The relation between half-life of a radio nuclide denoted by \( T_{\frac{1}{2}} \) and average life of a radio nuclide denoted by \( \tau \), is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( T_{\frac{1}{2}} = \tau \ln 2 \)
Both quantities are fixed by the decay constant \( \lambda \) of the radioactive decay law \( N = N_{0}e^{-\lambda t} \).
The half-life is the time in which \( N \) falls to \( N_{0}/2 \):
\( \displaystyle \frac{N_{0}}{2} = N_{0}e^{-\lambda T_{1/2}} \;\Rightarrow\; T_{1/2} = \frac{\ln 2}{\lambda} \)
The average (mean) life is
\( \displaystyle \tau = \frac{1}{\lambda} \)
Putting \( 1/\lambda = \tau \) in the half-life:
\( \displaystyle T_{1/2} = \tau \ln 2 \approx 0.693\,\tau \)
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. The other measure is the mean life τ, which is the time at which both N and R have been reduced to e–1 of their initial values. … We summarise these results with the following:
\( \displaystyle T_{1/2} = \frac{\ln 2}{\lambda} = \tau \ln 2 \qquad (13.18) \)”