Nuclei - Result Question 65
66. A mixture consists of two radioactive materials $A_1$ and $A_2$ with half lives of $20 s$ and $10 s$ respectively. Initially the mixture has $40 g$ of $A_1$ and $160 g$ of $A_2$. The amount of the two in the mixture will become equal after :
(a) $60 s$
(b) $80 s$
(c) $20 s$
(d) $40 s$
[2012]
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Answer:
Correct Answer: 66. (d)
Solution:
- (d) Let, the amount of the two in the mixture will become equal after $t$ years.
The amount of $A_1$, which remains after $t$ years
$N_1=\frac{N _{01}}{(2)^{t / 20}}$
The amount of $A_2$, which remains, after $t$ years
$N_2=\frac{N _{02}}{(2)^{t / 10}}$
According to the problem
$N_1=N_2$
$\frac{40}{(2)^{t / 20}}=\frac{160}{(2)^{t / 10}}$
$2^{t / 20}=2^{(\frac{t}{10}-2)}$
$\frac{t}{20}=\frac{t}{10}-2$
$\frac{t}{20}-\frac{t}{10}=2$
$\frac{t}{20}=2$
$t=40 s$
