To calculate the size of a hydrogen anion using the Bohr model, we assume that its two electrons move in an orbit such that they are always on diametrically opposite sides of the nucleus. With each electron having the angular momentum $\hbar=\mathrm{h} / 2 \pi$, and taking electron interaction into account the radius of the orbit in terms of the Bohr radius of hydrogen atom $\mathrm{a}_{\mathrm{B}}=\frac{4 \pi \varepsilon_{0} \mathrm{~h}^{2}}{\mathrm{me}^{2}}$ is

• A

  $a_{B}$
• Correct answer
  B

  $\frac{4}{3} a_{B}$
• C

  $\frac{2}{3} a_{B}$
• D

  $\frac{3}{2} \mathrm{a}_{\mathrm{B}}$

Solution

$\begin{array}{l} \mathrm{mvr}=\mathrm{n} \frac{\mathrm{h}}{2 \pi} \\ \& \frac{\mathrm{mv}^{2}}{\mathrm{r}}=\frac{\mathrm{ke}^{2}}{\mathrm{r}^{2}}-\frac{\mathrm{ke}^{2}}{(2 \mathrm{r})^{2}} \\ \frac{\mathrm{mv}^{2}}{\mathrm{r}}=\frac{3}{4} \frac{\mathrm{ke}^{2}}{\mathrm{r}^{2}} \\ \Rightarrow \mathrm{mv}^{2} \mathrm{r}=\frac{3}{4} \mathrm{ke}^{2} \end{array}$ Solving $(1)+(2)$ $\begin{array}{l} \mathrm{r}=\frac{4 \pi \varepsilon_{0} \mathrm{~h}^{2}}{\mathrm{me}^{2}} \times \frac{4}{3} \\ \mathrm{r}=\frac{4}{3} \mathrm{a}_{\mathrm{B}} \end{array}$

Hence, the correct answer is option: 'B'.