# Math Help - Plutonium Decay Problem

1. ## Plutonium Decay Problem

Plutonium,
239Pu, has a half-life of 24,100 years. What percent of a given amount remains after 100,000 years?
Is this done correctly?

$1/24100ln(1/2)$

k=0.00002876

$e^(0.00002876(100000))$
= 56.35% after 100,000 years

2. ## Nope

If it has a halflife of 24.000years how can there be more then half (over 50%) left after more then four half lifes?

3. Originally Posted by xterminal01
Plutonium,
239Pu, has a half-life of 24,100 years. What percent of a given amount remains after 100,000 years?
Is this done correctly?

$1/24100ln(1/2)$

k=0.00002876

$e^(0.00002876(100000))$
= 56.35% after 100,000 years

let $P$ = amount of plutonium at any time t

$P_0$ = amount of plutonium at t = 0

$P = P_0 e^{kt}$

$\frac{P}{P_0} = e^{kt}$

$\frac{1}{2} = e^{24100k}$

$\ln\left(\frac{1}{2}\right) = 24100k$

$k = -\frac{\ln{2}}{24100} \approx -2.876$

at $t = 100000$ years ...

$\frac{P}{P_0} = e^{100000k} \approx 0.056 = 5.6$%

4. ## I belive this is correct

Half-life = ln(2)/lambda

This gives lambda = 2.876*10^(-5)

Then N = N0*e^(-lambda*t)

Set N0 = 100% gives you:

100*e^(-2.876*10^-5*100000) = 5.9%
Sounds right 50;25;12.5;6.25 = 4 halflifes and gives you about 6% left

ops: Too slow, evenwithout this latex =)