Question

A neutral lithium atom has three electrons. As you will discover in Chapter 41 , two of these electrons form an "inner core," but the third-the valence electron-orbits at much larger radius. From the valence electron's perspective, it is orbiting a spherical ball of charge having net charge $+1e$ (i.e., the three protons in the nucleus and the two inner-core electrons). The energy required to ionize a lithium atom is$5.14\mathrm{eV}$. According to Rutherford's nuclear model of the atom, what are the orbital radius and speed of the valence electron?

Short answer

The orbital radius of the valence electron $r=1.4·{10}^{-10}\mathrm{m}$

The speed of the valence electron$v=1.34·{10}^6\frac{\mathrm{m}}{\mathrm{s}}$

Step-by-step solution

Given information

A spherical ball of charge having net charge $+1e$

The energy required to ionize a lithium atom is$5.14\mathrm{eV}$

Calculation

We can start solution with electron 's energy

$E=-\frac{e^2}{4\pi {ϵ}_{0}r}+\frac{1}{2}m{v}^2$

Electron has circular motion which follows from equilibrium between Coulomb force and centripetal force

$$\begin{gathered} F_C=F_{cp} \\ \frac{e^2}{4\pi {ϵ}_0{r}^2}=\frac{m{v}^2}{r} \\ \frac{e^2}{4\pi {ϵ}_0·2r}=\frac{m{v}^2}{2} \\ E=-\frac{e^2}{4\pi {ϵ}_{0}r}+\frac{e^2}{4\pi {ϵ}_0·2r} \\ E=-\frac{e^2}{8\pi {ϵ}_{0}r} \\ r=-\frac{e^2}{8\pi {ϵ}_{0}E} \\ r=-\frac{{\left(1.6·{10}^{-19}\right)}^2}{8\pi {ϵ}_0·\left(-5.14·1.6·{10}^{-19}\right)} \\ r=1.4·{10}^{-10}\mathrm{m} \\ v=\sqrt{\frac{e^2}{4\pi {ϵ}_{0}rm}} \\ v=\sqrt{\frac{-2E}{m}} \\ v=\sqrt{\frac{-2·\left(-5.14·1.6·{10}^{-19}\right)}{9.11·{10}^{-31}}} \\ v=1.34·{10}^6\frac{m}{s} \end{gathered}$$