in/ight/ DONNA BOGNER

Wichiia State University Wichita. KS 67208

Using Scientific Calculators in Teaching HaleLlfe David Olney Lexington High Schwl Lexington. MA 01273

In high school chemistry curricula, we often squeeze in a few days to discuss nuclear chemistry. A typical sort of problem-solving skill we demand is exemplified by this question: 1-131 has a half-life of 8.0 days. If we start with a sample of 40 g, how much remains 24 days later?

We show that the answer is 40('/~)(~/2)('/2) = 5.0 g because 24 days represents three half-life periods. And perhaps we noint out that this is the same as 40 times % abed . But what if the time interval chosen is not an ink& multiple of the half-life? What fraction remains after, say 20 days? By plot- ting the sample size versus time, we show exponential decay of the samole and can find our answer eraohicallv. That .. . same graph also shows that the answer to our new question is nut halfwav between 10 K and 5 e! Rut how doesone comoute it mathem&ically? In honors class, we might expect our students to cope with such a problem.

A correct solution is 40 times '1% raised to the 2.5th power! Indeed, more generally

where No is the initial sample size, T is the half-life of the isotope, t is the elapsed time, and Nis the sample size at that instant. In the example above, t /T = 2018 = 2.5.

The trick is in evaluating that expression. When I first began teaching, no convenient way to do so was available, so we used log tables-extrapolation and all-as we mastered this classic algorithm.

Doing all this, even usinga slide rule, was a true test of a high school student's mettle-just to plug into a formula!

I have discovered that in this era of hand-held scientific calculators, these prohlems are more easily tackled by sim- pler algorithms, discussed below.

Theory Since radioactive decay is first-order process with time,

which integrates to N = Nee-", which is the equivalent of

Taking the natural log of both sides yields

This reminds us that loes are exnonenta! If vou do not believeeq3 isvalid, try det&mining ;he valueof i.718 raised to the -0.693 Dower. It is 0.50. When one half-life has aone by, 1 = 7' and the fraction of atoms remaining is V,rwe should realize that the constant 2.303 is generated by the conversion from base e logs to base 10 logs (it is ln(10)), while -0.693 is the value of In (0.51. Much of our efforts used to be simply to convert between log bases! A scientific calculator has function keys that doallow such expressions to be quick- Iv worked out. Thev include LOG( 1. I.N( ). an INVFRSE .,, ... <ey, and a "yr" key:

The use of "v to the x" function kev on a calculator can he used directly to solve our problem. in our example, 2018 = 2.50. Punching in x as 0.5 and y as 2.5 will show a display of 0.177, the ratio of NtoNo; multiplying that by initial sample size. 40 e. reveals that 7.07 e remain in our samnle.

~n al<.rnative strategy isto use the INVERSE function in coniunction with the LN( ) function. It eenerates the value of draised to that power. We will use eq3 as our algorithm. On our calculators, multiply 0.693 by 20, divide hy 8, and change the sign. You will get -1.7325. Now tap INV, then LN( ), and you will see the display 0.177. This is the ratio of Nto No. As before, multiplying by 40 g will give our answer of 7.07 g. As always, the calculator spits out many more digits than these! A discussion of significant figures is certainly appropriate with our classes. But he warned that applying "rules" to numbers involving logarithms can get sticky.

Still another strategy is to use INV and the base 10 log function key. But in the prior multiplication one would use (-0.693)(20/(8)(2.303)).

Rearraneements of the decav law are also easilv done. ~onsiderti;is~uestion: How long does it take fora sarnpleof I-131 todecay ta 11100 ita initial sire?

Method 1 We shall use eq 4 to solve for t. On the calculator, in

sequence find the LN of 0.01, multiply by 8 (the number of days in one half-life), divide by 0.693, then change the sign. Your answer is on the display as 53.2 days. (That is reason- able; i t represents somewhat more than sin half-lives and (%) raised to the sixth power is 1/64 = 0.0156.)

Method 2 (Trial and Enor, But Qulck) Enter 0.5, hit the "y to the x" function key, then enter 6.

The display shows 0.0156, a bit too high. So try a value of x that is a little bigger, say 6.2 (yields 0.0136) or 6.5 (yields 0.011) or 6.65 (yields 0.00996). We have found it! Our answer must be (6.65)(8.0 days) or 53.2 days.

Volume 83 Number 10 October 1986 849

Still another variation would be to determine the half-life of a mystery isotope, given two sample sizes (or their activi- ties) at two different time intervals.

I have found it useful in the classroom to do such sample calculations for everyone to see by using the immediate mode of an Apple computer interfaced to a big-screen moni- tor. But one must remember here that in most BASIC lan- guages LOG( ) yields a natural log, not a base 10 log! And in place of a single "yz" key, one must enter in sequence y, t, x , = to get output.

The same arguments and methods could be used to study changes in concentrations of chemicals versus time that are involved in a first-order rate law eouation. "Half-Life" in this context means the time needed for the concentration of a reactant chemiral to be cut in half. However. if the chemi- cal in question is a product (initial concentration = O ) , we must change things a hit. Let No be the stoichiometric limit

of that product possible. Then,

gives the growth curve of that product. Finallv. I wish to acknowledge that the straiehtforward . . u

use of they' function key illustrated here was shown to him bs his students at Lexinrton H ~ e h School. and not vice versa! hey wanted to use the most ;ired system-base 0.5-to work in, and they were right! Its ease of operation on modern calculators and its intuitive hasis makes it quite useful. The classic algorithm noted a t the start has the advantage of being in the straight-line format of y = mx + b, hut its cumbersome use makes i t a less-than-satisfactory teaching device. We can use the powerful calculating abilities of com- puters and calculators to simplify the undertanding of our theories.

850 Journal of Chemical Education