Physics - Modern Physics Question with Solution | TestHub

The nuclear density $\rho$ is given by the ratio of the mass of the nucleus to its volume:

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

The radius of the nucleus is $R=R_0{A}^{1/3}$, which gives $R^3=R_0^{3}A$.

Substituting this into the density formula:

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

Rearranging for the mass number $A$:

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

Substituting the given values:

$A=\frac{3\times 19.926\times 10^{-27}}{12.56\times (1.2\times 10^{-15}{)}^3\times 2.29\times 10^{17}}$

$A=\frac{59.778\times 10^{-27}}{12.56\times 1.728\times 10^{-45}\times 2.29\times 10^{17}}$

$A=\frac{59.778\times 10^{-27}}{49.7\times 10^{-28}}$

$A=\frac{597.78}{49.7}\approx 12$

Answer: $12$