If 23.00 grams of a 45.00 gram sample of unknown isotope remain after 

17 days 12 hours, what is the half-life of the unknown isotope to the 

nearest minute?

The formula is 

               A = Pert

When time t=0 days, the amount A = 45g.  Substitute these two values

              45 = Per(0)

              45 = Pe0

              45 = P(1)

               P = 45 grams

so substitute 45 for P in A = Pert

               A = 45ert

When time t=17.5 days, the amount A = 23g.  Substitute these two values:

              23 = 45er(17.5)

              23 = 45e17.5r

Divide both sides by 45

           23/45 = e17.5r

     .5111111111 = e17.5r

Rewrite this equation using the fact that

" N = eM can be rewritten as M = ln(N) "

           17.5r = ln(.5111111111)

Use a calculator to find ln(.5111111111) = -.6711682738

           17.5r = -.6711682738

Solve for r

               r = -.6711682738/17.5

               r = -.0383524728

So substitute this value of r in A = 45ert

               A = 45e-.0383524728t

We wish to know how many minutes is required for it to become half

its original amount, or 1/2 of 45g, or 22.5 g.  So substitute this 

for A in A = 45e-.0394804867t

            22.5 = 45e-.0383524728t

Divide both sides by 45

         22.5/45 = e-.0383524728t

              .5 = e-.0383524728t

Rewrite this as 

          ln(.5) = -.0383524728t

Solve for t

    -.6931471806 = -.0383524728t

   -.0383524728t = -.6931471806

               t = -.6931471806/(-.0383524728)

               t = 18.07307665 days

To find hours, multiply .07307665 by 24

       18 days 1.753839658 hours

To find minutes, multiply .753839658 by 60

       18 days 1 hour 45.23037947 minutes or to nearest minute:

       18 days 1 hour, 45 minutes.

Edwin





 

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