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 Li⁺⁺ 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 -
$\frac{m_e{v}^2}{r}=\frac{Z{e}^2}{4\pi {ϵ}_0{r}^2}$
$m_e{v}^2=\frac{Z{e}^2}{4\pi {ϵ}_{0}r}$
$\Rightarrow \frac{1}{2}{m}_e{v}^2$= kinetic energy of the electron = $\frac{Z{e}^2}{8\pi {ϵ}_{0}r}=\frac{3e^2}{8\pi {ϵ}_{0}r}$.