An unknown nucleus has a nuclear density of $2.29 imes 10^{17} \mathrm{~kg} / \mathrm{m}^3$ and mass of $19.926 imes 10^{-27} \mathrm{~kg}$. Its mass number $A$ is approximately:

(Take $R_0=1.2 imes 10^{-15} \mathrm{~m}, 4 \pi=12.56$ )

Question

An unknown nucleus has a nuclear density of $2.29 	imes 10^{17} \mathrm{~kg} / \mathrm{m}^3$ and mass of $19.926 	imes 10^{-27} \mathrm{~kg}$. Its mass number $A$ is approximately:

(Take $R_0=1.2 	imes 10^{-15} \mathrm{~m}, 4 \pi=12.56$ )

Accepted Answer

A. 1. Treating the nucleus as a sphere, its **density** is defined as mass divided by volume, and the nuclear radius follows the standard formula $R=R_0A^{1/3}$.

$$
ho=\frac{M}{\frac{4}{3}\pi R^3}
$$

2. Substituting $R=R_0A^{1/3}$ into the volume term, the $A^{1/3}$ becomes $A$ after cubing, giving a direct relation between density, mass, and mass number.

$$
\frac{4}{3}\pi R_0^3 A=\frac{M}{ho}
$$

3. Rearranging to isolate $A$ and substituting all given numerical values.

$$
A=\frac{M}{ho}	imes\frac{3}{4\pi R_0^3}=\frac{19.926	imes10^{-27}	imes3}{2.29	imes10^{17}	imes12.56	imes(1.2	imes10^{-15})^3}
$$

4. Evaluating this expression numerically.

$$
A\approx12
$$

Hence, the correct answer is **Option A**.