Two nuclei have mass number in the ratio 1:27, then the ratio of thei…

The correct answer is Option (3) → 1 : 3 and 1 : 1

The nuclear radius grows as the cube root of the mass number, \( R = R_0 A^{1/3} \), so the volume is proportional to \( A \). The mass is also proportional to \( A \), so the nuclear density is the same for all nuclei.

    \( \displaystyle \frac{R_{1}}{R_{2}} = \left(\frac{A_{1}}{A_{2}}\right)^{1/3} = \left(\frac{1}{27}\right)^{1/3} = \frac{1}{3} \)

    \( \displaystyle \rho = \frac{m A}{\frac{4}{3}\pi R_0^{3} A} = \frac{3m}{4\pi R_0^{3}} \;\Rightarrow\; \rho_{1} : \rho_{2} = 1 : 1 \)

Here \( m \) is the mass of a nucleon. So the ratios are 1 : 3 and 1 : 1.

Note: This question is from NCERT (2026-27). NTA sets numerical questions from NCERT too, so do not ignore NCERT while practising numericals.

NCERT (2026-27):
Size of the Nucleus – “… It has been found that a nucleus of mass number A has a radius

    \( \displaystyle R = R_0 A^{1/3} \qquad (13.5) \)

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.”