Two radioactive substances X and Y simultaneously undergo decay in a vessel. The 50% decay of both X and Y occurs in 20 and 10 minutes, respectively. If the initial concentration of Y is 8 times that of X, how much time will it take for the concentration of X to be half of Y ?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 40 minutes
Radioactive decay is a first order process, so its half-life does not depend on how much substance is present. Every half-life halves the amount left. After a time t, the number of half-lives is n = t/t1/2, and:
\( \displaystyle [\mathrm{A}]=[\mathrm{A}]_0\left(\frac{1}{2}\right)^{n} \)
For the two substances (t1/2 of X = 20 min, t1/2 of Y = 10 min, [Y]0 = 8[X]0):
\( \displaystyle [\mathrm{X}]=[\mathrm{X}]_0\left(\frac{1}{2}\right)^{t/20} \)
\( \displaystyle [\mathrm{Y}]=8[\mathrm{X}]_0\left(\frac{1}{2}\right)^{t/10} \)
The condition is [X] = ½[Y]:
\( \displaystyle [\mathrm{X}]_0\left(\frac{1}{2}\right)^{t/20}=4[\mathrm{X}]_0\left(\frac{1}{2}\right)^{t/10} \)
\( \displaystyle \frac{2^{t/10}}{2^{t/20}}=4 \)
\( \displaystyle 2^{t/20}=2^{2} \)
\( \displaystyle \frac{t}{20}=2 \)
\( \displaystyle t=40\ \mathrm{min} \)
So X becomes half of Y after 40 minutes.
Shortcut: Count half-lives. In 40 minutes X passes 2 half-lives and falls to 1/4 of its starting amount. Y passes 4 half-lives and falls to 8 × 1/16 = 1/2. Since 1/4 is half of 1/2, 40 minutes is correct.
Note: It seems that there is a mistake in the original NTA paper. The question asked for the time “for the concentration of Y to be half of X”. As printed, the answer would be 80 minutes, which is not among the options. The answer given by NTA (40 minutes) is the time at which X becomes half of Y, so the question has been changed to read “the concentration of X to be half of Y”.