The half-life period of {}_{53}I^{125} is 60 | vedclass

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(B) Let the initial radioactivity be $N_0 = 100 \%$. Given, half-life period $(t_{1/2}) = 60$ days. Total time $(T) = 180$ days. The number of half-li

Vedclass · Accepted Answer

$12.5$

Vedclass · Answer

(B) Let the initial radioactivity be $N_0 = 100 \%$. Given, half-life period $(t_{1/2}) = 60$ days. Total time $(T) = 180$ days. The number of half-lives $(n)$ is calculated as: $n = \frac{T}{t_{1/2}} = \frac{180}{60} = 3$. The remaining radioactivity $(N)$ after $n$ half-lives is given by the formula: $N = N_0 	imes (\frac{1}{2})^n$. Substituting the values: $N = 100 \% 	imes (\frac{1}{2})^3 = 100 \% 	imes \frac{1}{8} = 12.5 \%$. Therefore, the radioactivity after $180$ days will be $12.5 \%$.

The half-life period of ${}_{53}I^{125}$ is $60$ days. The radioactivity after $180$ days will be (in $\%$)

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