Find the half-life (in hours) of a radioactive substance that is reduced by 35 percent in 15 hours.

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asked Feb 7, 2023 228k views

1 Answer

Jayakrishnan · Answer 1 · 2023-02-13T23:23:25+0000

Given that radioactive substance is reduced by 35 percent in 15 hours.

Let N(0) be the initial quantity and N(15) be the quantity of substance after 15 hours.

$N(15)=35\text{ \%of N(0)}$

$N(15)=(35)/(100)*\text{N(0)}$

We know that the formula to find the quantity of substance after t hours is

$N(t)=N(0)*(0.5)^{(t)/(T)}$

T is the half-life time.

$\text{Substitute }t=15,\text{ N(15)=}(35)/(100)N(0),\text{ we get}$

$(35)/(100)* N(0)=N(0)*(0.5)^{(15)/(T)}$

Cancel out N(0), we get

$(35)/(100)=(0.5)^{(15)/(T)}$

$0.35=(0.5)^{(15)/(T)}$

Taking log on both sides, we get

$In0.35=In(0.5)^{(15)/(T)}$
$\text{ Use }In(a^n)=n\text{ In(a).}$

$In0.35=(15)/(T)(In(0.5))^{}$

$T=15*(In\mleft(0.5\mright))/(In(0.35))$

Hence the half-life time is 10 hours.