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: (C)

${10}^{12}N$

Radius of a nucleus $=r_0{A}^{1/3}$

$A=$ number of the nucleus

$A=206$ (for lead)

the electrostatic force between two protons. In this nucleus $=?$

Radius of a nucleus of mass number $A$ is given by $R=R_0{A}^{1/3}$

where $R_0=1.2\times {10}^{-15}m$

This implies that the volume of the nucleus, which in proportional to $R^3$ is proportional to $A$.

Volume of nucleus $=\frac{4}{3}\pi r^3$

$=\frac{4}{3}\pi {\left(R_0{A}^{1/3}\right)}^3$

$=\frac{4}{3}\pi R_0^{3}A$

Density of nucleus $=\frac{mass}{volume}$

$=\frac{ma}{\frac{4}{3}\pi R_0^{3}A}=\frac{3m}{4\pi R_0^3}=3\times {10}^{12}N$