5.4 - Nuclear Decay

Learning objectives

5.4.133Nuclear binding energyUnderstand the concept of nuclear binding energy and be able to use the equation ΔE = c2Δm in calculations of nuclear mass (including mass deficit) and energy 5.4.134Atomic mass unitUse the atomic mass unit (u) to express small masses and convert between this and SI units 5.4.135Nuclear fusion, fission and binding energyUnderstand the processes of nuclear fusion and fission with reference to the binding energy per nucleon curve 5.4.136Fusion conditionsUnderstand the mechanism of nuclear fusion and the need for very high densities of matter and very high temperatures to bring about and maintain nuclear fusion 5.4.137Background radiationUnderstand that there is background radiation and how to take appropriate account of it in calculations 5.4.138Nuclear radiation propertiesUnderstand the relationships between the nature, penetration, ionising ability and range in different materials of nuclear radiations (alpha, beta and gamma)5.4.139Nuclear equationsBe able to write and interpret nuclear equations given the relevant particle symbols 5.4.140Core Practical 15 - gamma absorption by leadCORE PRACTICAL 15: Investigate the absorption of gamma radiation by lead 5.4.141Spontaneous and random nuclear decayUnderstand the spontaneous and random nature of nuclear decay 5.4.142Half-life and radioactive decay equationsDetermine half-life graphically and use A = λN, dN/dt = −λN, λ = ln 2/t½, N = N0e^(−λt), and A = A0e^(−λt), including the corresponding logarithmic equations.

Nuclear binding energy is the energy required to separate a nucleus completely into its individual protons and neutrons. The bound nucleus has less mass than those free nucleons, and this mass deficit corresponds to the binding energy.

$\Delta m=Zm_p+(A-Z)m_n-m_{nucleus}\qquad E_b=\Delta mc^2$

Use mutually consistent nuclear masses, calculate the positive mass deficit, convert it to kilograms if energy is required in joules, then multiply by $c^2$ . Binding energy per nucleon is the total binding energy divided by nucleon number $A$ .

A mass deficit of $0.528\,\mathrm{u}$ corresponds to about $492\,\mathrm{MeV}$ using $1\,\mathrm{u}c^2\approx931.5\,\mathrm{MeV}$ . For a 56-nucleon nucleus this is about $8.8\,\mathrm{MeV}$ per nucleon.

Binding energy is positive even though the bound nucleus has lower mass-energy than separated nucleons. Do not confuse total binding energy with binding energy per nucleon, and do not mix atomic masses containing electrons with bare nuclear masses without accounting consistently for electrons.

One atomic mass unit is defined as one twelfth of the mass of a neutral carbon-12 atom: $1\,\mathrm{u}=1.6605\times10^{-27}\,\mathrm{kg}$ . It is a mass unit, suited to atoms, nuclei and particles.

Conversion	Operation
u to kg	multiply by $1.6605\times10^{-27}$
kg to u	divide by $1.6605\times10^{-27}$
mass difference in u to energy	use $1\,\mathrm{u}c^2\approx931.5\,\mathrm{MeV}$

A mass of $4.00\,\mathrm{u}$ is $4.00(1.6605\times10^{-27})=6.64\times10^{-27}\,\mathrm{kg}$ . A mass loss of $0.0020\,\mathrm{u}$ releases about $0.0020(931.5)=1.86\,\mathrm{MeV}$ .

The symbol u is not an energy unit. Only after multiplying a mass by $c^2$ may its energy equivalent be stated in joules or electronvolts; keep enough significant figures until the final result.

The binding-energy-per-nucleon curve rises steeply for light nuclei, reaches a maximum near iron, then falls gradually for very heavy nuclei. A higher position means nucleons are more tightly bound on average.

Process	Nuclear change	Why energy is released
fusion	light nuclei combine into a heavier nucleus	products have greater binding energy per nucleon
fission	a very heavy nucleus splits into medium-mass nuclei	products have greater binding energy per nucleon

Calculate the total binding energy of reactants and products, not just their vertical coordinates. Energy released equals the increase in total binding energy; equivalently, the products have lower total mass and the mass difference appears as kinetic energy and radiation.

Movement towards the curve's peak can release energy; not every fusion or fission process does. A light, already tightly bound nucleus does not release energy by fission merely because it can be split conceptually.

Fusion joins positively charged light nuclei. Before the strong nuclear force can bind them, the nuclei must approach extremely closely despite electrostatic repulsion between their charges.

Condition	Causal role
very high temperature	gives nuclei high kinetic energy, so more collisions approach closely enough for the strong force to act
very high density	places many nuclei in a small volume, increasing collision rate
sustained confinement	prevents the hot plasma dispersing or cooling before enough fusion occurs

Fusion products can transfer energy to the surrounding plasma, helping maintain temperature, but a continuing reaction requires energy production to compete successfully with energy losses. In stars, gravitational pressure provides density and confinement.

High temperature does not remove electrostatic repulsion; it broadens the distribution of nuclear kinetic energies. High density alone raises collision frequency but cannot make low-energy collisions reach nuclear-force range.

A detector records ionising radiation even when the investigated source is absent. This background comes from the environment and detector surroundings, so a source measurement contains both source and background contributions.

$R_{source}=R_{measured}-R_{background}$

Measure background with the same detector, geometry and counting interval but without the source. Convert both readings to count rates before subtracting, or subtract counts measured for equal times. Repeat or count for longer to reduce the relative effect of random fluctuations.

If 1260 counts are recorded in $120\,\mathrm{s}$ with the source and 180 counts in $120\,\mathrm{s}$ without it, the corrected source rate is $(1260-180)/120=9.0\,\mathrm{s^{-1}}$ .

Do not add background or subtract a one-minute background count directly from a ten-minute source count. Background itself fluctuates, so a corrected value can carry uncertainty and should be based on a representative measurement.

Radiation	Nature and charge	Ionising ability	Range / penetration
alpha	helium nucleus, charge $+2e$	very strong	short range in air; stopped by paper or skin
beta-minus	electron, charge $-e$	moderate	several metres in air; stopped by thin aluminium
gamma	photon, no charge	weak per interaction	long range; intensity reduced by thick lead or concrete

A strongly ionising particle transfers energy frequently to matter, so it loses kinetic energy rapidly and has a short range. Gamma photons interact less frequently, so a beam is more penetrating; absorption is probabilistic and reduces intensity rather than giving every photon one fixed stopping depth.

Alpha is dangerous mainly when an emitter enters the body; gamma can irradiate from outside because it penetrates tissue. Beta lies between them. Shield choice must therefore match both the radiation and exposure geometry.

Penetrating does not mean non-ionising: gamma is ionising but interacts less often. Lead reduces gamma intensity but does not guarantee that all photons are stopped, and alpha's short range does not make an internal alpha source harmless.

In a nuclear equation, the sum of nucleon numbers $A$ and the sum of proton numbers $Z$ must each be the same before and after the reaction. Use the supplied particle symbols, including any coefficients, to identify a missing nuclide or emission.

Emission	Symbol	Daughter change
alpha	$^{4}_{2}\alpha$	$A-4$ , $Z-2$
beta-minus	$^{0}_{-1}e$	$A$ unchanged, $Z+1$
gamma	$^{0}_{0}\gamma$	$A$ and $Z$ unchanged

In $^{96}_{42}\mathrm{Mo}+{}^{2}_{1}\mathrm{H}\rightarrow{}^{95}_{43}\mathrm{Tc}+x{}^{1}_{0}n$ , nucleon number gives $96+2=95+x$ , so $x=3$ ; proton number already gives $42+1=43$ .

Balance $A$ and $Z$ separately; they are not ordinary algebraic subscripts. In beta-minus decay the emitted electron has $A=0$ and $Z=-1$ , allowing the daughter's proton number to rise by one while nucleon number stays fixed.

Clamp a sealed gamma source and detector at fixed separation and alignment. Measure background for a long interval, then place increasing measured thicknesses of lead between source and detector. Record counts for the same sufficiently long time at every thickness.

Control	Reason
fixed source-detector geometry	prevents inverse-square changes being mistaken for absorption
equal counting time and background correction	makes corrected count rates comparable
repeat readings / long intervals	reduces fractional random uncertainty
tongs, minimum handling time, maximum distance	applies time-distance-shielding radiation safety

$R=R_0e^{-\mu x}\qquad\ln R=\ln R_0-\mu x$

Subtract background before taking logarithms. Plot corrected $\ln R$ against lead thickness $x$ ; a straight line supports exponential attenuation and its gradient is $-\mu$ . The half-value thickness is the thickness that halves corrected count rate.

Do not move the detector as plates are added or take a logarithm of raw counts containing background. Gamma absorption is statistical, so thickness reduces intensity continuously rather than creating an exact all-or-none stopping point.

Radioactive decay is spontaneous: an unstable nucleus decays without needing an external trigger, and ordinary changes of temperature, pressure or chemical state do not control the event.

It is random: the exact nucleus that will decay next and its decay time cannot be predicted. Each undecayed nucleus of one isotope has the same constant probability of decay per unit time, independent of how long it has already survived.

Scale	What can be predicted
one nucleus	only a probability, not an exact decay time
many identical nuclei	average activity and exponential decrease
repeated short counts	values fluctuate statistically around a trend

Random does not mean the decay probability changes unpredictably or that the ensemble has no mathematical pattern. Spontaneous means untriggered, whereas random means individually unpredictable; the two words describe different features.

$A=\lambda N,\quad\frac{dN}{dt}=-\lambda N,\quad N=N_0e^{-\lambda t},\quad A=A_0e^{-\lambda t},\quad\lambda=\frac{\ln2}{t_{1/2}}$

The decay constant $\lambda$ is the probability per unit time for one nucleus; activity $A$ is decays per second in becquerels. Half-life is the time for $N$ or $A$ to fall to half its current value, not the time for all nuclei to decay.

Representation	Half-life or decay-constant evidence
corrected $A$ -against- $t$ curve	read several time intervals for successive halvings
$\ln N$ against $t$	$\ln N=\ln N_0-\lambda t$ ; gradient $=-\lambda$
$\ln A$ against $t$	$\ln A=\ln A_0-\lambda t$ ; gradient $=-\lambda$

For half-life 6.0 h, $\lambda=\ln2/6.0=0.116\,\mathrm{h^{-1}}$ . After 18 h, three half-lives have passed, so activity is $A_0/8$ , agreeing with $A=A_0e^{-\lambda t}$ .

Use consistent time units for $t$ and $\lambda$ , correct count data for background before graphical work, and use natural logarithms for the stated linear equations. Equal absolute activity decreases do not take equal times; equal fractional decreases do.

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5.4 - Nuclear Decay

Learning objectives

Binding energy comes from a nucleus's mass deficit

The atomic mass unit makes nuclear masses manageable

Fusion and fission release energy by moving up the binding curve

Fusion needs hot, dense matter to sustain nuclear collisions

Subtract background before attributing counts to a source

Ionisation strength controls radiation range and penetration

Nuclear equations conserve nucleon and proton numbers

Core Practical 15: measure gamma absorption by lead

Radioactive decay is spontaneous for a nucleus and random in time

Half-life and decay constant describe one exponential process

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