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Carbon Dating the Shroud of Turin


Date: 05/16/2000 at 21:45:08
From: Catherine Sullivan
Subject: radioactive decay

Please help me with the following:

The radioactive isotope carbon-14 is present in small quantities in 
all life forms, and it is constantly replenished until the organism 
dies, after which it decays to carbon-12 at a rate proportional to the 
amount of C-14 present, with a half-life of 5730 years. Suppose C(t) 
is the amount of C-14 at time t.

a) Find the value of the constant k in the differential equation:

     C" = -kC

b) In 1988, three teams of scientists found that the Shroud of Turin, 
which was reputed to be the burial cloth of Jesus, contained 91% of 
the amount of C-14 contained in freshly made cloth of the same 
material. How old is the Shroud according to the data?

Note: for part a, I know how to solve for k using the equation

     C = C*e^(kt)

but I need to do it from the differential equation. I took the 
integral of both sides but was left with a first derivative and too 
many unknowns. Please help!

Thanks


Date: 05/17/2000 at 07:56:48
From: Doctor Anthony
Subject: Re: radioactive decay

>a) Find the value of the constant k in the differential equation:
>
>     C" = -kC

I think you mean:

        C' = -kC

     dC/dt = -kC

      dC/C = -k.dt

     ln(C) = -kt + constant

         C = e^(-kt+constant)

         C = A.e^(-kt)   where A = e^constant

at t = 0, C = A
at t = 5730, C = A/2

        A/2 = A.e^(-5730k)

        1/2 = e^(5730k)

     -5730k = ln(1/2)

          k = 1.2097 x 10^(-4)

>b) In 1988, three teams of scientists found that the Shroud of Turin, 
>which was reputed to be the burial cloth of Jesus, contained 91% of 
>the amount of C-14 contained in freshly made cloth of the same 
>material. How old is the Shroud according to the data?

It is not necessary to use the formula derived above. We can say:

          C = C(0).(1/2)^(t/5730)

     C/C(0) = (1/2)^(t/5730)

      0.91  = (1/2)^(t/5730)

taking logs:

     (t/5730)ln(1/2) = ln(0.91)

              t/5730 = ln(0.91)/ln(1/2)  =  0.13606

                   t = 779.6  years

So the shroud is about 780 years old.
            
- Doctor Anthony, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Calculus
High School Physics/Chemistry

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