# The radius R of a nucleus of mass number A can be estimated

Question:

The radius $R$ of a nucleus of mass number $A$ can be estimated by the formula $R=\left(1.3 \times 10^{-15}\right) A^{1 / 3} \mathrm{~m}$. It follows that the mass density of a nucleus is of the order of :

$\left(M_{\text {prot. }} \cong M_{\text {neut. }} \simeq 1.67 \times 10^{-27} \mathrm{~kg}\right)$

1. (1) $10^{3} \mathrm{~kg} \mathrm{~m}^{-3}$

2. (2) $10^{10} \mathrm{~kg} \mathrm{~m}^{-3}$

3. (3) $10^{24} \mathrm{~kg} \mathrm{~m}^{-3}$

4. (4) $10^{17} \mathrm{~kg} \mathrm{~m}^{-3}$

Correct Option: , 4

Solution:

(4) Density of nucleus, $\rho=\frac{\text { Mass }}{\text { Volume }}=\frac{m A}{\frac{4}{3} \pi R^{3}}$

$\Rightarrow \rho=\frac{m A}{\frac{4}{3} \pi\left(R_{0} A^{1 / 3}\right)^{3}} \quad\left(\because R=R_{0} A^{1 / 3}\right)$

Here $m=$ mass of a nucleon

$\therefore \rho=\frac{3 \times 1.67 \times 10^{-27}}{4 \times 3.14 \times\left(1.3 \times 10^{-15}\right)^{3}}$ (Given, $R_{0}=1.3 \times 10^{-15}$ )

$\Rightarrow \rho=2.38 \times 10^{17} \mathrm{~kg} / \mathrm{m}^{3}$