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 is approximately

• A. $10$
• B. $0.98$
• C. $2N$
• D. $17N$

Answer 1

The correct option is B $0.98$

$R=r_{\circ }{A}^{\frac{1}{3}}$

For lead A=206 and radius R of the nucleus is given by,

$R=1.3\times {10}^{-13}\times {206}^{\frac{1}{3}}=7.68\times {10}^{-13}$

Now, if we consider that proton are present at diometrical end of the nucleus than the coloumbic force, between them is given by,

$F=\frac{9\times {10}^3\times {(1.6\times {10}^{-15})}^2}{{(7.68\times {10}^{-15}\times 2)}^2}=\frac{2.304\times {10}^{-28}}{2.359\times {10}^{-28}}F\simeq 0.976$