Plutonium Decay Problem

**xterminal01** · January 12th 2010, 01:12 PM

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

**Henryt999** · January 12th 2010, 01:44 PM

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

**skeeter** · January 12th 2010, 01:52 PM

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 ...

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

**Henryt999** · January 12th 2010, 01:55 PM

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 =)

Thread: Plutonium Decay Problem

LinkBack

Thread Tools

Display

Plutonium Decay Problem

Nope

I belive this is correct

Similar Math Help Forum Discussions

Decay of a Decay. Is the diff eqn correct

decay problem

Decay rates in exponential decay

Decay Problem

Radioactive decay Problem

Search Tags