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):

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