(B) The half-life $(T_{1/2})$ of the radioactive element is $10 \ h$.The total time elapsed $(t)$ is $40 \ h$.The number of half-lives $(n)$ is calcul

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(B) The half-life $(T_{1/2})$ of the radioactive element is $10 \ h$.The total time elapsed $(t)$ is $40 \ h$.The number of half-lives $(n)$ is calculated as $n = \frac{t}{T_{1/2}} = \frac{40 \ h}{10 \ h} = 4$.The fraction of the initial radioactivity remaining is given by the formula $\frac{N}{N_0} = (\frac{1}{2})^n$.Substituting the value of $n$,we get $\frac{N}{N_0} = (\frac{1}{2})^4 = \frac{1}{16}$.

Class 11 Chemistry Data Entry - Chemistry Level 0 of all question bank

The half-life of a radioactive element is $10 \ h$. The fraction of initial radioactivity of the element that will remain after $40 \ h$ is

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A
$\frac{1}{2}$
B
$\frac{1}{16}$
C
$\frac{1}{8}$
D
$\frac{1}{4}$

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The half-life of a radioactive element is $10 \ h$. The fraction of initial radioactivity of the element that will remain after $40 \ h$ is

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