Exponential Decay Section 9.2   Our last Exponential model was for Growth.  For radioactive decay, we also use an exponential model.  However, the rate is now negative to represent decay.   Example 1a:   If there were 20 grams of Iodine-131 8 days ago and now there is only 10 grams, write a decay model to represent this.   Q = aebt          Q = end amount         a = initial amount       b = rate     t = time (days)           10 = 20eb(8)                            Isolate e         .5  = e8b                                  Take ln of both sides         ln .5 = 8b                                Solve for b         ln.5 = b = - 0.086643397      Note: b is negative because we have decay.           8   Example 1b:   Predict how much Iodine-131 will be left in 3 weeks from day 0.           Q = 20e-.086643397(21) = 3.24 grams           if b > 0, there is exponential growth           if b < 0 , there is exponential decay.   Half-life:  the amount of time it takes for radioactive material to reduce to half its original amount. What is the half-life of Iodine-131?  8 days (See EX 1a)   Example 1c:   How much Iodine-131 is present after 16 days? 24 days?   Since 8 days is the half-life of Iodine-131, then in 16 days         ½ (10 grams) = 5 grams   In 24 days          ½ (5 grams) = 2.5 grams.   The half-life of a radioactive substance does not depend on its initial amount. Page 625, Figure 9.16.   EX 2:  Half-life of cobalt-60 is 5.3 years.  If you store 12.4 grams of cobalt-60 today, what will you have one year from now?  Since b is not given, use half-life information to derive model.         a/2 = aeb(5.3)                                      Isolate e         ½  = e5.3b                                          Take ln of both sides         ln ½ = 5.3b                                        Solve for b         ln ½  = b = -.130782487                   Store in memory          5.3           Q = 12.4e-.130782487(1) = 10.9  grams remaining after one year. Compare relative decay rate to actual rate using EX 2.         D Q        =  10.9 – 12.4 = -1.5 grams/1 yr.         D Time Relative Decay Rate:          -1.5 gr./1 yr/ 12.4 gr = - .12259 /yr = - 12.26%/year         -12.26 % vs. –13.08%   Radiocarbon Dating: Based on the fact that two types of carbon occur in nature; carbon-12 is stable, carbon-14 is radioactive.  When an organism dies, the carbon-14 begins to decay while the carbon-12 remains the same.  Using this ratio helps to determine the age of a fossil or artifact.   Example 3:  Determine the model for carbon-14.         a/2 = aeb(5730)                     ½  = e5730b         ln ½  = 5730b         b = -.000120968         Q = ae-.000120968t   Example 4:   Find the percentage of carbon-14 present in the Shroud of Turin that was determined to be dated in the 14th century.   Let          x = % of initial amount         ax = ae-.000120968(600)         x  = .92999999 = 93%   What percentage must be remaining for the Shroud to be considered authentic?         2001 – 33 = 1968 years         ax = ae-.000120968(1968)         x = .788 = 78.8% Back to Exponenials and Logarithms Main Page Back to the Survey of Math Ideas Home Page