Question

The radius $R$ of a nucleus of mass number $A$ can be estimated by the formula $R=(1.3\times {10}^{-15})A^{\frac{1}{3}}\text{m}.$ It follows that the mass density of a nucleus is of the order of: $M_{\text{prot.}}\cong M_{\text{neut.}}\simeq 1.67\times {10}^{-27}\text{kg}$

• A. ${10}^3{\text{kgm}}^{-3}$
• B. ${10}^{10}{\text{kgm}}^{-3}$
• C. ${10}^{24}{\text{kgm}}^{-3}$
• D. ${10}^{17}{\text{kgm}}^{-3}$

Answer 1

The correct option is D ${10}^{17}{\text{kgm}}^{-3}$

Density of nucleus,
$\rho =\frac{\text{Mass}}{\text{Volume}}=\frac{m_A}{\frac{4}{3}\pi R^3}$

$\Rightarrow \rho =\frac{m_A}{\frac{4}{3}\pi (R_0{A}^{\frac{1}{3}}{)}^3}$

Here, $m=\text{mass of a nucleon}$

$\Rightarrow \rho =\frac{3m}{4\pi R_0^3}$

Substituting the values,

$\Rightarrow \rho =\frac{3\times 1.67\times {10}^{-27}}{4\times 3.14\times (1.3\times {10}^{-15}{)}^3}$

$\Rightarrow \rho =2.38\times {10}^{17}{\text{kgm}}^{-3}$

Hence, option $\left(D\right)$ is correct.