(A) The formula for radioactive decay is $N_t = N_0 \times (\frac{1}{2})^n$,where $n$ is the number of half-lives.Given $N_0 = 100 \ mg$,$N_t = 25 \ m

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(A) The formula for radioactive decay is $N_t = N_0 \times (\frac{1}{2})^n$,where $n$ is the number of half-lives.Given $N_0 = 100 \ mg$,$N_t = 25 \ mg$,and $t_{1/2} = 5760 \ years$.$25 = 100 \times (\frac{1}{2})^n$$(\frac{1}{2})^n = \frac{25}{100} = \frac{1}{4} = (\frac{1}{2})^2$Therefore,$n = 2$.Total time $t = n \times t_{1/2} = 2 \times 5760 = 11520 \ years$.

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

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$^{14}C$ has a half-life of $5760 \ years$. $100 \ mg$ of a sample containing $^{14}C$ is reduced to $25 \ mg$ in ......... $years$.

A
$11520$
B
$2880$
C
$1440$
D
$17280$

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Rate of decay and Half-life

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$^{14}C$ has a half-life of $5760 \ years$. $100 \ mg$ of a sample containing $^{14}C$ is reduced to $25 \ mg$ in ......... $years$.

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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: