{4{\mathrm{\pi \epsilon }}_{\circ }}\frac{1}{a_{\circ }}\right)$

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

KE= E - PE

for role="math" localid="1659615925271" $E=-\frac{1}{2}\frac{e^2}{4{\mathrm{\pi \epsilon }}_{\circ }}\frac{1}{a_{\circ }}$androle="math" localid="1659615868874" $PE=-\frac{e^2}{4{\mathrm{\pi \epsilon }}_{\circ }}\frac{1}{a_{\circ }}$, we have-
role="math" localid="1659615966673" $\begin{array}{rcl}KE& =& \left(-\frac{1}{2}\frac{e^2}{4{\mathrm{\pi \epsilon }}_{\circ }}\frac{1}{a_{\circ }}\right)-\left(-\frac{e^2}{4{\mathrm{\pi \epsilon }}_{\circ }}\frac{1}{a_{\circ }}\right)\\ & =& \frac{1}{2}\frac{e^2}{4{\mathrm{\pi \epsilon }}_{\circ }}\frac{1}{a_{\circ }}\end{array}$

Therefore, the expectation value of kinetic energy is $\begin{array}{rcl}& & \frac{1}{2}\frac{e^2}{4{\mathrm{\pi \epsilon }}_{\circ }}\frac{1}{a_{\circ }}\end{array}$.