Question

Now suppose you were using Plutonium 239 in a nuclear fission reactor. How long would it take 20 g of radioactive Plutonium 239 decay down to a safe 2.5 g? SHOW WORK.

Answer 1

Answer:

It will take $7.23 \ 10^4$ years.

Explanation:

Plutonium 239 has a half life of $2.41 \ 10^4$ years.

We know, for radioactive decay, that the quantity of radioactive material N at time t can be obtained with the equation

$N(t) = N_0 \  (\frac{1}{2})^{\frac{t}{t_{\frac{1}{2}}}}$

where $N_0$ is the initial quantity of radioactive material and $t_{\frac{1}{2}}$ is the half life of the material.

Taking the values of our problem:

$N(t') = 2.5 \ g = 20 \ g \*  (\frac{1}{2})^{\frac{t'}{t_{\frac{1}{2}}}}$

$\frac{2.5 \ g}{20 \ g} =  \  (\frac{1}{2})^{\frac{t'}{t_{\frac{1}{2}}}}$

$log( 0.125) =    log((\frac{1}{2})^{\frac{t'}{t_{\frac{1}{2}}}} )$

$log( 0.125) =    {\frac{t'}{t_{\frac{1}{2}}}} * log((\frac{1}{2}) )$

$\frac{log( 0.125)}{log(\frac{1}{2})} =    \frac{t'}{t_{\frac{1}{2}}}}$

$t_{ \frac{1}{2} }  \frac{log( 0.125)}{log(\frac{1}{2})} =    t'  $

$2.41 \ 10^4 \years\ * 3 =  t'  $

$7.23 \ 10^4 \years\ =  t'  $