Skip to content
MathsGenie logo
Quick links
Open app

Course home

  1. A Level
  2. Physics Eduqas
  3. Revision guides

Nuclear Decay

What you'll learn

  • Why nuclear decay is spontaneous and random, but still predictable statistically.
  • How alpha, beta and gamma radiation differ in nature, penetration, ionisation and deflection.
  • How to correct count-rate measurements for background radiation.
  • How to use half-life, activity, decay constant and exponential decay equations.

Nuclear notation: describing a nucleus

A nucleus is made of protons and neutrons, collectively called nucleons. The two important numbers are:

  • Proton number, ZZZ: the number of protons in the nucleus. This determines the element.
  • Nucleon number, AAA: the total number of protons and neutrons.
Definition

Nuclide notation

A nucleus of element X is written as ZAX{}^A_Z\text{X}ZA​X, where AAA is the nucleon number and ZZZ is the proton number.

For example, 614C{}^{14}_6\text{C}614​C is carbon with 6 protons and 14 nucleons, so it has 8 neutrons.

In a decay equation, the original unstable nucleus is the parent nucleus. The nucleus produced after decay is the daughter nucleus.

Spontaneous nuclear decay

Some nuclei are unstable. They can become more stable by emitting radiation.

Definition

Spontaneous nuclear decay

Nuclear decay is spontaneous: it happens without any external trigger. For a particular isotope, decay is random for an individual nucleus but has a constant probability per unit time for each undecayed nucleus.

This means you cannot predict when one particular nucleus will decay, but you can predict the behaviour of a very large sample. Decay is not significantly affected by temperature, pressure, chemical state or physical state.

Definition

Ionising radiation

Ionising radiation has enough energy to remove electrons from atoms, forming ions. Alpha, beta and gamma radiation are all ionising.

Alpha, beta and gamma radiation

The three main types in this topic are:

  • Alpha radiation, α\alphaα: a helium nucleus, 24He{}^4_2\text{He}24​He, containing 2 protons and 2 neutrons.
  • Beta minus radiation, β−\beta^-β−: a fast-moving electron, −10e{}^0_{-1}\text{e}−10​e, emitted when a neutron changes into a proton.
  • Beta plus radiation, β+\beta^+β+: a fast-moving positron, +10e{}^0_{+1}\text{e}+10​e, emitted when a proton changes into a neutron.
  • Gamma radiation, γ\gammaγ: a high-energy electromagnetic photon emitted by an excited nucleus.

The comparison below links the nature of each radiation type to its penetration, ionising ability and deflection in an electric field.

Comparison of alpha, beta and gamma radiation, including penetration, ionising power and electric field deflection

Key Idea

Nature controls penetration

Alpha particles are massive and have charge +2, so they ionise strongly and lose energy quickly. Gamma photons are uncharged, so they ionise weakly and are much more penetrating. Beta particles are intermediate.

Nuclear transformation equations

In nuclear equations, the total nucleon number AAA and proton number ZZZ must balance on both sides.

Common decay equations are:

α:ZAX→Z−2A−4Y+24Heβ−:ZAX→Z+1AY+−10e+νˉβ+:ZAX→Z−1AY++10e+νγ:ZAX∗→ZAX+γ\begin{aligned} \alpha:\quad {}^A_Z\text{X} &\to {}^{A-4}_{Z-2}\text{Y} + {}^4_2\text{He} \\ \beta^-:\quad {}^A_Z\text{X} &\to {}^A_{Z+1}\text{Y} + {}^0_{-1}\text{e} + \bar{\nu} \\ \beta^+:\quad {}^A_Z\text{X} &\to {}^A_{Z-1}\text{Y} + {}^0_{+1}\text{e} + \nu \\ \gamma:\quad {}^A_Z\text{X}^{*} &\to {}^A_Z\text{X} + \gamma \end{aligned}α:ZA​Xβ−:ZA​Xβ+:ZA​Xγ:ZA​X∗​→Z−2A−4​Y+24​He→Z+1A​Y+−10​e+νˉ→Z−1A​Y++10​e+ν→ZA​X+γ​

Here, ν\nuν is a neutrino and νˉ\bar{\nu}νˉ is an antineutrino. In gamma decay, the star means the nucleus is in an excited state.

Tip

Balancing nuclear equations

Add the top numbers to conserve AAA, and add the bottom numbers to conserve ZZZ. Gamma emission changes neither AAA nor ZZZ.

Example

Writing an alpha decay equation

Radium-226 decays by alpha emission. Write the nuclear equation.

  1. Alpha emission releases 24He{}^4_2\text{He}24​He, so the daughter nucleus has nucleon number 226−4=222226 - 4 = 222226−4=222.

  2. The proton number decreases by 2, so the daughter has proton number 88−2=8688 - 2 = 8688−2=86.

  3. Element 86 is radon, so the equation is:

88226Ra→86222Rn+24He {}^{226}_{88}\text{Ra} \to {}^{222}_{86}\text{Rn} + {}^4_2\text{He} 88226​Ra→86222​Rn+24​He

Distinguishing alpha, beta and gamma radiation

You can identify radiation experimentally using:

  • Absorbers: paper stops alpha; a few millimetres of aluminium stops beta; thick lead or concrete reduces gamma.
  • Range in air: alpha travels a few centimetres, beta a few metres, gamma many metres.
  • Electric or magnetic fields: charged alpha and beta deflect in opposite directions; gamma is undeflected.
  • Cloud chamber tracks: alpha gives short, thick, straight tracks; beta gives thinner, more irregular tracks; gamma mainly produces secondary tracks.
Common Mistake

Gamma is reduced, not completely stopped

Do not say “lead stops gamma completely”. Gamma intensity is reduced by shielding, but some photons may still pass through.

Background radiation

Background radiation is ionising radiation always present in the environment, from sources such as cosmic rays, rocks, radon gas and medical sources.

A detector such as a Geiger-Müller tube records a count rate, usually in counts per second. To find the count rate due to your source, subtract the background count rate.

corrected count rate=measured count rate−background count rate\text{corrected count rate} = \text{measured count rate} - \text{background count rate}corrected count rate=measured count rate−background count rate
Example

Correcting a count rate for background

A source gives 1500 counts in 60 s. Background radiation gives 240 counts in 120 s. Find the corrected count rate.

  1. Convert the source measurement into a rate:
150060 s=25.0 counts s−1 \frac{1500}{60\ \text{s}} = 25.0\ \text{counts s}^{-1} 60 s1500​=25.0 counts s−1
  1. Convert the background measurement into a rate:
240120 s=2.00 counts s−1 \frac{240}{120\ \text{s}} = 2.00\ \text{counts s}^{-1} 120 s240​=2.00 counts s−1
  1. Subtract the rates:
25.0 counts s−1−2.00 counts s−1=23.0 counts s−1 25.0\ \text{counts s}^{-1} - 2.00\ \text{counts s}^{-1} = 23.0\ \text{counts s}^{-1} 25.0 counts s−1−2.00 counts s−1=23.0 counts s−1
Common Mistake

Subtract rates, not unmatched counts

If the time intervals are different, do not subtract the raw counts directly. Convert both measurements to count rates first.

Because decay is random, counts fluctuate. For a count nnn, the approximate uncertainty is n\sqrt{n}n​, so longer counting times reduce the percentage uncertainty.

Half-life, activity and decay constant

Definition

Half-life

The half-life, T1/2T_{1/2}T1/2​, is the mean time taken for the number of undecayed nuclei, or the activity, to fall to half its original value.

Definition

Activity and becquerel

The activity, AAA, is the rate at which nuclei decay. The unit is the becquerel, Bq, where one becquerel means one decay per second.

Activity is related to the number of undecayed nuclei by:

A=λNA = \lambda NA=λN

where λ\lambdaλ is the decay constant. It is the probability per unit time that an undecayed nucleus will decay, so its unit is s−1\text{s}^{-1}s−1 if time is in seconds.

Common Mistake

Activity is not the same as count rate

Activity is the number of decays per second in the source. Count rate is what the detector records, and depends on background, detector efficiency, distance and shielding.

Example

Calculating activity from a decay constant

A sample contains 8.0×10128.0 \times 10^{12}8.0×1012 undecayed nuclei. Its decay constant is 4.0×10−8 s−14.0 \times 10^{-8}\ \text{s}^{-1}4.0×10−8 s−1. Find its activity.

  1. Choose the activity equation:
A=λN A = \lambda N A=λN
  1. Substitute the values:
A=(4.0×10−8 s−1)(8.0×1012) A = \left(4.0 \times 10^{-8}\ \text{s}^{-1}\right)\left(8.0 \times 10^{12}\right) A=(4.0×10−8 s−1)(8.0×1012)
  1. Calculate and give the unit:
A=3.2×105 Bq A = 3.2 \times 10^5\ \text{Bq} A=3.2×105 Bq

Exponential decay law

Radioactive decay follows an exponential pattern because the same fraction of the remaining nuclei decays in each equal time interval.

N=N0e−λtN = N_0 e^{-\lambda t}N=N0​e−λt

and, because A=λNA = \lambda NA=λN,

A=A0e−λtA = A_0 e^{-\lambda t}A=A0​e−λt

You can also use the half-life form:

N=N02xN = \frac{N_0}{2^x}N=2xN0​​ A=A02xA = \frac{A_0}{2^x}A=2xA0​​

where xxx is the number of half-lives elapsed. It does not have to be a whole number.

The graph below shows the exponential shape and the straight-line form obtained by plotting ln⁡N\ln NlnN against time.

Exponential radioactive decay graph with half-life markings and linear ln N against time inset

Key Idea

Equal times, equal fractions

In radioactive decay, equal time intervals remove equal fractions, not equal amounts. That is why the graph curves towards zero instead of being a straight line.

Example

Using a non-integer number of half-lives

A sample has initial activity 960 Bq and half-life 4.0 h. Find its activity after 10 h.

  1. Find the number of half-lives:
x=10 h4.0 h=2.5 x = \frac{10\ \text{h}}{4.0\ \text{h}} = 2.5 x=4.0 h10 h​=2.5
  1. Use the half-life form:
A=A02x=960 Bq22.5 A = \frac{A_0}{2^x} = \frac{960\ \text{Bq}}{2^{2.5}} A=2xA0​​=22.5960 Bq​
  1. Calculate the activity:
A=1.7×102 Bq A = 1.7 \times 10^2\ \text{Bq} A=1.7×102 Bq

Deriving the link between half-life and decay constant

At one half-life, t=T1/2t = T_{1/2}t=T1/2​ and N=N02N = \frac{N_0}{2}N=2N0​​.

N02=N0e−λT1/212=e−λT1/2ln⁡2=λT1/2λ=ln⁡2T1/2\begin{aligned} \frac{N_0}{2} &= N_0 e^{-\lambda T_{1/2}} \\ \frac{1}{2} &= e^{-\lambda T_{1/2}} \\ \ln 2 &= \lambda T_{1/2} \\ \lambda &= \frac{\ln 2}{T_{1/2}} \end{aligned}2N0​​21​ln2λ​=N0​e−λT1/2​=e−λT1/2​=λT1/2​=T1/2​ln2​​

This same relationship works for activity because activity is proportional to NNN.

Example

Finding half-life from a log graph

A graph of ln⁡(A/Bq)\ln(A / \text{Bq})ln(A/Bq) against time has gradient −3.5×10−4 s−1-3.5 \times 10^{-4}\ \text{s}^{-1}−3.5×10−4 s−1. Find the half-life.

  1. For a log graph of activity against time, the gradient is −λ-\lambda−λ, so:
λ=3.5×10−4 s−1 \lambda = 3.5 \times 10^{-4}\ \text{s}^{-1} λ=3.5×10−4 s−1
  1. Use the half-life relationship:
T1/2=ln⁡2λ T_{1/2} = \frac{\ln 2}{\lambda} T1/2​=λln2​
  1. Substitute and calculate:
T1/2=0.6933.5×10−4 s−1=2.0×103 s T_{1/2} = \frac{0.693}{3.5 \times 10^{-4}\ \text{s}^{-1}} = 2.0 \times 10^3\ \text{s} T1/2​=3.5×10−4 s−10.693​=2.0×103 s

Specified practical work

Dice analogy for radioactive decay

This models random decay without using radioactive material.

  1. Start with a large number of dice. Each die represents an undecayed nucleus.
  2. Roll all the dice. Choose one face, such as a 6, to represent “decayed”.
  3. Remove all dice showing that face. Count and record the remaining dice, NNN.
  4. Repeat for many rolls and plot NNN against roll number.
  5. Repeat the experiment or combine class data to reduce random fluctuations.

For one “decay face”, the probability of decay per roll is 16\frac{1}{6}61​, so the expected fraction remaining after each roll is 56\frac{5}{6}65​. The model predicts:

N=N0(56)rN = N_0\left(\frac{5}{6}\right)^rN=N0​(65​)r

where rrr is the number of rolls.

Common Mistake

Limits of the dice model

Dice decay happens in discrete rolls, while real nuclear decay is continuous in time. The model is useful because it shows randomness, constant probability and exponential decrease.

Gamma intensity with distance

For a point gamma source, intensity decreases with distance because the radiation spreads out over a larger area. After subtracting background, the count rate is approximately proportional to 1r2\frac{1}{r^2}r21​ if absorption in air is negligible.

The practical layout below shows how to vary distance, subtract background and test for an inverse-square relationship.

Gamma radiation intensity with distance practical setup and corrected count-rate graphs

Good practice includes:

  • Measure background count rate with no source present, then subtract it from every reading.
  • Keep the source, detector and ruler aligned.
  • Count for long enough to reduce percentage uncertainty.
  • Repeat readings, especially at large distances where count rates are low.
  • Plot corrected count rate against 1r2\frac{1}{r^2}r21​; a straight line supports an inverse-square relationship.
Tip

Radiation safety

Use tongs, maximise distance, minimise exposure time, use suitable shielding, and store sources in their containers when not in use.

Exam technique

In the exam

  1. Balance nuclear equations by checking both AAA and ZZZ on each side before naming the daughter nucleus.
  2. For count-rate data, correct for background before using half-life or exponential decay equations.
  3. Keep time units consistent: if λ\lambdaλ is in s−1\text{s}^{-1}s−1, use time in seconds.
  4. For graph questions, remember that a plot of ln⁡N\ln NlnN or ln⁡A\ln AlnA against time has gradient −λ-\lambda−λ.
Self review

Check yourself

  • Why can we predict the half-life of a large sample but not the decay time of one nucleus?
  • What changes happen to AAA and ZZZ in alpha, beta minus, beta plus and gamma decay?
  • How would you use absorber data and background correction to identify an unknown radiation source?
PreviousNext

How was this guide?

Teach Genie

Review Nuclear Decay by teaching Genie

Teach it back in your own words, spot gaps, and remember it better.

Start teaching
Genie and Baby Genie

Lesson

Recap your knowledge with an interactive lesson

8 minute activity

Start lesson

A nucleus is described by proton number ZZZ and nucleon number AAA. In 614C{}^{14}_{6}\text{C}614​C, the 6 identifies carbon, and the number of neutrons is 14−6=814 - 6 = 814−6=8.

Some nuclei are unstable and decay spontaneously, which means no outside trigger is needed. The exact decay time of one nucleus is random, but for a large sample the overall decay pattern is predictable because each undecayed nucleus has a constant probability of decaying per unit time.

Alpha, beta, and gamma are all ionising radiation, so they can remove electrons from atoms and form ions. Temperature, pressure, and chemical state do not significantly change the decay rate.

Flashcards

Remember key concepts with flashcards

4 flashcards

Practice flashcards

What does it mean for nuclear decay to be spontaneous?

Nuclear Decay Revision Guide

  1. A Level
  2. /Physics
  3. /Nuclear Decay