Sheegog, Tara asks:

CATAGORY: Mechanics QUESTION: A 3.00-g sample of an unknown type of radioactive material has decreased after two days to the point that 2.52g remain. (a) What is the half life of this material?

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The decay of a radioactive isotope is given by the formula

(1)          M = Mo exp(-kt)

Where Mo is the initioal mass at time t=0 and M is the mass at time t.
We can solve this equation for the decay rate (which is not the half
life but we need it) k. Duvide by Mo and take the natural logarithim
of both sides of (1) to give us

(2)         ln(M/Mo) =-kt 


(3)         k = -ln(M/Mo)/t =ln(Mo/M)/t

In our case M = 2.52g, Mo = 3.00g and t =2 days so 

(3)         k = ln(3.00/2.52)/2days = .0872 days^-1

Now return to (1).  The half life the time is when M/Mo = .5
so (1) becomes

(4)        .5 =exp(-.0872 t(half))

Where t(half) is the half life. Solving for t(half)

(5)       ln(.5) =  -(.0872 days^-1) t(half)


(6)       t(half) = -ln(.5)/.0872 days = ln(2)/.0872 days
                  = 7.95 days
          *                         *
          *   t(half)  = 7.95 days  *
          *                         *

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