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