Q151 / 170 · Manipal (MET) 2015 · Physics
Order of magnitude of density of uranium nuclus is ( $m_p=1.6 \times 10^{-27} \mathrm{~kg}$ )
- A$10^{20} \mathrm{~kg} / \mathrm{m}^3$
- Correct answerB$10^{17} \mathrm{~kg} / \mathrm{m}^3$
- C$10^{14} \mathrm{~kg} / \mathrm{m}^3$
- D$10^{11} \mathrm{~kg} / \mathrm{m}^3$
Solution
Radius of a nucleus is given by
$\begin{aligned}
R & =R_0 A^{1 / 3}\left(\text { where, } R_0=1.25 \times 10^{-15} \mathrm{~m}\right) \\
& =1.25 \times 10^{-15} A^{1 / 3} \mathrm{~m}
\end{aligned}$
Here, $A$ is the mass number and mass of the uranium nucleus will be
$\begin{aligned} m & \approx A m_p, m_p=\text { mass of proton } \\ & =A\left(1.67 \times 10^{-27} \mathrm{~kg}\right)\end{aligned}$
$\therefore \quad$ Density $=\frac{\text { Mass }}{\text { Volume }}=\frac{m}{\frac{4}{3} \pi R^3}$
$=\frac{A\left(1.67 \times 10^{-27}\right)}{A \times \frac{4}{3} \times \frac{22}{7} \times\left(1.25 \times 10^{-15}\right)^3}$
$\Rightarrow \quad \rho=2 \times 10^{17} \mathrm{~kg} / \mathrm{m}^3$
$\therefore$ order is $10^{17}$
Hence, the correct answer is option: 'B'.