The radius of a nucleus is given by $\mathrm{r}_{0} \mathrm{~A}^{1 / 3}$ where $\mathrm{r}_{0}=1.3 \times 10^{-15} \mathrm{~m}$ and $\mathrm{A}$ is the mass number of the nucleus, the Lead nucleus has $\mathrm{A}=206$. the electrostatic force between two protons in this nucleus is approximately

• Correct answer
  A

  $10^{2} \mathrm{~N}$
• B

  $10^{7} \mathrm{~N}$
• C

  $10^{12} \mathrm{~N}$
• D

  $10^{17} \mathrm{~N}$

Solution

$\begin{aligned} \mathrm{r}=\mathrm{r}_{0} \mathrm{~A}^{1 / 3} \quad & \mathrm{r}_{0}=1.3 \times 10^{-15} \\ & \mathrm{~F}=\frac{1}{4 \pi \in_{0}} \times \frac{\mathrm{q}_{1} \mathrm{q}_{2}}{\mathrm{r}^{2}} \end{aligned}$
$F=\frac{9 \times 10^{9} \times 1.6 \times 10^{-19} \times 1.6 \times 10^{-19}}{\mathrm{r}_{0}^{2} \mathrm{~A}^{2 / 3}}$ $\mathrm{~F}=\frac{9 \times 10^{9} \times 1.6 \times 10^{-19} \times 1.6 \times 10^{-19}}{(1.3)^{2} \times 10^{-30} \times(206)^{2 / 3}}$ $=\frac{23.04 \times 10^{39} \times 10^{-38}}{(1.69) \times 34.81}$ $=\frac{23.04 \times 10}{1.69 \times 34.81}$ $=3.91 \mathrm{Newton}$ $=0.039 \times 10^{2} \mathrm{Ans} .(\mathrm{A})$

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