Using scientific calculators in teaching half-life

Lexington High Schwl. Lexington. MA 01273. In high school chemistry curricula, we often squeeze in a few days to discuss nuclear chemistry. A typical ...
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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 % a b e d . But what if the time interval chosen is not an ink& multiple of the half-life? What fraction remains after, say 20 days? By plotting 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. '%raised to the 2.5th power! A correct solution is 40 times 1 Indeed, more generally

where Nois 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.

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 quickIv worked out. Thev include LOG(.,,1. I.N(... ). an INVFRSE