Question

Physicists first attempted to understand the hydrogen atom by applying the laws of classical physics. Consider an electron of mass m and charge -e in a circular orbit of radius r around a proton of charge +e.

a. Use Newtonian physics to show that the total energy of the atom is $\mathrm{E}=-\frac{{\mathrm{e}}^2}{8{\mathrm{\pi ϵ}}_0\mathrm{r}}$

b. Show that the potential energy is -2 times the electron’s kinetic energy. This result is called the virial theorem.

c. The minimum energy needed to ionize a hydrogen atom (i.e., to remove the electron) is found experimentally to be 13.6 eV. From this information, what are the electron’s speed and the radius of its orbit?

Short answer

(a) The total energy of the atom is ${\mathrm{E}}_{\mathrm{net}}=-\frac{1}{8{\mathrm{\pi ϵ}}_0}\frac{{\mathrm{e}}^2}{\mathrm{r}}$

(b) The Resultant equation is $\frac{\mathrm{U}}{\mathrm{K}}=-2$

(c) The electron’s speed and the radius of its orbit is$\mathrm{v}=2.19.{10}^6\frac{\mathrm{m}}{\mathrm{s}}$

Step-by-step solution

Part (a) Step 1: Given information

The total energy of the atom is $\mathrm{E}=-\frac{{\mathrm{e}}^2}{8{\mathrm{\pi \epsilon }}_0\mathrm{r}}$

Part (a) Step 2: Determine the electron's total energy

We can begin by establishing a solution equilibrium between the centripetal and Coulomb forces$\begin{array}{r}{\mathrm{E}}_{\mathrm{net}}={\mathrm{E}}_{\mathrm{k}}+\mathrm{U}\\ {\mathrm{E}}_{\mathrm{net}}=\frac{{\mathrm{m}}_{\mathrm{e}}{\mathrm{v}}^2}{2}-\frac{1}{4{\mathrm{\pi ϵ}}_0}\frac{{\mathrm{e}}^2}{\mathrm{r}}\text{(expressions for kinetic and potential energy)}\\ {\mathrm{E}}_{\mathrm{net}}=\frac{1}{4{\mathrm{\pi ϵ}}_0}\frac{{\mathrm{e}}^2}{2\mathrm{r}}-\frac{1}{4{\mathrm{\pi ϵ}}_0}\frac{{\mathrm{e}}^2}{\mathrm{r}}\text{(substitute}{\mathrm{m}}_{\mathrm{e}}{\mathrm{v}}^2\text{)}\\ {\mathrm{E}}_{\text{net}}=-\frac{1}{8{\mathrm{\pi ϵ}}_0}\frac{{\mathrm{e}}^2}{\mathrm{r}}\end{array}$

The total energy of the electron must now be determined.

$$\begin{gathered} {\mathrm{E}}_{\mathrm{net}}={\mathrm{E}}_{\mathrm{k}}+\mathrm{U} \\ {\mathrm{E}}_{\text{net}}=\frac{{\mathrm{m}}_{\mathrm{e}}{\mathrm{v}}^2}{2}-\frac{1}{4{\mathrm{\pi ϵ}}_0}\frac{{\mathrm{e}}^2}{\mathrm{r}}(\mathrm{expressions}\mathrm{for}\mathrm{kinetic}\mathrm{and}\mathrm{potential}\mathrm{energy}) \\ {\mathrm{E}}_{\mathrm{net}}=\frac{1}{4{\mathrm{\pi ϵ}}_0}\frac{{\mathrm{e}}^2}{2\mathrm{r}}-\frac{1}{4{\mathrm{\pi ϵ}}_0}\frac{{\mathrm{e}}^2}{\mathrm{r}} \\ {\mathrm{E}}_{\text{net}}=-\frac{1}{8{\mathrm{\pi ϵ}}_0}\frac{{\mathrm{e}}^2}{\mathrm{r}} \end{gathered}$$

Part (b) Step 2: Given information

The potential energy of an electron is -2 times its kinetic energy.

Part (b) Step 2: Finding ratio of K.E and P.E.

The ratio of potential and kinetic energy can be calculated easily:

$$\begin{gathered} \frac{\mathrm{U}}{\mathrm{K}}=\frac{-\frac{{\mathrm{e}}^2}{4{\mathrm{\pi ϵ}}_0\mathrm{r}}}{\frac{{\mathrm{m}}_{\mathrm{e}}{\mathrm{v}}^2}{2}}\text{(substitute expressions for energies)} \\ \frac{\mathrm{U}}{\mathrm{K}}=\frac{-\frac{{\mathrm{e}}^2}{4{\mathrm{\pi ϵ}}_0\mathrm{r}}}{\frac{{\mathrm{e}}^2}{2\cdot 4{\mathrm{\pi ϵ}}_0\mathrm{r}}}\left(\text{substitute}{\mathrm{m}}_{\mathrm{e}}{\mathrm{v}}^2\right) \\ \frac{\mathrm{U}}{\mathrm{K}}=-2 \end{gathered}$$

Part(c) Step 1: Given information

Ionization of a hydrogen atom requires at least 13.6 eV of energy.

Part (c) Step 2: Finding the electron’s speed and the radius of its orbit

We can begin with expression ${\mathrm{E}}_{\mathrm{ion}}=\left|{\mathrm{E}}_{\mathrm{net}}\right|$

$$\begin{gathered} {\mathrm{E}}_{\mathrm{ion}}=\frac{1}{8{\mathrm{\pi ϵ}}_0}\frac{{\mathrm{e}}^2}{\mathrm{r}}\left(\text{substitute}{\mathrm{E}}_{\mathrm{net}}\right) \\ \mathrm{r}=\frac{1}{8{\mathrm{\pi ϵ}}_0}\frac{{\mathrm{e}}^2}{{\mathrm{E}}_{\mathrm{ion}}}\left(\mathrm{express}\mathrm{r}\right) \\ \mathrm{r}=\frac{1}{8{\mathrm{\pi ϵ}}_0}\frac{{\left(1.6\cdot {10}^{-19}\right)}^2}{13.6\cdot 1.6\cdot {10}^{-19}}\left(\text{substitute}\right) \\ \mathrm{r}=5.3\cdot {10}^{-11}\mathrm{m}=5.3\mathrm{pm} \\ {\mathrm{E}}_{\text{ion}}=\frac{1}{2}\cdot \frac{1}{4{\mathrm{\pi ϵ}}_0}\frac{{\mathrm{e}}^2}{\mathrm{r}} \\ {\mathrm{E}}_{\text{ion}}=\frac{1}{2}\cdot {\mathrm{m}}_{\mathrm{e}}{\mathrm{v}}^2\left(\text{substitute}{\mathrm{m}}_{\mathrm{e}}{\mathrm{v}}^2\right) \\ \mathrm{v}=\sqrt{\frac{2{\mathrm{E}}_{\mathrm{ion}}}{{\mathrm{m}}_{\mathrm{e}}}}\left(\mathrm{express}\mathrm{v}\right) \\ \mathrm{v}=\sqrt{\frac{2\cdot 13.6\cdot 1.6\cdot {10}^{-19}}{9.11\cdot {10}^{-31}}}\text{(substitute)} \\ \mathrm{v}=2.19\cdot {10}^6\frac{\mathrm{m}}{\mathrm{s}} \end{gathered}$$