Question:

Two radioactive substances X and Y originally have $N_1$ and $N_2$ nuclei respectively. Half life of X is half of the half life of Y. After three half lives of Y, number of nuclei of both are equal. The ratio $N_1/N_2$ will be equal to :

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Remaining nuclei $N = N_0(1/2)^n$, where $n$ is the number of half-lives. A substance with a shorter half-life decays more times in the same interval.
Updated On: Jan 9, 2026
  • 3/1
  • 1/3
  • 8/1
  • 1/8
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The Correct Option is C

Solution and Explanation

Step 1: Let $T_Y = T$. Then $T_X = T/2$.
Step 2: In time $t = 3T$, substance Y undergoes 3 half-lives. $N_Y = N_2 / 2^3 = N_2 / 8$.
Step 3: In the same time $t = 3T$, substance X undergoes $t / (T/2) = 6$ half-lives. $N_X = N_1 / 2^6 = N_1 / 64$.
Step 4: Given $N_X = N_Y \Rightarrow N_1 / 64 = N_2 / 8$.
Step 5: $N_1 / N_2 = 64 / 8 = 8/1$.
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