A radioactive substance has an average life o | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The average life (mean life) $	au$ is given as $5 	ext{ hours}$.The decay constant $\lambda$ is related to the average life by $	au = \frac{1}{

Vedclass · Accepted Answer

More than half of the active nuclei decay

Vedclass · Answer

(C) The average life (mean life) $	au$ is given as $5 	ext{ hours}$.The decay constant $\lambda$ is related to the average life by $	au = \frac{1}{\lambda}$.The number of radioactive nuclei remaining after time $t$ is given by $N(t) = N_0 e^{-\lambda t}$.Substituting $t = 	au = \frac{1}{\lambda}$,we get $N(	au) = N_0 e^{-\lambda(1/\lambda)} = N_0 e^{-1} = \frac{N_0}{e} \approx \frac{N_0}{2.718} \approx 0.368 N_0$.The number of nuclei that have decayed is $N_{decayed} = N_0 - N(	au) = N_0 - 0.368 N_0 = 0.632 N_0$.This means $63.2\%$ of the nuclei have decayed,which is more than half $(50\%)$.Therefore,the correct option is $C$.

$A$ radioactive substance has an average life of $5$ hours. In a time of $5$ hours,

Similar Questions

The half-life of a radioactive sample is $T$. What fraction of the substance will remain after a time $T/2$?

The half-life of a radioactive substance is $T$. The time taken for disintegrating $\frac{7}{8}$th part of its original mass will be

$A$ radioactive nucleus $A$ with a half-life $T$ decays into a nucleus $B$. At $t = 0$,there is no nucleus $B$. At some time $t$,the ratio of the number of $B$ to that of $A$ is $0.3$. Then,$t$ is given by

Two radioactive samples $A_1$ and $A_2$ have decay constants $10\lambda_0$ and $\lambda_0$ respectively. If they initially contain the same number of nuclei,find the ratio of the number of undecayed nuclei after time $t = 1/(9\lambda_0)$.

Two radioactive materials $M_1$ and $M_2$ have decay constants $9 \lambda$ and $\lambda$ respectively. Initially,they have the same number of nuclei. The ratio of the number of nuclei of $M_1$ to that of $M_2$ will be $\frac{1}{e}$ after a time interval of:

Vedclass Test Series

Exam Paper Generator

Online Exam Module