The ratio of the Nuclear density of the gold isotope \( {}^{197}_{79}\mathrm{Au} \) and the silver isotope \( {}^{107}_{47}\mathrm{Ag} \) is:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( 1 : 1 \)
The radius of a nucleus of mass number \( A \) is \( R = R_0 A^{1/3} \), so its volume is proportional to \( A \). Its mass is about \( A \) times the mass \( m \) of a nucleon.
\( \displaystyle \rho = \frac{A\,m}{\frac{4}{3}\pi R_0^{3} A} = \frac{3m}{4\pi R_0^{3}} \)
The mass number cancels, so the nuclear density is the same for all nuclei. The ratio for gold and silver is \( 1 : 1 \).
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.”