The half life of a radioactive substance is 20 s. The decay constant…

The correct answer is Option (1) → \( 0.0346\,\mathrm{s^{-1}} \)

The half-life \( T_{1/2} \) is the time in which half the nuclei decay, and it is related to the decay constant \( \lambda \) by \( T_{1/2} = \ln 2/\lambda \).

    \( \displaystyle \lambda = \frac{0.693}{T_{1/2}} = \frac{0.693}{20\,\mathrm{s}} = 0.03465\,\mathrm{s^{-1}} \approx 0.0346\,\mathrm{s^{-1}} \)

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 T_1/2 of a radionuclide, which is the time at which both N and R have been reduced to one-half their initial values. … To relate T_1/2 to the disintegration constant λ, we put R = (1/2) R₀ and t = T_1/2 in Eq. (13.15) and solve for T_1/2, we find

    \( \displaystyle T_{1/2} = \frac{\ln 2}{\lambda} = \frac{0.693}{\lambda} \qquad (13.17) \)”