Question

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

• A. $10^{2}$ N
• B. $10^7$ N
• C. $10^{12}$ N
• D. $10^{17}$ N

Answer 1

Answer: (A)

Taking protons at dimetrically opposite points of nucleus,

Force of electrostatic repulsion,
$F=\frac{k(e)(e)}{r^{2}}=\frac{k e^{2}}{\left(r_{0} A^{1 / 3}\right)^{2}}=\frac{k \cdot e^{2}}{r_{0}^{2} \cdot A^{2 / 3}}$
Substituting values in above equation, we get
$F=\frac{9 \times 10^{9} \times\left(1.6 \times 10^{-19}\right)^{2}}{\left(13 \times 10^{-15}\right)^{2}(206)^{2 / 3}}$
$=0.039 \times 10^{2}$N