# Math Help - Half-Life

1. ## Half-Life

The level of radioactivity on the site of a nuclear explosion is decaying exponentially. The level measured in 1980 was found to be 0.8 times the level measured in 1970. What is the half-life.

2. Originally Posted by lilikoipssn
The level of radioactivity on the site of a nuclear explosion is decaying exponentially. The level measured in 1980 was found to be 0.8 times the level measured in 1970. What is the half-life.
Let L(0) denote the initial level of radiation and t the passed time. Then the level after t years is calculated by:

$L(t)=L(0)\cdot e^{k\cdot t}$ with k a constant typical for the decaying process.

From L(0) = 1 and L(10) = 0.8 you can derive k:

$0.8 = 1\cdot e^{k\cdot 10}~\implies~ k=\frac1{10} \cdot \ln(0.8)~\approx~-0.02231...$

Now calculate the time which is necessary to reach only the half of the radioactivity:

$0.5 = e^{\frac1{10} \cdot \ln(0.8) \cdot t}~\implies~ t\approx 31.06\ yrs$

Btw: That is a very optimistical result

3. Originally Posted by lilikoipssn
The level of radioactivity on the site of a nuclear explosion is decaying exponentially. The level measured in 1980 was found to be 0.8 times the level measured in 1970. What is the half-life.
$A(t)=A_0\left( \frac{1}{2}\right)^{\frac{t}{h}}$

let 1970 be time t=0

Then

$A(10)=.8A_0=A_0\left( \frac{1}{2}\right)^{\frac{10}{h}}$

$.8=\left( \frac{1}{2}\right)^{\frac{10}{h}}$

$\ln(.8)=\frac{10}{h}\ln\left( \frac{1}{2}\right)$

$h=\frac{10 \ln \left( \frac{1}{2}\right)}{\ln(.8)} \approx 31.06$

I hope this helps.