91Ó°ÊÓ

We can start solution by determining speed from equilibrium between Coulomb and centripetal force :

$\begin{array}{rl}{F}_{cp}=F_q& \\ \frac{m_e{v}^2}{r}=\frac{1}{4\mathrm{Ï€}{\mathrm{Ï\mu }}_0}\frac{q_1{q}_2}{r^2}\text{(substitute expressions for forces)}& \\ \frac{m_e{v}^2}{r}=\frac{1}{4\mathrm{Ï€}{\mathrm{Ï\mu }}_0}\frac{e^2}{r^2}\text{(substitute electron and proton charge)}& \\ v=\sqrt{\frac{e^2}{4\mathrm{Ï€}{\mathrm{Ï\mu }}_{0}r{m}_e}}& \text{(express}v)\end{array}$

Now we can use expression for speed in circular motion with radius $r$and period $T$

localid="1650782065477" $$\begin{gathered} v=\frac{2r\mathrm{Ï€}}{T} \\ v=2r\mathrm{Ï€}f \\ r=\frac{v}{2\mathrm{Ï€}f} \\ c=\mathrm{λ}f \\ f=\frac{c}{\mathrm{λ}} \\ \text{(express}f\text{)} \\ r=\frac{\sqrt{\frac{e^2}{4\mathrm{Ï€}{\mathrm{Ï\mu }}_{0}r{m}_e}}}{2\mathrm{Ï€}\frac{c}{\mathrm{λ}}} \\ {r}^3=\frac{e^2{\mathrm{λ}}^2}{4{\mathrm{Ï€}}^3·4{\mathrm{Ï\mu }}_0{m}_e{c}^2} \\ \text{(substitute}v\text{and}f\text{)} \\ \text{(square and edit)} \\ r={\left(\frac{{\left(1.6·{10}^{-19}\right)}^2·{\left(600·{10}^{-9}\right)}^2}{4·{\mathrm{Ï€}}^3·4{\mathrm{Ï\mu }}_0·9.11·{10}^{-31}·{\left(3·{10}^8\right)}^2}\right)}^{\frac{1}{3}} \\ r=2.95·{10}^{-10}\mathrm{m}=0.295\mathrm{nm} \\ \left(\text{express}r\right) \end{gathered}$$

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Now we must determine total energy of the electron:

$m_e{v}^2=\frac{1}{4\mathrm{Ï€}{\mathrm{Ï\mu }}_0}\frac{e^2}{r}$express $m_e{v}^2$from beginning of part (a)) $E_{net}=E_k+U$$E_{net}-\frac{m_e{v}^2}{2}-\frac{1}{4\mathrm{Ï€}{\mathrm{Ï\mu }}_0}\frac{e^2}{r}$(expressions for kinetic and potential energy) $E_{\text{net}}=\frac{1}{4\mathrm{Ï€}{\mathrm{Ï\mu }}_0}\frac{e^2}{2r}-\frac{1}{4\mathrm{Ï€}{\mathrm{Ï\mu }}_0}\frac{e^2}{r}$(substitute $\left.m_e{v}^2\right)$$E_{net}=-\frac{1}{8\mathrm{Ï€}{\mathrm{Ï\mu }}_0}\frac{e^2}{r}$$E_{\text{net}}=-\frac{1}{8\mathrm{Ï€}{\mathrm{Ï\mu }}_0}\frac{{\left(1.6·{10}^{-19}\right)}^2}{2.95·{10}^{-10}}$ $E_{net}=-3.91·{10}^{-19}\mathrm{J}=-2.44\mathrm{eV}$