The radius of a nucleus of mass number \(A\) is \(R=R_0A^{1/3}\), where \(R_0=1.2\ \mathrm{fm}\). The number of nucleons in each cubic femtometre of a nucleus is about
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(0.14\ \mathrm{fm^{-3}}\)
The volume of a nucleus is \( \frac{4}{3}\pi R^{3}=\frac{4}{3}\pi R_0^{3}A \). So the volume grows in step with \( A \), and each nucleon takes up the same space. The number of nucleons in unit volume is \( A/V \).
\( \begin{aligned} n &= \frac{A}{V} \\ &= \frac{A}{\frac{4}{3}\pi R_0^{3}A} \\ &= \frac{3}{4\times3.14\times(1.2)^{3}} \\ &= 0.14\ \mathrm{fm^{-3}} \end{aligned} \)
From the above, \( A \) does not appear in the result. Hence, every nucleus has about 0.14 nucleons in each cubic femtometre.
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. …”