(A) The number of nuclei remaining after time $t$ is given by $N(t) = N_0 e^{-\lambda t}$.The fraction that decays is $f = \frac{N_0 - N(t)}{N_0} = 1

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(A) The number of nuclei remaining after time $t$ is given by $N(t) = N_0 e^{-\lambda t}$.The fraction that decays is $f = \frac{N_0 - N(t)}{N_0} = 1 - e^{-\lambda t}$.For one mean life,$t = \tau = \frac{1}{\lambda}$. Thus,$f_1 = 1 - e^{-\lambda(1/\lambda)} = 1 - e^{-1} \approx 1 - 0.368 = 0.632$.For one half-life,$t = T_{1/2} = \frac{\ln 2}{\lambda}$. Thus,$f_2 = 1 - e^{-\lambda(\ln 2 / \lambda)} = 1 - e^{-\ln 2} = 1 - 0.5 = 0.5$.Comparing the two,$0.632 > 0.5$,therefore $f_1 > f_2$.

Class 12 Physics Nuclei Law of Radioactivity by Rutherford and Soddy and Half Life and Mean Life

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$A$ fraction $f_1$ of a radioactive sample decays in one mean life,and a fraction $f_2$ decays in one half-life.

A
$f_1 > f_2$
B
$f_1 < f_2$
C
$f_1 = f_2$
D
May be $(A), (B)$ or $(C)$ depending on the values of the mean life and half life

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Law of Radioactivity by Rutherford and Soddy and Half Life and Mean Life

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$A$ fraction $f_1$ of a radioactive sample decays in one mean life,and a fraction $f_2$ decays in one half-life.

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