(C) Given: Total time $T = 50 \ days$,Final amount $N = \frac{1}{32} N_0$.Using the radioactive decay formula: $N = N_0 \times (\frac{1}{2})^n$,where

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(C) Given: Total time $T = 50 \ days$,Final amount $N = \frac{1}{32} N_0$.Using the radioactive decay formula: $N = N_0 \times (\frac{1}{2})^n$,where $n$ is the number of half-lives.$\frac{1}{32} N_0 = N_0 \times (\frac{1}{2})^n \implies (\frac{1}{2})^5 = (\frac{1}{2})^n \implies n = 5$.Since $n = \frac{T}{t_{1/2}}$,we have $5 = \frac{50}{t_{1/2}}$.Therefore,$t_{1/2} = \frac{50}{5} = 10 \ days$.

Class 12 Chemistry Nuclear Chemistry Rate of decay and Half-life

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The number of $\alpha$-particles emitted per second by a radioactive element falls to $1/32$ of its original value in $50 \ days$. The half-life period of this element is .......... $days$.

A
$5$
B
$15$
C
$10$
D
$20$

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Rate of decay and Half-life

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The number of $\alpha$-particles emitted per second by a radioactive element falls to $1/32$ of its original value in $50 \ days$. The half-life period of this element is .......... $days$.

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The decay profiles of three radioactive species $A$,$B$ and $C$ are given below:
These profiles imply that the decay constants $k_A$,$k_B$ and $k_C$ follow the order: