Episode 516: Exponential and logarithmic equations

Students may find this mathematical section difficult. It is worth pointing out that they have already covered the basic ideas of radioactive decay in the earlier episodes.

Summary

• Discussion: The exponential decay equation (15 minutes)
• Student question: An example using the equation (20 minutes)
• Discussion: The logarithmic form of the equation (15 minutes)
• Worked example: Using the log equation (20 minutes)
• Student questions: Practice calculations (60 minutes)
• Discussion (optional): Using Lilley’s formula (15 minutes)

Discussion: The exponential decay equation
Explain that the equation N = N0 e-lt can be used to generate an exponential decay graph. Work through a numerical example, perhaps related to the dice-throwing analogue (N0 = 100; l = 1/6). Make sure that your students know how to use the ex key on their calculators.

Emphasise that similar equations apply to activity A (A = A0 e-lt) and count rate C (C = C0 e-lt).

(Not all the radiation emitted in all directions by a source will collected by a detector lined up in one direction from a source)

Student question: An example using the equation
Set students the task of drawing a graph for a lab source, e.g. Co-60 (λ = 0.132 y-1, C0 = 200 counts s-1). They should first calculate and tabulate values of C at intervals of 1 year, and then draw a graph. From the graph, deduce half-life. Does this agree with the value from

T1/2 = ln 2/ λ = 0.693/ λ?

Discussion: The logarithmic form of the equation
Point out that a straight line graph is usually more useful than a curve, particularly when dealing with experimental data. Introduce the equation lnN = lnN0 - lt. Emphasise that this embodies the same relationship as the exponential equation. Use a sketch graph to how its relationship to the straight line equation y = mx + c (intercept = lnN0, gradient = −l).

Worked example: Using the log equation
Start with some experimental data (e.g. from the decay of protactinium), draw up a table of ln (count rate) against time. Draw the log graph and deduce l (and hence half-life). The experimental scatter should be obvious on the graph, and hence the value of a straight line graph can be pointed out.

You will need to ensure that your students can find natural logs using their calculators.

Student questions: Practice calculations
Your students should now be able to handle a range of questions involving both ex and ln functions. It is valuable to link them to some of the applications of radioactive materials (e.g. dating of rocks or ancient artefacts, diagnosis and treatment in medicine, etc).

Episode 516-1: Decay in theory and practice (Word, 42 KB)

Episode 516-2: Radioactive decay with exponentials (Word, 77 KB)

Episode 516-3: Radioactive decay used as a clock (Word, 48 KB)

Episode 516-4: Two important dating techniques (Word, 29 KB)

Discussion (optional): Using Lilley’s formula
Some students may benefit from a simpler approach to the mathematics of radioactive decay, using ‘Lilley’s formula’ f = (1/2)n.

When first introduced at pre-16 level, radioactivity calculations are limited to integral number of half lives. After 1, 2, 3, …, half-lives, 1/2, 1/4, 1/8, … remains. The pattern here is that after n half lives, a fraction f = (1/2)n remains to decay.

This formula works for non integral values of n; i.e. it also gives the fraction remaining yet to decay after any non-whole number of half-lives (e.g. 2.4, or 3.794). To use this formula, a little skill with a calculator is all that is required.

For example: The T1/2 of 14C is 5730 yrs. What fraction f of a sample of 14C remains after 10 000 years? Answer:

f = (1/2)n

The number of half lives n = 10 000/5730 = 1.745

Thus the fraction remaining f = (1/2)1.745

And using the yxbutton on a calculator gives f = 0.298.

If students know how to take logs (or can lean the log version of the formula), they can solve other problems:

How many years will it take for 99% of 60Co to decay if its half life is 5.23 yr?

The fraction remaining = 1%, so f = 0.01.

Hence 0.01 = (1/2)n

Taking logs of both sides gives ln (0.01) = n ln (1/2)

giving n = 6.64 half lives, and so the number of years = 6.64 ´ 5.23 = 34.7 years.