First slide

Size of nucleus

Question

Radius of ${}_{2}^{4}{He}$ nucleus is 3 Fermi. The radius of ${}_{82}^{206}{Pb}$ nucleus will be

Moderate

A

5 Fermi
B

6 Fermi
C

11.6 Fermi
D

8 Fermi

Solution

$\begin{array}{l} We have, r \propto A^{1 / 3} \Rightarrow \frac{r_{2}}{r_{1}} = {(\frac{A_{2}}{A_{1}})}^{1 / 3} = {(\frac{206}{4})}^{1 / 3} \\ ∴ r_{2} = 3 {(\frac{206}{4})}^{1 / 3} = 11 .6 Fermi . \end{array}$

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