What is the half-life period of a nucleus in | vedclass

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(C) The decay equation is given by $N = N_0 e^{-\lambda t}$ or $\lambda = \frac{2.303}{t} \log \frac{N_0}{N}$.Given $t = 4.2 \ days$ and $N = 0.798 \

Vedclass · Accepted Answer

$12.83$

Vedclass · Answer

(C) The decay equation is given by $N = N_0 e^{-\lambda t}$ or $\lambda = \frac{2.303}{t} \log \frac{N_0}{N}$.Given $t = 4.2 \ days$ and $N = 0.798 \ N_0$,we have $\frac{N_0}{N} = \frac{1}{0.798} \approx 1.253$.$\lambda = \frac{2.303}{4.2} \log(1.253) \approx 0.5483 	imes 0.0979 \approx 0.0537 \ days^{-1}$.The half-life $t_{1/2}$ is given by $t_{1/2} = \frac{0.693}{\lambda}$.$t_{1/2} = \frac{0.693}{0.0537} \approx 12.9 \ days$.Comparing with the given options,the closest value is $12.83$ $days$.

What is the half-life period of a nucleus in $days$ if at the end of $4.2$ $days$,$N = 0.798 \ N_0$?

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