Physics Nuclear Physics ETEA 2018
PMDC Verified Question 58 of 98
Two radioactive samples, \(S_1\) and \(S_2\) have half-live 3 hours and 7 hours respectively. If they have the same activity at certain instant t, what is the ratio of the number of atoms of \(S_1\) to \(S_2\) at instant t?
A
9 : 49
B
49 : 9
C
3 : 7
D
7 : 3
Tap any option to test your recall and reveal the step-by-step Propolis autopsy.

Propolis Cognitive Error Autopsy

Official Correct Choice:
Option C: 3 : 7


Concept:

Activity (A) is defined as \( A = \lambda N \). If two samples have equal activities, their quantities must be inversely proportional to their decay constants.

Solution:

  • We are given \( A_1 = A_2 \).


  • Expand activity: $$ \lambda_1 N_1 = \lambda_2 N_2 $$


  • Since \( \lambda = \frac{0.693}{T_{1/2}} \), substitute \(\lambda\): $$ \frac{0.693}{T_{1/2(1)}} N_1 = \frac{0.693}{T_{1/2(2)}} N_2 $$


  • Cancel \(0.693\) and rearrange to find the ratio \(N_1 / N_2\): $$ \frac{N_1}{N_2} = \frac{T_{1/2(1)}}{T_{1/2(2)}} $$


  • Substitute the given half-lives: $$ \frac{N_1}{N_2} = \frac{3}{7} $$


Why other options are incorrect:

Option D flips the ratio. Options A and B mistakenly square the values, falsely assuming an inverse-square law relation.

Quality & Fidelity Assurance: Every question on BeambePrep is rigorously curated against the official PMDC syllabus with zero filler, zero out-of-syllabus content, and zero typos. When an authentic past paper originally contained a historical mistake or ambiguity from the examining board (such as UHS or NUMS), BeambePrep faithfully reflects the original paper while detailing the nuance and scientific consensus in the autopsy above.

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