E.3 Radioactive decay

Define an isotope

Isotopes are atoms of the same element with the same number of protons but different numbers of neutrons. They share chemical identity but can have different physical properties.

Read the numbers

The proton number $Z$ stays fixed within an element. Different isotopes have different nucleon numbers $A$, so their neutron numbers $N=A-Z$ differ.

Common trap

Do not define isotopes only as atoms with different A and Z. The same proton number is essential; otherwise the atoms are different elements.

Define mass defect

A bound nucleus has less mass than the separated protons and neutrons that form it. The missing mass is the mass defect $\Delta m$, associated with the energy released when the nucleus forms.

Convert mass to binding energy

Use $E_b=\Delta mc^2$. If $\Delta m$ is in unified atomic mass units, the convenient conversion is approximately $931.5\,\mathrm{MeV}/c^2$ per u, giving energy directly in MeV.

Worked example — mass defect

If separated nucleons have total mass $4.0320\,\mathrm{u}$ and the nucleus has mass $4.0015\,\mathrm{u}$, then $\Delta m=0.0305\,\mathrm{u}$. Hence $E_b=(0.0305)(931.5)=28.4\,\mathrm{MeV}$. The positive result is the energy needed to separate the nucleus, and the same energy magnitude was released when it formed.

Interpret the sign

Binding energy is the energy required to separate the nucleons completely, and the same amount is released when the bound nucleus forms. It is positive as a required or released energy magnitude.

Common trap

Do not multiply a mass difference in u by c² again after using 931.5 MeV per u; that conversion already includes the mass–energy relation.

Read the curve

Binding energy per nucleon rises for light nuclei, reaches a broad maximum for medium-mass nuclei, then decreases gradually for very heavy nuclei. The curve compares average nuclear stability per nucleon, not total binding energy.

Predict energy release

Fusion of light nuclei can move products upward toward the maximum. Fission of very heavy nuclei can also move products upward. In either case, the increase in binding energy per nucleon corresponds to released energy.

Sketch the trend

Show a rise from the light-nucleus region, a maximum between roughly A=50 and A=100, and a slow decline for larger A. Exact numerical values are not required for the qualitative graph.

Common trap

Do not claim that the heaviest nucleus is most stable simply because it has the largest total binding energy. Use binding energy per nucleon to compare stability.

Use E=mc²

A change in rest mass corresponds to energy through $E=mc^2$. In a nuclear reaction, compare the total mass before and after to find the mass converted into released or absorbed energy.

Worked example — energy from a mass decrease

For $\Delta m=2.0\times10^{-12}\,\mathrm{kg}$, $\Delta E=(2.0\times10^{-12})(3.00\times10^8)^2=1.8\times10^5\,\mathrm{J}$. A smaller total rest mass of the products means this energy is released.

Compare energy yields

Energy released per reaction is proportional to mass converted. Energy released per unit mass also depends on the converted fraction: divide the energy from one reaction by the mass of fuel involved.

Track the system

Mass–energy equivalence applies to the mass difference of the defined reaction system. Do not compare only the total mass of the reactants without accounting for products.

Common trap

Do not confuse a large energy per reaction with a large energy per unit mass. The question’s denominator determines the comparison.

Describe the force

The strong nuclear force is attractive between nucleons at nuclear separations and has a very short range. It can bind protons and neutrons despite the electrostatic repulsion between protons.

Explain stability

At short distances the strong force can dominate, while the electromagnetic force is repulsive and long range. A stable nucleus requires the attractive nuclear interaction to overcome proton repulsion within the nucleus.

Keep the range distinction

The strong force does not act as a long-range force between separated nuclei. Its short range is why increasing nuclear size makes stability more difficult.

Common trap

Do not call the strong force repulsive between nucleons in the binding explanation, and do not confuse it with the weak nuclear interaction.

Treat each nucleus independently

Radioactive decay is spontaneous and random: the exact nucleus and instant of decay cannot be predicted. For a large sample, however, the fraction decaying per unit time follows a stable statistical law.

Separate random from law-like

Random decay does not mean the activity is random noise. The expected number of decays is predictable from the number of undecayed nuclei and the decay constant.

Apply the statistical model

You cannot identify which nucleus will decay next. But if two large samples contain the same nuclide and the second has twice as many undecayed nuclei, its expected activity is twice as large. Individual unpredictability and ensemble predictability coexist.

Common trap

Do not claim that randomness prevents prediction of half-life or activity. It prevents prediction of an individual decay, not the ensemble behaviour.

Alpha decay

Alpha decay emits a ${}^{4}_{2}\mathrm{He}$ nucleus. The parent’s nucleon number decreases by 4 and proton number decreases by 2.

Beta decay

In beta-minus decay, a neutron becomes a proton and an electron is emitted, so $A$ is unchanged and $Z$ increases by 1. In beta-plus decay, a proton becomes a neutron and a positron is emitted, so $A$ is unchanged and $Z$ decreases by 1.

Gamma decay

Gamma emission changes the nucleus from an excited state to a lower energy state. Neither $A$ nor $Z$ changes.

Common trap

Do not change A during beta decay, and do not treat gamma emission as a change of element.

Balance alpha decay

Write ${}^{A}_{Z}X\rightarrow{}^{A-4}_{Z-2}Y+{}^{4}_{2}\mathrm{He}$. Check both A and Z on the two sides.

Balance beta decay

For beta-minus use $^{A}_{Z}X\rightarrow{}^{A}_{Z+1}Y+{}^{0}_{-1}e+\bar{\nu}_e$. For beta-plus use $^{A}_{Z}X\rightarrow{}^{A}_{Z-1}Y+{}^{0}_{+1}e+\nu_e$. Gamma emission adds $^{0}_{0}\gamma$ after an excited daughter.

Balance a reaction

Conserve total nucleon number and charge. For uranium-235 absorbing a neutron and producing xenon-140 and strontium-94, the remaining nucleon number identifies the emitted neutrons.

Common trap

Do not omit the neutrino or antineutrino when the syllabus asks for a complete beta-decay equation, and do not balance A while leaving charge unbalanced.

Choose the correct particle

Beta-minus decay emits an electron and an electron antineutrino: $n\rightarrow p+e^-+\bar{\nu}_e$. Beta-plus decay emits a positron and an electron neutrino: $p\rightarrow n+e^++\nu_e$ (inside a nucleus).

Check lepton number

An electron has lepton number $+1$, so the accompanying antineutrino has $-1$. A positron has $-1$, so the accompanying neutrino has $+1$. Each beta reaction therefore keeps the initial total lepton number at zero.

Common trap

Do not swap the beta partners: $\beta^-$ pairs with $\bar{\nu}_e$, while $\beta^+$ pairs with $\nu_e$. The continuous beta spectrum is treated separately in the HL objective E.3.18.

Alpha radiation

Alpha particles are heavy and doubly charged. They interact strongly with matter, so they are highly ionizing but have low penetration and a short range in air.

Beta radiation

Beta particles are much lighter and singly charged. They are moderately ionizing and more penetrating than alpha particles, but can be deflected by electric and magnetic fields.

Gamma radiation

Gamma photons are neutral and travel at the speed of light in vacuum. They are weakly ionizing compared with alpha and beta, but have the greatest penetration.

Common trap

Do not rank penetration and ionization in the same order. The usual qualitative order is alpha > beta > gamma for ionization and gamma > beta > alpha for penetration.

Define activity

Activity is the number of nuclear decays per unit time, measured in becquerels: one Bq is one decay per second. As the number of undecayed nuclei falls, activity falls.

Use half-life steps

After each half-life, half of the remaining nuclei survive: $N=N_0(1/2)^n$, where $n=t/T_{1/2}$ is the number of half-lives elapsed.

Track count rate

If detector efficiency and background are unchanged, count rate is proportional to activity. Apply the same half-life scaling to the net count rate.

Common trap

Do not halve the original amount repeatedly without using the remaining amount, and do not confuse count rate with the number of nuclei when background is present.

Use integer half-lives

If activity changes from $A_0$ to $A$, use $A/A_0=(1/2)^n$ to find the number of half-lives $n$. For example, a fall to one-eighth means three half-lives.

Worked example — integer half-lives

A net count rate falls from $640\,\mathrm{s^{-1}}$ to $80\,\mathrm{s^{-1}}$ in 18 h. Since $80/640=1/8=(1/2)^3$, three half-lives elapsed. Therefore $T_{1/2}=18/3=6.0\,\mathrm{h}$.

Find the half-life

Once $n$ is known, divide the elapsed time by $n$: $T_{1/2}=t/n$. This is often quicker and clearer than starting with the exponential form.

Check the direction

A decay interval must reduce activity or count rate. If the calculated half-life or number of half-lives implies growth, revisit the ratio.

Common trap

Do not call a drop to one-eighth “one half-life”; half-life is the time for one factor of one-half.

Separate sample and background

A detector count rate can include decays from the sample plus background radiation. The measured rate is $R_{measured}=R_{sample}+R_{background}$.

Subtract before analysing

Estimate the background count rate with the source absent or from the long-time plateau, then calculate $R_{net}=R_{measured}-R_{background}$. Use the net rate for half-life comparisons.

Worked example — subtract, decay, restore

A detector reads $260\,\mathrm{Bq}$ with a $20\,\mathrm{Bq}$ background. The initial net rate is $240\,\mathrm{Bq}$. After four half-lives it is $240/16=15\,\mathrm{Bq}$, so the detector reads $15+20=35\,\mathrm{Bq}$.

Interpret a non-zero limit

If the measured rate approaches a non-zero constant, the remaining signal may be background radiation or a systematic detector contribution. The sample activity itself may have continued toward zero.

Common trap

Do not fit a half-life directly to a count rate that still contains background; the offset distorts the decay curve.

Use nuclear stability as evidence

Protons repel electrically, yet stable nuclei exist. This requires an additional attractive interaction between nucleons that is strong enough at nuclear distances.

Use scattering evidence

At high energies, deviations from Rutherford scattering show that the electrostatic model is incomplete at close range. The change is evidence for the strong interaction becoming relevant.

State the evidence precisely

Evidence supports a short-range strong force; it does not by itself provide a complete potential-energy curve or a long-range attraction between nuclei.

Common trap

Do not use “the nucleus is stable” as a complete explanation. State which observed fact requires an attractive force and how its range differs from electromagnetic repulsion.

Light stable nuclei

For small proton numbers, stable nuclei tend to have similar numbers of neutrons and protons, so $N\approx Z$.

Heavy stable nuclei

As $Z$ increases, proton–proton electromagnetic repulsion grows. Stable heavy nuclei therefore need extra neutrons to add strong-force binding without adding proton repulsion, so $N>Z$.

Read the stability band

The line of stable nuclides bends above $N=Z$ at larger $Z$. Nuclei on either side can decay toward the band, often through beta decay.

Common trap

Do not say every stable nucleus has more neutrons than protons. The approximation $N\approx Z$ is useful for light nuclei.

Read the heavy-nucleus trend

Above approximately $A\approx60$, binding energy per nucleon is broadly similar but slowly decreases as nucleon number increases. The increasing proton repulsion makes very heavy nuclei less tightly bound per nucleon.

Use the approximation carefully

“Approximately constant” does not mean identical for every nuclide. Use the trend to compare regions and to explain why fission of very heavy nuclei can release energy.

Common trap

Do not turn the broad plateau into a new maximum at large A. The main maximum is in the medium-mass region, followed by a gradual decline.

Use nuclear spectra

Alpha and gamma radiation can contain discrete energies. Since $E=hf$, fixed photon frequencies correspond to fixed energy differences between nuclear states.

Infer nuclear quantization

A line spectrum means the nucleus changes between allowed, discrete energy levels rather than a continuous range. Different transitions produce different alpha or gamma energies.

Use multiple routes

If two decay routes lead to the same final state, their energy relationships can reveal shared intermediate nuclear levels. Treat the routes as evidence about the level structure.

Common trap

Do not infer continuous nuclear energies from a continuous beta spectrum; beta continuity has a different explanation involving the neutrino.

Read the beta spectrum

Beta particles from one radioactive transition are emitted with a continuous range of kinetic energies, from nearly zero up to a maximum.

Use energy sharing

The beta particle and neutrino share the decay energy in variable proportions. The neutrino therefore explains why the beta particle does not always receive one fixed energy.

Common trap

Do not attribute the continuous spectrum to a continuous set of nuclear levels. Alpha and gamma line spectra show the contrasting discrete-level behaviour.

Use the exponential law

The number of undecayed nuclei after time $t$ is $N=N_0e^{-\lambda t}$. The same factor applies to the remaining mass when each daughter product is stable and the sample starts pure.

Worked example — arbitrary time

For $N_0=1.0\times10^{10}$, $\lambda=0.0126\,\mathrm{s^{-1}}$ and $t=60\,\mathrm{s}$, $N=(1.0\times10^{10})e^{-(0.0126)(60)}=4.70\times10^9$ nuclei. About $5.30\times10^9$ parent nuclei have decayed.

Find daughter amount

If every parent decay produces one daughter nucleus, the number formed is $N_{daughter}=N_0-N$. Define whether the question asks for remaining parent or accumulated daughter before substituting.

Control units

Use seconds when $\lambda$ is in $\mathrm{s^{-1}}$. Convert minutes, days or years before evaluating the exponential.

Common trap

Do not use $N_0e^{-\lambda t}$ for daughter amount directly; it gives the parent nuclei remaining.

Define lambda

The decay constant $\lambda$ is the probability per unit time that an individual undecayed nucleus will decay, in the small-time interval sense. Its unit is inverse time.

Use the approximation

When $\lambda\Delta t$ is very small, $\lambda\Delta t$ approximates the probability that a particular nucleus decays during $\Delta t$. The exact exponential law applies over longer intervals.

Separate lambda from activity

$\lambda$ describes a property of the nuclide. Activity $A$ describes the whole sample and depends on how many nuclei remain: $A=\lambda N$.

Common trap

Do not call lambda the number of decays per second of the whole sample; that is activity.

Use the activity relation

Activity is the decay rate: $A=\lambda N$. Combining this with the decay law gives $A=\lambda N_0e^{-\lambda t}$.

Worked example — number of nuclei

If $A=2.5\times10^5\,\mathrm{Bq}$ and $\lambda=1.8\times10^{-6}\,\mathrm{s^{-1}}$, then $N=A/\lambda=(2.5\times10^5)/(1.8\times10^{-6})=1.4\times10^{11}$ undecayed nuclei.

Find N first

For a sample mass $m$, find the number of nuclei using $N=(m/M)N_A$ before multiplying by $\lambda$. Use the isotopic molar mass and consistent units.

Track time dependence

Activity falls with the same exponential factor as the number of undecayed nuclei. If $t=0$, use $A_0=\lambda N_0$.

Common trap

Do not multiply lambda by sample mass directly. Convert mass to a number of nuclei first.

Use the half-life relation

Half-life and decay constant are related by $T_{1/2}=\frac{\ln2}{\lambda}$. A larger decay constant means a shorter half-life.

Worked example — convert time first

For $T_{1/2}=6.0\,\mathrm{h}=2.16\times10^4\,\mathrm{s}$, $\lambda=0.693/(2.16\times10^4)=3.21\times10^{-5}\,\mathrm{s^{-1}}$. The inverse-second unit matches a probability rate.

Convert units first

If half-life is given in days, hours or years but lambda is required in $\mathrm{s^{-1}}$, convert the time to seconds before dividing $\ln2$ by it.

Check the scale

The product $\lambda T_{1/2}$ should equal approximately 0.693. Use this as a quick unit and order-of-magnitude check.

Common trap

Do not use $1/\lambda$ as the half-life; it is the characteristic time and differs by the factor $\ln2$.