Question

The nuclear radius is given by $R=r_0{A}^{1/3}$, where $r_0$ is constant and A is the atomic mass number. Then, the nuclear mass density of $U^{238}$ is

• A. twice that of $S{n}^{119}$
• B. thrice that of $S{n}^{119}$
• C. same as that of $S{n}^{119}$
• D. half that of $S{n}^{119}$

Answer 1

Answer: (C)

Given, nuclear radius is

$R=r_0{A}^{\frac{1}{3}}$

Here, atomic mass number of nucleus $=A$

$\therefore$ Nuclear density $d$ is given by

$d=\frac{\text{Mass number}}{\text{Volume}}$

$\Rightarrow d=\frac{A}{\frac{4}{3}\pi R^3}$

$=\frac{A}{\frac{4}{3}\pi (r_0{A}^{\frac{1}{3}}{)}^3}$

$\Rightarrow d=\frac{A}{\frac{4}{3}\pi r_0^3\cdot A}$

$=\frac{3}{4\pi r_0^3}$

As $r_0=a$ constant, so nuclear density is a constant quantity.

$\therefore$ Nuclear mass density of $U^{238}$ is same as that of $S{n}^{119}$