JEE PYQ: Nuclear Physics Question 11
Question 11 - 2020 (03 Sep Shift 1)
In a radioactive material, fraction of active material remaining after time $t$ is 9/16. The fraction that was remaining after $t/2$ is:
(1) $\frac{4}{5}$
(2) $\frac{3}{5}$
(3) $\frac{3}{4}$
(4) $\frac{7}{8}$
Show Answer
Answer: (3)
Solution
As we know, for first order decay, $N(t) = N_0 e^{-\lambda t}$ According to question, $\frac{N(t)}{N_0} = \frac{9}{16} = e^{-\lambda t}$ After time $t/2$: $N(t/2) = N_0 e^{-\lambda(t/2)}$ $\frac{N(t/2)}{N_0} = \sqrt{e^{-\lambda t}} = \sqrt{\frac{9}{16}}$ $\therefore N(t/2) = \frac{3}{4} N_0$
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