Question

A classical atom orbiting at frequency f ould emit electromagnetic waves of frequency f because the electron's orbit, seen edge-on, looks like an oscillating electric dipole.

a. At what radius, in $\mathrm{nm}$, would the electron orbiting the proton in a hydrogen atom emit light with a wavelength of $600\mathrm{nm}$?

b. What is the total mechanical energy of this atom?

Short answer

$$\begin{gathered} a)r=2.95·{10}^{-10}\mathrm{m}=0.295\mathrm{nm} \\ b)E_{\text{net}}=-3.91·{10}^{-19}\mathrm{J}=-2.44\mathrm{eV} \end{gathered}$$

Step-by-step solution

Part (a) Step 1:Soluction

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

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

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

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

:

Part (b) Step 2: solution

Now we must determine total energy of the electron:

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