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.

Lesson 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^{-λ t}

can be used to generate an exponential decay graph. Work through a numerical example, perhaps related to the dice-throwing analogue (N₀ = 100 ; λ = 16). Make sure that your students know how to use the e^x key on their calculators.

Emphasise that similar equations apply to activity

A = A₀ × e^{-λ t}

and count rate

C = C₀ × e^{-λ t}

(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

T _½  = ln(2)λ

or

T _½  = 0.693λ

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 ln N = ln (N₀) − λ t. 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 = ln(N₀) , gradient = −l).

Worked examples: 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 e^x 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)

Radio carbon dating

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

fraction =  ½ 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 =  ½ 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 T _½ of 146C is 5730 years. What fraction of a sample of 146C remains after 10 000 years? Answer:

fraction remaining =  ½ ⁿ

The number of half lives,

n = 10 0005730

n = 1.745

Thus the

fraction remaining =  ½ ^1.745

And using the y^x button on a calculator gives fraction = 0.298.

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

How many years will it take for 99% of 6027Co to decay if its half life is 5.23 yr?

The fraction remaining is 1%, so

fraction = 0.01.

Hence

0.01 =  ½ ⁿ

Taking logs of both sides gives

ln(0.01) = n × ln( ½ )

giving

n = 6.64 half lives,

and so the

number of years = 6.64 × 5.23 years

number of years = 34.7 years.