Lesson 5 of 10
Exponential decay and data
I can define activity and decay constant and use A = lambda N.Syllabus
Syllabus 9478, 20(h). Define activity and decay constant and use A = lambda N.
I can infer and sketch exponential decay and use x = x0 exp(-lambda t) for an appropriate population, activity or received source rate.Syllabus
Syllabus 9478, 20(i). Infer and sketch exponential decay and use x = x0 exp(-lambda t) for an appropriate population, activity or received source rate.
I can use lambda = ln(2)/half-life.Syllabus
Syllabus 9478, 20(k). Use lambda = ln(2)/half-life.
The decay constant describes the chance of decay per nucleus per unit time. A large population follows a predictable expected trend even though each individual decay remains random.
Let N be the number of undecayed parent nuclei and λ the decay constant. For a sufficiently short interval Δt, the probability that one parent nucleus decays is approximately λΔt. The unit of λ is s-1 when time is in seconds.
N = N0e-λt
A = A0e-λt
Activity is the expected decay rate, in Bq: one decay per second. A constant λ means a constant fractional decay rate, not a constant number of decays per second. As N decreases, A decreases. The cumulative number that have decayed is N0 - N, which increases.
These equations describe one parent radionuclide with constant decay constant. They do not make its individual nuclei decay at predetermined times, and a mixture need not have one exponential total activity.
Connect decay constant and half-life
At t = t1/2, half the original parent population remains:
ln(1/2) = -λt1/2
λ = ln(2)/t1/2
A half-life of order 100 s suggests λ of order 10-2 s-1. For the supplied 120 s half-life, λ = 0.00577623 s-1. Convert minutes to seconds before using that value with a time.
Worked activity and detector rate
Decay rate is a property of the source
A separate sample initially contains N0 = 6.00 × 105 parent nuclei and has half-life 120 s:
= 3465.74 Bq ≈ 3.47 × 103 Bq
After 240 s, two half-lives have elapsed. N = 1.50 × 105 and A = 866.434 Bq, approximately 866 Bq.
Suppose a supplied calibration for this unchanged setup counts 5.00% of decays, with one relevant emission per decay. Its initial net detector rate is then:
= 173.287 counts/s
This is a different source from the 80 counts/s model below. The fraction detected is supplied calibration information, not a universal conversion between counts/s and Bq. Detector geometry, response and attenuation matter.
Use ratios and logarithms
With constant detection conditions and correctly subtracted background, the source count rate is proportional to activity:
ln(r/r0) = -λt
The logarithm's argument is a dimensionless ratio. A plot of ln(r/r0) against t has gradient -λ. For the 120 s model, the time to reach 10.0% is:
= 398.631 s ≈ 399 s
A logarithm turns the supplied fractional model into a straight line
This graph is the supplied model ln(r/r0) = -ln(2)t/120, with time in seconds. Its gradient is about -0.00578 per second. The logarithm acts on a dimensionless positive ratio; the zero ordinate means r = r0, not r = 0.
Zero and negative corrected rates have no real logarithm. A positive value very close to background can also have large relative uncertainty; being mathematically loggable does not make it a reliable fitting point.
Try it step by step
Work it out in steps: Decay constant lambda = ln 2 / t1/2 in s-1, then Activity A = lambda N in Bq.
Show the answer
Problem 1.
Decay constant lambda = ln 2 / t1/2 in s-1: t1/2 = 8.0 x 86400 = 6.91 x 105 s, so lambda = 0.693 / 6.91 x 105 = 1.00 x 10-6 s-1
Activity A = lambda N in Bq: A = 1.00 x 10-6 x 2.0 x 1015 = 2.0 x 109 Bq
Problem 2.
Decay constant lambda = ln 2 / t1/2 in s-1: t1/2 = 5.3 x 3.16 x 107 = 1.67 x 108 s, so lambda = 0.693 / 1.67 x 108 = 4.14 x 10-9 s-1
Activity A = lambda N in Bq: A = 4.14 x 10-9 x 1.0 x 1020 = 4.14 x 1011 Bq
Problem 3.
Decay constant lambda = ln 2 / t1/2 in s-1: t1/2 = 6.0 x 3600 = 21600 s, so lambda = 0.693 / 21600 = 3.21 x 10-5 s-1
Activity A = lambda N in Bq: A = 3.21 x 10-5 x 5.0 x 1012 = 1.60 x 108 Bq
Graph reader
Find a half-life from measured counts
A detector records the total count rate from a source plus background. Move the time cursor to read the graph.
Check your understanding
A sample initially contains 6.00 x 105 parent nuclei and has half-life 120 s. With lambda = ln(2)/120 s-1, what is its activity at 240 s?
Show the answer
About 866 Bq.
Two half-lives leave N = 1.50 x 105. A = lambda N = (ln2/120)(1.50 x 105) = 866 Bq. A count rate need not equal the activity because geometry and registration matter.
Paper 4 practice: spreadsheet
Preserve the counts, then calculate rates
Open the counts and decay workbook, or import the raw count CSV. These are constructed teaching records. The working workbook contains formulas and charts; inspect and reproduce the calculations rather than treating the file as an experiment you performed.
For CSV import, choose comma separation and keep the start, duration and count columns numeric. Create sheets named Rates and Decay. Import the complete CSV into Rates starting at A10. If it opens in A1:F16 on a separate raw-import sheet, preserve that sheet and copy the entire block to Rates A10:F25. This places the headers and records at the cells used below.
You can also type the supplied records into those same cells manually. Retain the record IDs, series and data-status labels alongside the numbers. The Motion data method explains numeric import, formula copying and XY plots.
On Rates, headers are in A10:F10: record ID, series, start time in s, duration in s, total count and data status. Source-plus-background records S01-S05 occupy rows 11-15; background records B01-B03 occupy 16-18; decay records D01-D07 occupy 19-25. Each series uses its own time origin.
Keep these raw columns intact. A blank is not a measured zero. Resolve missing readings or non-positive durations before calculating or fitting; the stated ranges do not automatically expand when records are added.
Pool counts using the actual duration
The S records contain 4000 counts in 100 s; the B records contain 720 counts in 180 s. These separate stable-rate series give 40.0 total, 4.0 background and 36.0 net counts/s. The unweighted mean of the five S rates is 40.1533 counts/s, which gives equal weight to unequal counting durations.
- Rates B4: pooled background rate
=SUM(E16:E18)/SUM(D16:D18)- Rates B5: pooled total rate
=SUM(E11:E15)/SUM(D11:D15)- Rates B6: net stable-source rate
=B5-B4- G11: interval midpoint, in s
=C11+D11/2- H11: total rate, in counts/s
=E11/D11- I11: background-subtracted rate
=IF(B11="background","",H11-$B$4)
Copy G11:I11 down through row 25. The fixed reference $B$4 keeps the same background rate. Do not average the entire decaying series into one rate to estimate its time dependence.
Make the logarithmic selection explicit
On Decay, B4 contains reference rate 80 counts/s, B5 contains expected half-life 120 s, and B6 contains =LN(2)/B5. Label A10:F10 as record ID, midpoint, net rate, ratio, logarithm and status. Rows 11-17 link to the seven D records:
- A11: record ID; B11: midpoint
=Rates!A19=Rates!G19- C11: corrected rate; D11: dimensionless ratio
=Rates!I19=C11/$B$4- E11: logarithm only for positive rate
=IF(C11>0,LN(D11),"")- F11: retain the reason for an absent logarithm
=IF(C11>0,"positive","not logged")
Copy A11:F11 through row 17. The first five D records are rounded expected counts over finite windows; the final two deliberately illustrate fluctuations near background.
| Record | Midpoint / s | Net / counts s-1 |
|---|---|---|
| D01 | 2 | 79 |
| D02 | 120 | 40 |
| D03 | 240 | 20 |
| D04 | 360 | 10 |
| D05 | 480 | 5 |
| D06 | 960 | -0.1 |
| D07 | 1200 | 0 |
Keep all seven records. The last two are background-subtracted estimates, not negative or zero claims about the source's expected activity. Their logarithm cells remain blank, rather than being replaced with numerical zero.
- Decay E4: fitted decay constant, in s-1
=-SLOPE(E11:E15,B11:B15)- E5: free intercept of the log fit
=INTERCEPT(E11:E15,B11:B15)- E6: fitted half-life, in s
=LN(2)/E4
This specified fit uses D01-D05 only. It does not force the intercept to zero. For genuine records, preserve all readings and justify the range using the method and uncertainty; do not remove points merely to make a line straighter.
Compare the expected and fitted trends
A22:A82 contains model times 0 to 1200 s in 20 s steps. Use these formulas and copy B22:D22 through row 82:
- B22: expected net rate
=$B$4*EXP(-$B$6*A22)- C22: expected total including background
=B22+Rates!$B$4- D22: net rate from the free-intercept fit
=$B$4*EXP($E$5-$E$4*A22)
Use an XY chart with numeric midpoint time in s horizontally and corrected count rate in counts/s vertically. Show all seven records as individual markers, with the expected and fitted trends separately labelled. A second XY chart plots only E11:E15 against B11:B15, with dimensionless ln(r/r0) vertically and the fitted straight line. Do not join the record markers into an apparently observed continuous path.
Display the fitted numeric equation beside the log graph, using the intercept in E5 and negative gradient -E4. The working workbook does this in H43. For the supplied records it is approximately ln(r/r0) = -0.00061647 - 0.005774516t, with t in seconds and gradient unit s-1. Keep the underlying coefficient cells at full precision.
Compare the calculated results
The fitted λ is 0.005774516 s-1, the log intercept is -0.00061647, and the fitted half-life is 120.0355 s. These are close to the supplied generating values, but not identical.
A finite-window rate approximates the instantaneous midpoint rate. In this model, the 20 s early counting window changes that relation by about 0.05562%. Together with rounding to integer counts, this explains the small fitted difference; it is not evidence of a faulty real detector.
Keep full stored precision and format only the displayed values. Changing a supplied input should update its derived midpoint, rate and fit; preserve an original copy before exploring such changes. Long windows collect more counts but blur a rapidly changing rate.
With actual records, also consider random counting scatter, uncertainty or drift in background, timing, detector dead time, geometry and efficiency changes, and more than one radionuclide. Agreement with a constructed model is a calculation check. The measurement-record method explains how to preserve actual readings, instrument information and justified exclusions when analysing an investigation.
Practise an exam question
Write your answer first, then tick the mark points it makes.
Exam-style question
Data-based question: A radioisotope generator on a space probe [20]
Space probes sent far from the Sun cannot rely on solar panels. Instead they carry a radioisotope generator. It contains plutonium-238 (proton number 94), which decays by alpha emission to uranium-234 with a half-life of 87.7 years. The alpha particles are absorbed inside the fuel and its casing, which become hot. Thermocouples convert part of this heat into electrical energy; the rest is radiated away by cooling fins.
| mass of plutonium-238 at launch | 4.0 kg |
| molar mass of plutonium-238 | 0.238 kg mol-1 |
| kinetic energy of each alpha particle | 5.59 MeV |
| efficiency of conversion from heat to electrical energy | 6.0% |
| output voltage | 28 V |
| 1 year | 3.16 x 107 s |
| time after launch / years | 0 | 5 | 10 | 15 | 20 |
|---|---|---|---|---|---|
| power / W | 136 | 122 | 110 | 99 | 89 |
- (a) Write the nuclear equation for the decay of plutonium-238. [2]
- (b) Explain why an alpha emitter with a half-life of several decades is chosen for this use. [3]
- (c) Show that the decay constant of plutonium-238 is about 2.5 x 10-10 s-1. [1]
- (d) Calculate the number of plutonium-238 nuclei at launch. [1]
- (e) Calculate the activity at launch and the rate at which heat is produced. [3]
- (f) State one assumption you made in (e). [1]
- (g) Calculate the electrical power at launch and the current supplied at 28 V. [2]
- (h) Calculate the electrical power expected after 20 years if only radioactive decay reduced it. [2]
- (i) Use Table 2 to evaluate whether radioactive decay alone explains the fall in output. [3]
- (j) If the cooling fins failed, all the heat would go into the 10 kg fuel capsule, of average specific heat capacity 500 J kg-1 K-1. Calculate the initial rate of rise of its temperature. [2]
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