91Ó°ÊÓ

We can begin the solution with the pulse's total energy.

We can begin by calculating speed using the equilibrium between Coulomb and centripetal force as a starting point.

$$\begin{gathered} {\mathrm{F}}_{\mathrm{cp}}={\mathrm{F}}_{\mathrm{q}} \\ \frac{{\mathrm{m}}_{\mathrm{e}}{\mathrm{v}}^2}{\mathrm{r}}=\frac{1}{4{\mathrm{Ï€Ï\mu }}_0}\frac{{\mathrm{q}}_1{\mathrm{q}}_2}{{\mathrm{r}}^2}\text{(substitute expressions for forces)} \\ \frac{{\mathrm{m}}_{\mathrm{e}}{\mathrm{v}}^2}{\mathrm{r}}=\frac{1}{4{\mathrm{Ï€Ï\mu }}_0}\frac{{\mathrm{e}}^2}{{\mathrm{r}}^2} \\ \mathrm{v}=\sqrt{\frac{{\mathrm{e}}^2}{4{\mathrm{Ï€Ï\mu }}_0{\mathrm{rm}}_{\mathrm{e}}}}\text{(substitute electron and proton charge)} \end{gathered}$$

We can now utilise the expression for circular motion speed with radius r and period T.

$$\begin{gathered} \mathrm{v}=\frac{2\mathrm{°ùÏ€}}{\mathrm{T}} \\ \mathrm{v}=2\mathrm{°ùÏ€´Ú}\text{(definition of frequency)} \\ \mathrm{r}=\frac{\mathrm{v}}{2\mathrm{Ï€´Ú}}\text{(express}\mathrm{r}\text{)} \\ \mathrm{c}=\mathrm{λ´Ú}\begin{array}{r}\text{(expression for speed}\end{array} \\ \begin{array}{rrr}\mathrm{f}& =\frac{\mathrm{c}}{\mathrm{λ}}& \text{(express}\mathrm{f})\end{array} \\ \mathrm{r}=\frac{\sqrt{\frac{{\mathrm{Ï\mu }}^2}{4{\mathrm{Ï€Ï\mu }}_0{\mathrm{rm}}_{\mathrm{e}}}}}{2\mathrm{Ï€}\frac{\mathrm{c}}{\mathrm{λ}}}\text{(substitute}\mathrm{v}\text{and}\mathrm{f}\text{)} \\ {\mathrm{r}}^3=\frac{{\mathrm{e}}^2{\mathrm{λ}}^2}{4{\mathrm{Ï€}}^3â‹\dots 4{\mathrm{Ï\mu }}_0{\mathrm{m}}_{\mathrm{e}}{\mathrm{c}}^2}(\mathrm{square}\mathrm{and}\mathrm{edit}) \\ \begin{array}{r}\mathrm{r}={\left(\frac{{\left(1.6â‹\dots {10}^{−19}\right)}^2â‹\dots {\left(600â‹\dots {10}^{−9}\right)}^2}{4â‹\dots {\mathrm{Ï€}}^3â‹\dots 4{\mathrm{Ï\mu }}_0â‹\dots 9.11â‹\dots {10}^{−31}â‹\dots {\left(3â‹\dots {10}^8\right)}^2}\right)}^{\frac{1}{3}}\\ \text{(express}\mathrm{r}\text{)}\end{array} \\ \mathrm{r}=2.95â‹\dots {10}^{−10}\mathrm{m}=0.295\mathrm{nm} \end{gathered}$$

The total mechanical energy of this atom$E_{net}=−3.91â‹\dots {10}^{−19}\mathrm{J}=−2.44\mathrm{eV}$

The total energy of the electron must now be determined:

$$\begin{gathered} \begin{array}{rl}{\mathrm{m}}_{\mathrm{e}}{\mathrm{v}}^2& =\frac{1}{4{\mathrm{Ï€Ï\mu }}_0}\frac{{\mathrm{e}}^2}{\mathrm{r}}\end{array} \\ \begin{array}{rl}{\mathrm{E}}_{\mathrm{net}}& ={\mathrm{E}}_{\mathrm{k}}+\mathrm{U}\end{array} \\ {\mathrm{E}}_{\text{net}}={\mathrm{E}}_{\mathrm{k}}+\mathrm{U} \\ {\mathrm{E}}_{\mathrm{net}}=\frac{{\mathrm{m}}_{\mathrm{e}}{\mathrm{v}}^2}{2}−\frac{1}{4{\mathrm{Ï€Ï\mu }}_0}\frac{{\mathrm{e}}^2}{\mathrm{r}}\text{(expressions for kinetic and potential energy)} \\ {\mathrm{E}}_{\mathrm{net}}=\frac{1}{4{\mathrm{Ï€Ï\mu }}_0}\frac{{\mathrm{e}}^2}{2\mathrm{r}}−\frac{1}{4{\mathrm{Ï€Ï\mu }}_0}\frac{{\mathrm{e}}^2}{\mathrm{r}}\left(\text{substitute}{\mathrm{m}}_{\mathrm{e}}{\mathrm{v}}^2\right) \\ {\mathrm{E}}_{\mathrm{net}}=−\frac{1}{8{\mathrm{Ï€Ï\mu }}_0}\frac{{\mathrm{e}}^2}{\mathrm{r}}\text{(edit)} \\ {\mathrm{E}}_{\mathrm{net}}=−\frac{1}{8{\mathrm{Ï€Ï\mu }}_0}\frac{{\left(1.6â‹\dots {10}^{−19}\right)}^2}{2.95â‹\dots {10}^{−10}}\left(\mathrm{substitute}\right) \\ {\mathrm{E}}_{\text{net}}=−3.91â‹\dots {10}^{−19}\mathrm{J}=−2.44\mathrm{eV} \end{gathered}$$