The half-life of iodine-131 is 7.2 days. How long will it take for a sample of this substance to decay to 30% of its original amount?

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Answer 1 · 2017-12-11T12:56:29

To decay $30 %$ $30%$ of original amount it will take $12.51$ $12.51$ days.

Half life of iodine -131 is $t = 7.2$ $t=7.2$ days

We know $p (t) = p (0) \cdot e^{k t} or e^{k t} = \frac{p (t)}{p (0)} = \frac{1}{2} = 0.5$ $p(t)=p(0)*e^(kt) or e^(kt) = (p(t))/(p(0))= 1/2=0.5$

Taking natural log on both sides we get,

$k t = ln (0.5) or 7.2 k = ln (0.5) or k = \frac{ln (0.5)}{7.2} \approx - 0.09627$ $kt= ln(0.5) or 7.2k = ln(0.5) or k=ln(0.5)/7.2 ~~-0.09627$

When $(p(t_0.3))/(p(0))=0.3 :. e^(kt_0.3) = (p(t_0.3))/(p(0))=0.3$ or

$k*t_0.3= ln(0.3):. 0.09627*t_0.3=ln(0.3)$ or

$t_0.3=ln(0.3)/-0.09627~~12.51(2dp)$ days

To decay $30%$ of original amount it will take $12.51$ days. [Ans]