Question

The radius of the $2H{e}^4$ nucleus is 3 fermi. The radius of ${}_{82}P{b}^{206}$ nucleus will be

• A. 5 fermi
• B. 6 fermi
• C. 11.16 fermi
• D. 8 fermi

Answer 1

Answer: (C)

We have, $r\propto A^{1/3}$ $\Rightarrow \frac{r_2}{r_1}={\left[\frac{A_2}{A_1}\right]}^{1/3}={\left[\frac{206}{4}\right]}^{1/3}$ $\therefore r_2=3{\left[\frac{206}{4}\right]}^{1/3}=11.16$ fermi