Order of magnitude of density of uranium nucleus is $\left(m_p=1.67 \times 10^{-27} kg\right)$
Answer & explanation
Correct answer: option 2
The order of magnitude of mass and volume of uranium nucleus will be $m ≃ A\left(1.67 × 10^{-27} kg\right)$
$V=\frac{4}{3} \pi r^3 \simeq \frac{4}{3} \pi\left[\left(1.25 \times 10^{-15} \mathrm{~m}\right) A^{1 / 3}\right]^3 \simeq\left(8.2 \times 10^{-45} \mathrm{~m}^3\right) \mathrm{A}$
Hence, $\rho=\frac{m}{V}=\frac{A\left(1.67 \times 10^{-27} \mathrm{~kg}\right)}{\left(8.2 \times 10^{-45} \mathrm{~m}^3\right) \mathrm{A}} \simeq 2.0 \times 10^{17} \mathrm{~kg} / \mathrm{m}^3$.