The half-life period of a radioactive element | vedclass

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(C) The radioactive decay law is given by $N(t) = N_0(1/2)^{t/T_{1/2}}$,where $N(t)$ is the amount remaining at time $t$,$N_0$ is the initial amount,a

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 $N(t)$ is the amount remaining at time $t$,$N_0$ is the initial amount,and $T_{1/2} = 10$ days.If $90\%$ of the element disintegrates,the amount remaining is $N(t) = 10\%$ of $N_0 = 0.1 N_0$.Substituting this into the decay equation: $0.1 N_0 = N_0(1/2)^{t/10}$.$0.1 = (1/2)^{t/10}$.Taking the logarithm on both sides: $\log_{10}(0.1) = (t/10) \log_{10}(0.5)$.$-1 = (t/10) 	imes (-0.3010)$.$t/10 = 1 / 0.3010 \approx 3.322$.$t \approx 33.22$ days.Rounding to the nearest integer,the time taken is $33$ days.

The half-life period of a radioactive element is $10$ days. How long does it take for $90\%$ of a given mass of this element to disintegrate?

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