Half-life of a radioactive element is 10 , da | vedclass

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(C) The radioactive decay law is given by $N(t) = N_0 (1/2)^{t/T_{1/2}}$,where $T_{1/2} = 10 \, days$.We want to find $t$ such that $N(t) = N_0 / 10$.

Vedclass · Accepted Answer

$33$

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(C) The radioactive decay law is given by $N(t) = N_0 (1/2)^{t/T_{1/2}}$,where $T_{1/2} = 10 \, days$.We want to find $t$ such that $N(t) = N_0 / 10$.Substituting the values: $N_0 / 10 = N_0 (1/2)^{t/10}$.$1/10 = (1/2)^{t/10}$.Taking the logarithm on both sides: $\log_{10}(1/10) = (t/10) \log_{10}(1/2)$.$-1 = (t/10) 	imes (-0.3010)$.$t = 10 / 0.3010 \approx 33.22 \, days$.Therefore,the time is approximately $33 \, days$.

Half-life of a radioactive element is $10 \, days$. The time during which the quantity remains $1/10$ of the initial mass will be ......... $days$.

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