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 $Sn^{119}$
• B. thrice that of $Sn^{119}$
• C. same as that of $Sn^{119}$
• D. half that of $Sn^{119}$

Answer 1

Answer: (C)

Given, nuclear radius is
$R= r_0A^{\frac{1}{3}}$
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} $
$= \frac{A}{\frac{4}{3} \pi (r_0A^{\frac{1}{3}})^3}$
$\Rightarrow d = \frac{A}{\frac{4}{3}\pi r_0^3 \cdot A}$
$ = \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 $Sn^{119}$