Question

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

Answer 1

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.

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

$N(15)=\frac{35}{100}\times\text{N(0)}$

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

$N(t)=N(0)\times(0.5)^{\frac{t}{T}}$

T is the half-life time.

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

$\frac{35}{100}\times N(0)=N(0)\times(0.5)^{\frac{15}{T}}$

Cancel out N(0), we get

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

$0.35=(0.5)^{\frac{15}{T}}$

Taking log on both sides, we get

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

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

$T=15\times\frac{In\mleft(0.5\mright)}{In(0.35)}$

$T=15\times\frac{-0.6931}{-1.0498}=9.9033$$T=9.9=10$

Hence the half-life time is 10 hours.