
It has been determined that the rate of radioactive decay is first order. We can apply our knowledge of first order kinetics to radioactive decayto determine rate constants, original and remaining amounts of radioisotopes,half-lives of the radioisotopes, and apply this knowledge to the datingof archeological artifacts through a process known ascarbon-14 dating.
Since the rate of radioactive decay is first order we can say: r = k[N]1, where r is a measurement of the rate of decay, kis the first order rate constant for the isotope, and N is the amount ofradioisotope at the moment when the rate is measured. The rate ofdecay is often referred to as the activity of the isotope and is oftenmeasured in Curies (Ci), one curie = 3.700 x 1010 atoms thatdecay/second. By knowing the amount of radioisotope and the activityof the sample, the rate constant can be determined.
Example: A 1.00 g sample of cobalt-60 (59.92 g/mol) hasan activity of 1.1 x 103 Ci. Determine the rate constant.
(1.1 x 103 Ci)(3.700 x 1010 atoms/s/Ci)= 4.1 x 1013 atoms that decay/sec
k = 4.1 x 10-9 s-1
Example: Determine the activity (in Curies) of a 2.00 mgsample of Cobalt-60.
Using the equation for first-order kinetics, the following equationcan be derived:
ln(N/No) = -kt
where "N" is the amount of radioisotope remaining after time "t" haselapsed. "No" is the initial amount of radioisotope atthe beginning of the period, and "k" is the rate constant for the radioisotopebeing studied. In this equation, the units of measure for N and Nocan be in grams, atoms, or moles. It does not matter as long as theyare like measures. The units of measure for time are dependent uponthe unit of measure for the rate constant. The ratio of "N/No"gives the percentage activity as compared to the activity at time zero. This equation has a variety of applications.
Example: How much of a 2.00 g sample of radioisotope(k = 0.15 min-1) will remain after 20 minutes?
ln(N/2.00g) = -(0.15 min-1)(20 min)
ln(N/2.00 g) = -0.30
N/2.00g = e-0.30 = 0.741
N = 1.48 g
Example: A sample of radioisotope has an activity of 450microcuries after 2 days. If the rate constant for the isotope is0.056 d-1, what was the activity of the sample 2 days ago?
ln(450 uCi/No) = -(0.056 d-1)(2 d)
ln(450 uCi/No) = -0.112
450 uCi/No = e-0.112 = 0.894
No = 503 uCi
Example: A sample is 70 % as active after 20 hours haveelapsed. Determine the value of the rate constant.
ln(70/100) = -k(20 h)
-0.357 = -k(20 h)
k = 0.018 h-1
Example: How long will it take a radioisotope to decayto 30% of its original activity if the rate constant for the isotope is0.055 s-1?
ln(30/100) = -(0.055 s-1)t
-1.20 = -(0.055 s-1)t
t = 22 s
Example: What is the half-life of a radioisotope that hasa rate constant of 0.225 d-1?
Half-life (t1/2) is the time for the radioisotope to reach50% of its original amount.
ln(50/100) = -(0.225 d-1)t1/2
-0.6931 = -(0.225 d-1)t1/2
t1/2 = 3.08 days
Example: How much of a 60 Ci sample of radioisotope willremain after 10 minutes if the half-life of the radioisotope is 2.75 min?
ln(50/100) = -k(2.75 min)
k = 0.252 min-1
ln(N/60 Ci) = -(0.252 min-1)(10 min)
N = 4.8 Ci
Carbon-14 is a radioisotope formed in our atmosphere by the bombardmentof nitrogen-14 by cosmic rays. The amount of carbon-14 in the atomosphereis, on an average, relatively constant. Plants take in carbon-14through the process of photosynthesis. Animals eat the plants sothey too have carbon-14 in their tissues. Carbon-14 is decaying constantlywith a half-life of 5720 years. As long as the organism is alive,the amount of carbon-14 remains relatively constant. However, whenthe organism dies, the amount will decrease over time. By comparingthe activity of an archeological artifact to that of a sample of the livingorganism one can estimate the age of the artifact.
Example: A sample of wood taken from an ancient tomb hadan activity of 7.0 counts per minute (decays per minute). A similarsample of freshly cut wood of the same type of tree had an activity of15.3 cpm. Estimate the age of the wood taken from the tomb.
ln(50/100) = -k(5720 y)
k = 1.212 x 10-4 y-1
ln(7.0 cpm/15.3 cpm) = -(1.212 x 10-4 y-1)t
t = 6982 y
