The half lives of 2 radioactive specimen are \( T_1 = 5 \) years and \( T_2 = 7.5 \) years. The ratio of their disintegration constants \( \lambda_1 \) and \( \lambda_2 \) is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( \displaystyle\frac{3}{2} \)
The half-life and the disintegration (decay) constant are inversely related.
\( \displaystyle T_{1/2} = \frac{\ln 2}{\lambda} = \frac{0.693}{\lambda} \;\Rightarrow\; \lambda \propto \frac{1}{T_{1/2}} \)
\( \displaystyle \frac{\lambda_1}{\lambda_2} = \frac{T_2}{T_1} = \frac{7.5\,\mathrm{years}}{5\,\mathrm{years}} = \frac{3}{2} \)
The specimen with the shorter half-life decays faster, so it has the larger decay constant.
Note: This question is from the pre-rationalised NCERT book (2022-23). The topic was removed from the latest NCERT book (2026-27), and it also appears to be out of the CUET 2026 syllabus.
NCERT (Pre-rationalised edition, 2022-23):
Law of Radioactive Decay – “… One measure is the half-life T1/2 of a radionuclide, which is the time at which both N and R have been reduced to one-half their initial values. … To relate T1/2 to the disintegration constant λ, we put R = (1/2) R0 and t = T1/2 in Eq. (13.15) and solve for T1/2, we find
\( \displaystyle T_{1/2} = \frac{\ln 2}{\lambda} = \frac{0.693}{\lambda} \qquad (13.17) \)”