It is observed that only 25% of a given radioactive sample is left undecayed after 16 days. What is the decay constant of the sample in \( \mathrm{day^{-1}} \) ?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 0.087
A radioactive sample follows the decay law \( N = N_0 e^{-\lambda t} \), where \( \lambda \) is the decay constant. When 25% is left, \( N/N_0 = 1/4 \).
\( \displaystyle \frac{1}{4} = e^{-16\lambda} \;\Rightarrow\; 16\lambda = \ln 4 = 1.386 \)
\( \displaystyle \lambda = \frac{1.386}{16} \approx 0.087\,\mathrm{day^{-1}} \)
Shortcut: A quarter is left after two half-lives, so \( T_{1/2} = 16/2 = 8 \) days. Then \( \lambda = 0.693/T_{1/2} = 0.693/8 \approx 0.087\,\mathrm{day^{-1}} \).
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 – “… Setting t0 = 0 and rearranging Eq. (13.12) gives us
\( \displaystyle \ln\frac{N}{N_0} = -\lambda t \qquad (13.13) \)
which gives
\( \displaystyle N(t) = N_0\, e^{-\lambda t} \qquad (13.14) \)
… The decay of radionuclides follows quite a different law, the law of radioactive decay represented by Eq. (13.14).”