We are given the half-life of PU-239 as 24360 years. We are asked to find, to the nearest year, the time it will take a 10 gram sample to decay to 1 gram.
This is an example of exponential decay. A model for exponential decay can be written as A(t)=P(r)^(t) where A(t) represents the amount at time t, P is the original amount, r is the decay rate, and t is the time.
Here we are looking for t in years, A(t) will be 1 gram, P is 10 grams, and r=.5; we note that it takes 24360 years to undergo a decay of 1/2 of the present material.
We need to solve the following equation for t:
1=10(.5)^(t/24360) SO:
.5^(t/24360)=.1
Taking a logarithm of both sides (we can use a logarithm of any base -- we will use the natural logarithm base e) we get:
ln[.5^(t/24360)]=ln(.1)
A property of logs yields:
(t/24360)ln(.5)=ln(.1)
t/24360=(ln(.1))/(ln(.5))
t=(ln(.1))/(ln(.5))*24360~~80922
Thus it will take approximately 80922 years for a 10 gram sample to decay to 1 gram.
( A quick check for reasonableness: there are 80922/24360 or approximately 3.32 halving periods; 10 -->5, 5-->2.5, 2.5--> 1.25 and then the final 1/3 of 24360 years to get from 1.25 grams to 1 gram.)
A graph of the model starting with 10 grams (time in 10000 year increments):
http://mathworld.wolfram.com/ExponentialDecay.html
http://mathworld.wolfram.com/ExponentLaws.html
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