Background radiation: sources and how to correct a count rate for it

The nature, ionising ability, penetration and range of alpha, beta and gamma radiation

Nuclear equations for α, β⁻, β⁺ and γ emission

The spontaneous and random nature of decay, activity and the decay constant

Half-life, A = λN, N = N₀e^(-λt), λ= ln 2/t_(1/2) and the log forms

Carbon dating and medical tracers as contexts

Core Practical 15: the absorption of gamma radiation by lead

1. Every calculation on this page comes from one statement, that each unstable nucleus has the same constant probability of decaying per second, which gives A = λN, a constant half-life and exponential decay.

2. Always subtract the background count rate (measured with the source removed) before using a count rate.

3. Alpha: strongly ionising, a few cm of air, stopped by paper. Beta: moderately ionising, stopped by a few mm of aluminium. Gamma: weakly ionising, only reduced by several cm of lead.

4. In every nuclear equation nucleon numbers and proton numbers balance. α: A − 4, Z − 2. β⁻: Z + 1. β⁺: Z − 1. γ: no change.

5. Decay is random (no way to tell which nucleus decays next) and spontaneous (unaffected by temperature, pressure or chemistry).

6. N = N₀e^(-λt) with t_(1/2) = ln 2/λ; a graph of ln A against t is a straight line of gradient -λ.

Natural sources: radon gas from rocks and soil (often the largest single contribution indoors), rocks and building materials such as granite, cosmic rays, and radioactive isotopes in food, drink and our bodies (potassium-40, carbon-14). Artificial sources: medical X-rays and treatments, and small contributions from past weapons tests and the nuclear industry.

A Geiger–Müller (GM) tube counts background events as well as those from your source. The background fluctuates randomly too, so measure it for several minutes and use the mean rate. Then

Not in the formulae list (learn it). Both rates must be in the same unit (s⁻¹ or min⁻¹).

Correct every reading before you halve it, plot it or take its log; it matters most when the source count rate is small.

WORKED EXAMPLE · QUESTION

With no source present, a GM tube records 486 counts in 20.0 minutes. With a source in front of the tube, 1840 counts are recorded in 2.00 minutes. (a) Calculate the corrected count rate due to the source, in s⁻¹. (b) Explain why the background was measured for much longer.

SOLUTION

(a) Background = 486 ÷20.0 = 24.3 min⁻¹; with source = 1840 ÷2.00 = 920 min⁻¹. Corrected: 920 - 24.3 = 896 min⁻¹: 14.9 s⁻¹.

(b) Background counts fluctuate randomly; a longer count averages out the fluctuations and gives a more reliable mean rate.

Unstable nuclei become more stable by emitting radiation. What each kind is decides how strongly it ionises and how far it travels.

Ionising vs penetrating: each ionisation takes energy. An α particle (large charge, slow, massive) ionises thousands of atoms per mm of air and runs out of energy within a few cm; an uncharged γ photon interacts only occasionally, so it travels much further.

Hazard: outside the body α cannot reach living cells, so β and γ are the concern. Inside the body an α source is the most dangerous, because all its energy is deposited in a small volume of tissue.

Identifying a source: add absorbers between it and a detector. A large drop with paper means α; a further drop with a few mm of aluminium means β; what remains is γ.

A nuclide is written ᴬ_ZX: A is the nucleon (mass) number, Z the proton (atomic) number. In every nuclear equation the total nucleon number and the total proton number (charge) are the same on both sides.

The (anti)neutrino has no charge and almost no mass, so it does not affect the balance, but include it: it is needed to conserve energy, momentum and lepton number, and it explains why β particles from one isotope have a range of energies.

WORKED EXAMPLE · QUESTION

(a) Polonium-210 (²¹⁰₈₄Po) is an α emitter. (b) Strontium-90 (⁹⁰₃₈Sr) is a β⁻ emitter. (c) Fluorine-18 (¹⁸₉F) is a β⁺ emitter. Write each decay equation. (d) Thorium-232 (²³²₉₀Th) decays through α and β⁻ emissions to stable lead-208 (²⁰⁸₈₂Pb). How many of each are emitted?

SOLUTION

(a) ²¹⁰₈₄Po →²⁰⁶₈₂Pb + ⁴₂α

(b) Element 39 is yttrium: ⁹⁰₃₈Sr →⁹⁰₃₉Y + ⁰₋₁β+ ν̄ₑ

(c) Element 8 is oxygen: ¹⁸₉F →¹⁸₈O + ⁰₊₁β+ νₑ

(d) Only α changes A: fall in A = 232 − 208 = 24, so 6 α decays, which alone would take Z to 90 − 12 = 78. Each β⁻ raises Z by 1: 82 − 78 = 4. 6 α and 4 β⁻.

Random: it is impossible to predict which nucleus will decay next, or when a particular nucleus will decay. Every nucleus of the same isotope has the same probability of decaying in the next second, however long it has existed.

Spontaneous: decay is not triggered by anything outside the nucleus; its rate is unaffected by temperature, pressure, chemical combination or physical state.

Evidence of randomness: repeated equal-time counts from a long-lived source scatter about a mean.

If each of N nuclei has probability λΔt of decaying in a short time Δt, the expected number decaying is λNΔt: the rate of decay is proportional to the number left.

For large N the fluctuations average out and the sample follows a smooth exponential; for a few nuclei the pattern is ragged. A detector catches only a fraction of the emissions, so the count rate is smaller than the activity but proportional to it for a fixed arrangement.

Both are in the formulae list; the minus sign shows that N decreases. With A in Bq, λ must be in s⁻¹, so convert a half-life in hours, days or years to seconds first.

These are in the formulae list. Put N = N₀/2 at t = t_(1/2): ½ = e^(-λt_(1/2)), so λt_(1/2) = ln 2. After n half-lives the fraction left is (½)ⁿ (not in the list: learn it). The same equations work for N, A, the corrected count rate and the mass of isotope left, since all are proportional to N.

To find a half-life from a graph, read the time for the activity to fall from any value to half of it; repeat from two or three starting points and average.

WORKED EXAMPLE · QUESTION

A patient is injected with technetium-99m, a gamma emitter with a half-life of 6.0 hours; the activity at injection is 400 MBq. (a) Calculate the decay constant in s⁻¹. (b) Calculate the number of nuclei injected. (c) Calculate the activity 24 hours later. (d) Calculate the time for the activity to fall to 1.0 MBq.

SOLUTION

(a) λ= 0.693 ÷(6.0 ×3600): 3.21 ×10⁻⁵ s⁻¹.

(b) N = A/λ= 4.00 ×10⁸ ÷3.21 ×10⁻⁵: 1.2 ×10¹³ (1.25 × 10¹³).

(c) 24 h is exactly 4 half-lives: 400 ×(½)⁴: 25 MBq.

(d) t = ln(A₀/A)/λ= ln(400 ÷1.0) ÷3.21 ×10⁻⁵ = 1.87 ×10⁵ s: 52 hours. Shortcut: t = t_(1/2) ×ln(A₀/A)/ln 2 = 6.0 ×8.64 h.

WORKED EXAMPLE · QUESTION

A sample contains 2.0 µg of iodine-131 (atom mass 131 u, half-life 8.0 days; 1 u = 1.66 × 10⁻²⁷ kg). (a) Calculate the activity. (b) Calculate how long it takes for the activity to fall to 5.0% of this value.

SOLUTION

(a) N = 2.0 ×10⁻⁹ ÷(131 ×1.66 ×10⁻²⁷) = 9.20 ×10¹⁵; λ= 0.693 ÷(8.0 ×24 ×3600) = 1.00 ×10⁻⁶ s⁻¹; A = λN: 9.2 ×10⁹ Bq (mass given to 2 s.f.).

(b) t = t_(1/2) ×ln(1/0.050)/ln 2 = 8.0 ×ln 20 ÷0.693: 35 days.

An exponential curve is hard to judge by eye. Taking natural logs turns it into a straight line:

These are in the formulae list too (they are the natural logs of the exponential forms). Compared with y = mx + c, plot ln A (or ln C) against t: gradient = -λ, intercept = ln A₀. A straight line confirms exponential decay. Label the axis ln (C / s⁻¹), since a log has no unit.

WORKED EXAMPLE · QUESTION

Corrected count rates from a short-lived sample, every 40 s:

(a) Explain why ln C against t should be a straight line. (b) Use the graph to determine the decay constant and half-life. (c) Determine the initial count rate.

SOLUTION

(a) Taking logs of C = C₀e^(-λt) gives ln C = -λt + ln C₀, the form y = mx + c with constant m = -λ.

(b) Gradient = -1.80 ÷200 = -9.0 ×10⁻³ s⁻¹, so λ= 9.0 ×10⁻³ s⁻¹; t_(1/2) = 0.693 ÷9.0 ×10⁻³: 77 s.

(c) Intercept ln C₀ = 4.16, so C₀ = e^(4.16): 64 s⁻¹.

Check: the table halves from 64.0 to about 32 s⁻¹ in about 80 s, consistent with 77 s.

Carbon dating

Cosmic rays produce carbon-14 in the upper atmosphere (β⁻, half-life 5730 years), which mixes into atmospheric carbon dioxide. Living things keep exchanging carbon with their surroundings, so the activity per gram of their carbon stays steady. After death no new carbon-14 is taken in and the activity per gram falls exponentially; comparing an old sample with living material gives the time since death.

Limitations: it assumes past atmospheric carbon-14 levels were the same (in practice corrected with samples of known age); very old samples have so little activity that background and random fluctuations swamp it, limiting the method to roughly 50 000 years; contamination with modern carbon gives wrong ages.

WORKED EXAMPLE · QUESTION

Carbon from living wood has an activity of 0.25 Bq per gram; carbon from a wooden tool has 0.084 Bq per gram. Half-life of carbon-14 = 5730 years. (a) Calculate the age of the tool. (b) The sample contains 5.0 g of carbon. Explain why measuring its activity accurately is difficult.

SOLUTION

(a) λ= 0.693 ÷5730 = 1.21 ×10⁻⁴ year⁻¹ (no need for seconds). t = ln(0.25 ÷0.084) ÷1.21 ×10⁻⁴ = 1.091 ÷1.21 ×10⁻⁴: 9.0 ×10³ years.

(b) Activity = 5.0 ×0.084 = 0.42 Bq: fewer than one count per second even if every emission were detected, no more than a typical background rate. The corrected count is small with a large random uncertainty, so long counting times and shielding are needed.

Medical tracers

A tracer is a radioactive isotope attached to a chemical the body takes to the organ of interest; a camera outside the body builds an image from the radiation.

Gamma emitter: γ leaves the body and, being weakly ionising, does least damage on the way. An α emitter would be absorbed inside and do most harm.

Half-life of a few hours: long enough to reach the organ and complete the scan, short enough that the activity soon becomes negligible. Technetium-99m (6.0 h) is the standard example.

The daughter should be stable or very long-lived and the chemical non-toxic. Positron emitters such as fluorine-18 are used in PET scans: the two gamma photons from each annihilation travel in opposite directions and locate the source.

Core Practical 15 finds how the gamma count rate through lead depends on the thickness of lead.

With the source locked away, record the background over several minutes (e.g. 10 min).

Clamp the GM tube and set the sealed source a fixed distance d from its window. Record the count over a fixed time (e.g. 100 s) with no lead.

Measure each sheet's thickness with a micrometer at several points and average.

Add sheets one at a time, recording the count for each total thickness x; repeat and average.

Correct every count rate for background, then plot ln C against x.

Control variables: same source, distance (gamma intensity also falls with distance), counting time and GM tube settings. Longer counts give more reliable rates; steepest and least steep acceptable lines give the gradient's uncertainty.

Safety: handle the source with tongs, pointing away from people, and return it to its lead-lined box promptly; keep exposure short and distant; lead is toxic, so wear gloves or wash hands.

If each millimetre of lead absorbs the same fraction of the photons reaching it, the corrected count rate falls exponentially: C = C₀e^(-μx), with μ a constant for lead and this gamma energy. It is the same mathematics as decay with time: ln C against x has gradient -μ, and the half-thickness ln 2/μ plays the part of the half-life.

WORKED EXAMPLE · QUESTION

Background = 0.45 s⁻¹; counts taken for 100 s at each thickness.

(a) Show how C for 12 mm was calculated. (b) Use the ln graph to determine μ. (c) Calculate the thickness that halves the count rate. (d) Estimate the thickness needed to reduce C to 1.0% of its value with no lead.

SOLUTION

(a) C = 2122 ÷100 - 0.45: 20.8 s⁻¹.

(b) Gradient = -1.16 ÷20 = -0.058 mm⁻¹, so μ= 0.058 mm⁻¹.

(c) x_(1/2) = 0.693 ÷0.058: 12 mm.

(d) x = ln(100)/μ= 4.61 ÷0.058: 79 mm (about 8 cm). The table agrees: every 12 mm roughly halves C (42.4 → 20.8 → 10.6 s⁻¹).

3 Binding energy, fission and fusion, and the standard model (Unit 4)

Why nuclei are unstable and the energy released in a decay come from mass deficit and ΔE = c²Δm. In β⁻ decay a down quark becomes an up quark (udd → uud); neutrinos conserve lepton number; a positron annihilates with an electron to give two gamma photons.

Capacitors (Unit 4)

Discharge follows the same mathematics: Q = Q₀e^(-t/RC), time to halve RCln 2, and ln Q against t is a straight line, just as t_(1/2) = ln 2/λ here.

Photons (Unit 2) and practical skills (Unit 6)

Gamma photon energies follow E = hf and are quoted in MeV; the log graphs, gradients and uncertainties of Core Practical 15 are exactly the skills assessed in practical work.

1. What fraction of a radioactive isotope remains after three half-lives?

2. An isotope has a half-life of 30 minutes. Calculate its decay constant in s⁻¹.

3. A nucleus emits a β⁺ particle. What happens to its nucleon number and its proton number?

4. A source has an activity of 3.0 × 10⁵ Bq and a half-life of 2.5 days. What is its activity after 10 days?

5. Why does the count rate from a long-lived source vary from one 10 s count to the next?

1. (½)³ = ⅛ = 0.125 of the original nuclei.

2. λ= 0.693 ÷(30 ×60) = 3.85 ×10⁻⁴ s⁻¹.

3. The nucleon number is unchanged and the proton number decreases by 1: a proton has changed into a neutron, emitting a positron and an electron neutrino.

4. 10 days = 4 half-lives, so A = 3.0 ×10⁵ ×(½)⁴ = 1.9 ×10⁴ Bq (1.88 × 10⁴ Bq).

5. Decay is random: the number of nuclei decaying in each interval fluctuates about the mean, even though the probability of decay per nucleus is constant. The background fluctuates in the same way.