Question

According to Rutherford's model, what will be the kinetic energy of an electron of mass $m_e$ at a distance r from the center of a $L{i}^{++}$ ion? (The Lithium nucleus has three protons.)

• A. $\frac{3e^2}{8\pi {ϵ}_{0}r}$
• B. $\frac{3e^2}{8\pi {ϵ}_0{r}^2}$
• C. $\frac{3e^2}{4\pi {ϵ}_{0}r}$
• D. $\frac{3e^2}{4\pi {ϵ}_0{r}^2}$

Answer 1

The correct option is A

$\frac{3e^2}{8\pi {ϵ}_{0}r}$

The atomic number of Lithium is Z=3, which means it has three protons in the nucleus, having a total positive charge of (+Ze). An electron's charge is (-e). Since the centripetal acceleration is provided by the electrostatic force,
we can write - $\frac{m_e{V}^2}{r}=\frac{Z{e}^2}{4\pi {ϵ}_0{r}^2}\Rightarrow m_e{v}^2=\frac{Z{e}^2}{4\pi {ϵ}_{0}r}\Rightarrow \frac{m_e{v}^2}{2}=\text{kinetic energy of the electron}=\frac{Z{e}^2}{8\pi {ϵ}_{0}r}=\frac{3e^2}{8\pi {ϵ}_{0}r}.$