Two radioactive elements \( E_1 \) and \( E_2 \) with their half lives \( T_1 \) and \( T_2 \) have undecayed atoms \( n_1 \) and \( n_2 \) respectively present at a given instant. What will be the ratio of their activities at this instant?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( n_1T_2 : n_2T_1 \)
The activity of a sample is its rate of decay, \( A = \lambda n \), where \( \lambda \) is the decay constant and n the number of undecayed atoms. The decay constant is related to the half-life by \( \lambda = \frac{\ln 2}{T} \).
\( \displaystyle A_1 = \frac{\ln 2}{T_1}\,n_1, \qquad A_2 = \frac{\ln 2}{T_2}\,n_2 \)
\( \displaystyle \frac{A_1}{A_2} = \frac{n_1/T_1}{n_2/T_2} = \frac{n_1T_2}{n_2T_1} \)
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 – “… We can now write Eq. (13.10) in terms of the decay rate R of the sample as
\( \displaystyle R = \lambda N \qquad (13.16) \)
where R and the number of radioactive nuclei that have not yet undergone decay must be evaluated at the same instant. The total decay rate R of a sample of one or more radionuclides is called the activity of that sample.”
Law of Radioactive Decay – “… 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
\( \displaystyle T_{1/2} = \frac{\ln 2}{\lambda} = \frac{0.693}{\lambda} \qquad (13.17) \)”