The activity of a radioactive sample decreases to $(1/3)$ of its original value in $3\, \text{days}$. Then, in $9\, \text{days}$, its activity will become:
- A$(1/27)$ of the original value
- B$(1/9)$ of the original value
- C$(1/18)$ of the original value
- D$(1/3)$ of the original value
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If $20 \, g$ of a radioactive substance reduces to $10 \, g$ due to radioactive decay in $4 \, minutes$, then in what time will $80 \, g$ of the same substance reduce to $10 \, g$?
Medium
View SolutionThe activity of a radioactive sample is measured as $N_0$ counts per minute at $t = 0$ and $N_0/e$ counts per minute at $t = 5\, \text{minutes}$. The time (in minutes) at which the activity reduces to half of its initial value is
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View SolutionTwo radioactive materials $X_1$ and $X_2$ have decay constants $10 \lambda$ and $\lambda$ respectively. If initially they have the same number of nuclei, then the ratio of the number of nuclei of $X_1$ to that of $X_2$ will be $1 / e$ after a time:
The natural logarithm of the activity $R$ of a radioactive sample varies with time $t$ as shown. At $t=0$,there are $N_0$ undecayed nuclei. Then,$N_0$ is equal to [Take $e^2=7.5$].
Two radioactive samples $A$ and $B$ have half-lives $T_1$ and $T_2$ $(T_1 > T_2)$ respectively. At $t=0$,the activity of $B$ was twice the activity of $A$. Their activity will become equal after a time:
Medium
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