RADIOACTIVE DECAY CALCULATIONS WITHOUT CALCULUS WILLIAM B. GUENTHER University of the South, Sewanee, Tennessee
THE increased space and importance allotted to radiochemistry in recent elementary textbooks do not seem to have engendered a rethinkinn of the calculations possible in this fascinating topicWforhigh school and general chemistry students. If any problems a t all are given on the amount of radioactive decay or age determinations, only those with whole numbers of half-lives are used. The student is probably left with the impression that exact calculation between half-lives i s difficult and mysterious. The first order differential equation and its integration are still beyond these students. However, there is a simple method of deriving the exact equation for radioactive decay without integration. With this the student can then solve problems on decay and age determination. The method is, in essence, applicable t o any first order process. The usual derivation is based on the experimental rate law, a differential equation. One obtains the same result starting from the equally fundamental fact that the half-life is a constant for a certain pure, radioactive element. This derivation proceeds as follows: Start with A. grams of a radioactive element. Let A , be the amount left after time x. So, after one, two, and three half-lives, there will be left: A , = '/SAC, '/zX '/do,or l/2 X X 1 / Z A O .SO, in general, after any number, n, of half-lives, A, = ('/*)"Aa
(1)
This simple equation is the one often used for approximate answers by interpolation between half-lives when n is not an integer. This equation is not limited t o integral values of n. To derive the general equation in t.erms of times, substitute n = tz/ty,, because the number of half-lives must equal the time elapsed, t,, divided by the length of one half-life, LIE, in the same units. Now, rearrange equation (I), take the logarithm of each side, and insert the new term for n: ASIA. =
('/n)"
I o d A J A d = n log('/d
(2)
log(A,/Ad = (t./tl/dog('/d
(3)
I t can be shown that this is the same result that is obtained from the usual differential equation:'
2.303 log (A,/Ao) = -kt,
(4) where k is the radioaotive dewy constant. After one half-life the time.. t,- = /I/,. -. and A,. = '/.An. . - 8a: ~
.
2.303 log(%) = -kty,, and k = -(2.303/11).)l0g(~/~) (5) When this value of k is put into equation (4), equation (3) is obtained.
Equations ( 2 ) and ( 3 ) allow the student t o do problems on the amount of radioactive decay, by getting A,, or the per cent decayed, (A,/Ao)lOO, and on age determination by solving for t,. For the general chemistry course, recent publications suggest many problems of interest in geochemistry2 and on the age of the earth and solar ~ y s t e m . ~ , ~ . ~ An example will serve to compare the result of exact calculation by equation (3) and the approximate result of interpolation with equation (1). Problem A hospital has 2.00 mg. of Com for radiation treatments. How much will be left after 25 months of use? The half-life of Corn is 5.2 years. (Students should be encouraged to start with equation (1) each time rather than memorizing (31.) Solution A, is sought in the derived equation (3).
log (A.lZ.00) = (-0.301/5.2 yrs.)(25 mo./12 mo./pr.) A, = 1.5, mg. Approzimate One milligram decays in the first half-life. So, interpolate between zero time and one half-life:
A',
=
(25 mo./5.2 X 12 mo./yr.) 1.00 mg. =, 0.40 mg. gone A, = 2.00 - 0.40 = 1.60 mg. left
The error of 7% in the A , value gives some idea of the accuracy expected with the approximate method. Estimates of the amount left, A,, from ( I ) and the correct values from ( 3 ) usually differ by less than 10%. DANIELS,F., AND R. A. ALBEETY,"~h&a1 Chemistry," -' -t. John Wiley & Sons, Inc., New York, 1955, ppi.6067. FRIEDLANDER, G., AND J. W. KENNEDY, "Z?ucIear and Radiochemistry," John Wiley & Sons, Inc., New Ykrk, 1955, pp. 38993. a BROWN, H., Sei. American, 196, 8044 (1957). ' LINNENBAUM, V. J., J. CHEM.EDUC.,32, 58-68 (1955). KNOPF,ADOLPH, Sci. Monlhly, 85,225-36 (1957).
For example, a t n = 0.5, one would estimate A, = 0.75 Ao,while equation (3) gives 0.707 Ao. Similarly, a t n = 1.5, the error is also found t o be about 6'%. Clearly there is no error when n is an integer. Also rigorously correct would he the result of extracting the proper root of as an alternative to logarithms. For example, if n = 1.5,A,/Ao = = 0.354,as compared to the approximate value, 0.375,half the linear distance between one and two half-lives. There follows an example with time as the unknown term, in which the error of the approximate method is much larger. Problem What is the age of a uranium ore containing 1.66 g. of Pba' and 10.0 g. of U"? (Half-life, 4.51 X 10gyears.) (Explain why the hold-upof the elements in secular equilibrium with the U"* has no effect on the problem because the longest lived memher of the U'M- Pbmeseries is UZ3'of half-life 105 years, only about 1/10,000 of the U4' half-life.)
- -?JOW,35,
NO. 8, AUGUST, 1958
Solution 1.66 g. Pb came originally from 1.66 (238/206) = 1.92 g. U2' so that A , was 10.0 1.92 = 11.92a. U2'. Then, in equation (3) solved for t,, 1, = log(lO.O/ll.92) X 4.51 X IOs/log 1% = 1.14 X 10gyears
+
Approzimate In one half-life 5.96 g. of the 11.92 g. of U '" would decay; so: (1.92/5.96) X 4.51 X loQ= 1.45 X lo9 years
Thus, an error of about 30% occurs in t by the crude linear method. Other examules of recent interest can easily be uropounded. The use of CL4(half-life 5570 years) for archeological dating.? The use of H3 (12.46 years) for the time of isolation of water sample^.^ The decay time of Srgo(19.9 years) in plants, milk, etc., from homh fall-out.
