The order of magnitude of the density of a uranium nucleus is \( \left(m_p=1.67\times10^{-27}\,\mathrm{kg}\right) \)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 10^{17}\,\mathrm{kg\,m^{-3}} \)
The radius of a nucleus of mass number \( A \) is \( R=R_0A^{1/3} \), with \( R_0=1.2\times10^{-15}\,\mathrm{m} \). So its volume is proportional to \( A \). Its mass is about \( A\,m_{\mathrm{p}} \), because a neutron has nearly the mass of a proton. Therefore, the density of nuclear matter is the same for all nuclei, uranium included.
\( \begin{aligned} \rho &= \frac{A\,m_{\mathrm{p}}}{\frac{4}{3}\pi R_0^{3}A} \\ &= \frac{3m_{\mathrm{p}}}{4\pi R_0^{3}} \\ &= \frac{3\times1.67\times10^{-27}\,\mathrm{kg}}{4\times3.14\times(1.2\times10^{-15}\,\mathrm{m})^{3}} \\ &\approx 2.3\times10^{17}\,\mathrm{kg\,m^{-3}} \end{aligned} \)
So the order of magnitude of the density is \( 10^{17}\,\mathrm{kg\,m^{-3}} \).
NCERT (2026-27):
Size of the Nucleus – “It has been found that a nucleus of mass number A has a radius
\( R=R_0A^{1/3} \qquad (13.5) \)
where R0 = 1.2 × 10–15 m (=1.2 fm; 1 fm = 10–15 m). This means the volume of the nucleus, which is proportional to R3 is proportional to A. Thus the density of nucleus is a constant, independent of A, for all nuclei. Different nuclei are like a drop of liquid of constant density. The density of nuclear matter is approximately 2.3 × 1017 kg m–3.”