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23. Nuclear physics A2

Syllabus
9702–2028–2029
Section
23
Level
A2

23.1 Mass defect and nuclear binding energy

Syllabus
9702–2028–2029
Topic
23.1
Level
A2

Mass and energy are equivalent: use E=mc² with consistent units

Mass is a form of stored energy. A system of mass m has rest energy E=mc²; when its mass changes by Δm, the corresponding energy change is ΔE=Δmc².

E=mc2,m=E/c2,c=3.00×108ms−1UseminkgforEinJ.E=mc², m=E/c², c=3.00×10⁸ m s⁻¹ Use m in kg for E in J.

For 1.00 u=1.66×10⁻²⁷ kg, E=(1.66×10⁻²⁷)(3.00×10⁸)²=1.49×10⁻¹⁰ J=931 MeV.

E=mc² is rest-energy equivalence, not the classical kinetic-energy formula. Convert g or u to kg before using SI c, unless using the established 1 u≈931 MeV/c² bridge.

Balance nuclear equations by conserving nucleon number A and proton number Z

In nuclide notation ᴬ_ZX, A is nucleon number and Z is proton number. Across a nuclear reaction, the total A and total Z are each conserved.

Write two balances: ΣA(left)=ΣA(right) and ΣZ(left)=ΣZ(right). Solve both before identifying the missing nuclide or particle from its A and Z.

¹74N+24He→817O+11HA:14+4=17+1;Z:7+2=8+1¹⁴₇N + ⁴₂He → ¹⁷₈O + ¹₁H A: 14+4=17+1; Z: 7+2=8+1

Useful entries: neutron ¹₀n, proton ¹₁p, beta-minus ⁰₋₁e, beta-plus ⁰₊₁e and gamma ⁰₀γ.

Do not conserve the number of written symbols or electrons as in chemistry. A beta-minus particle contributes A=0 and Z=−1 to the equation.

Mass defect becomes the binding energy that holds a nucleus together

Mass defect Δm is the mass of the constituent nucleons when separated to infinity minus the mass of the bound nucleus.

Binding energy is the minimum energy required to separate all nucleons of the nucleus to infinity. The same energy is released when the nucleus forms from those separated nucleons.

Δm=Zmp+(A−Z)mn−mnucleusEbinding=Δmc2;bindingenergypernucleon=Ebinding/AΔm=Zmp+(A−Z)mn−mnucleus E_binding=Δmc²; binding energy per nucleon=E_binding/A

A positive mass defect means the bound state has lower mass-energy. More binding energy must be supplied to dismantle that nucleus into the same free nucleons.

Mass is not lost from conservation laws: the lower rest mass corresponds to energy released. State the infinitely-separated reference state in definitions.

Sketch binding energy per nucleon: steep rise, iron peak, slow decline

Horizontal axis: nucleon number A, from about 1 to 250. Vertical axis: binding energy per nucleon, usually in MeV.

Starting near A=1, draw a steep rise to one broad maximum near A≈56 (iron/nickel region), then a much shallower continuous decrease toward heavy nuclei.

The right-hand branch stays well above zero at A≈250. Light fusion candidates lie left of the peak; heavy fission or alpha-decay candidates lie right of it.

Greater binding energy per nucleon means nucleons are, on average, more tightly bound and more energy per nucleon is required to separate the nucleus.

Do not draw a symmetric bell curve or return the heavy-nucleus end to zero. The graph is binding energy per nucleon, not total binding energy.

Fusion joins light nuclei; fission splits a heavy nucleus

Nuclear fusion is the joining of two light nuclei to form one heavier nucleus, usually requiring very high temperature so nuclei can approach despite electrostatic repulsion.

Nuclear fission is the splitting of one heavy nucleus into two or more lighter nuclei, commonly after the heavy nucleus absorbs a neutron; further neutrons may be released.

Fusion: light + light → heavier. Fission: heavy → lighter + lighter (and often neutrons). Both conserve total nucleon number and proton number.

Energy release is a consequence, not the definition. Do not call alpha decay fission merely because a heavy nucleus emits a small particle.

Fission and fusion release energy by moving nuclei toward greater binding per nucleon

A reaction releases energy when its products have greater total binding energy than its reactants. The products then have lower total rest mass; the difference appears as released energy.

Fusion starts with light nuclei left of the A≈56 peak. Joining them moves the product toward the peak, increasing binding energy per nucleon and total binding energy.

Fission starts with a very heavy nucleus right of the peak. Splitting it into medium-mass products also moves nucleons toward the peak and increases average binding.

releasedenergy=totalbindingenergy(products)−totalbindingenergy(reactants)wheretotalbindingenergy=A×(bindingenergypernucleon)foreachnucleusreleased energy = total binding energy(products) − total binding energy(reactants) where total binding energy = A×(binding energy per nucleon) for each nucleus

Compare totals when nuclei have different A. Energy is released because binding energy increases and mass-energy decreases—not because binding energy is consumed.

Calculate nuclear energy release from the reaction mass decrease

Δm=Σmreactants−ΣmproductsEreleased=Δmc2Δm=Σm_reactants−Σm_products E_released=Δmc²

For ²₁H+²₁H→⁴₂He with masses 2.013553 u and 4.001505 u: Δm=2(2.013553)−4.001505=0.025601 u.

Convert Δm=(0.025601)(1.66×10⁻²⁷)=4.25×10⁻²⁹ kg, then E=(4.25×10⁻²⁹)(3.00×10⁸)²=3.82×10⁻¹² J per helium nucleus.

For 1.00 mol of helium nuclei, multiply by 6.02×10²³: E=2.30×10¹² J. For reaction rate R, power P=RE_reaction.

If binding energies are given, calculate Ereleased=ΣBEproducts−ΣBEreactants. If binding energy per nucleon is given, multiply each value by that nucleus's A before summing.

Include every coefficient in the mass sum. A positive reactant-minus-product Δm means release; keep per-reaction, per-nucleus, per-mole and per-second quantities distinct.

23.2 Radioactive decay

Syllabus
9702–2028–2029
Topic
23.2
Level
A2

Count-rate fluctuations reveal the random timing of nuclear decays

Measure counts in many equal time intervals under unchanged conditions: the results fluctuate above and below a mean or smooth decay trend rather than repeating exactly.

These irregular fluctuations are evidence that individual decay events occur at unpredictable times. Only the expected behavior of a large population is stable.

Longer counting intervals usually collect more events, so the fractional fluctuation is smaller even though decay remains random.

Scatter is not automatically evidence that λ or the detector changes. Compare repeated equal intervals and allow for background count before interpreting the trend.

Radioactive decay is random and spontaneous—but the terms mean different things

Random: it is impossible to predict which particular unstable nucleus will decay next or exactly when that nucleus will decay.

Spontaneous: decay occurs without being triggered and its probability is unaffected by external or environmental factors such as temperature, pressure and chemical state.

Although one event is unpredictable, each nucleus of the same isotope has the same constant decay probability per unit time, so a large sample has a predictable statistical law.

Spontaneous does not mean immediate, and random does not mean there is no stable probability or no predictable average behavior.

Activity is the sample decay rate; decay constant is probability per unit time

Activity A is the number of nuclear disintegrations per unit time. Its unit is the becquerel: 1 Bq=1 s⁻¹.

Decay constant λ is the probability per unit time that one undecayed nucleus decays. Its unit is reciprocal time, such as s⁻¹ or min⁻¹.

A=λN=−dN/dtA=λN=−dN/dt

If λ=2.0×10⁻⁴ s⁻¹ and N=3.0×10¹², then A=6.0×10⁸ Bq. As N falls, A falls in the same proportion.

Activity is the source disintegration rate; received count rate can be smaller because detector efficiency and geometry are not 100%. Match the time unit of λ to the required rate unit.

Half-life is the constant time for a decay quantity to halve

Half-life t½ is the time taken for the number of undecayed nuclei—or equivalently the activity—to decrease to half its current value.

afternhalf−lives:x=x0(1/2)n1→1/2→1/4→1/8→…after n half-lives: x=x₀(1/2)ⁿ 1 → 1/2 → 1/4 → 1/8 → …

For a particular isotope, each halving interval has the same duration because the decay constant is constant; the absolute amount lost in each interval becomes smaller.

Half-life is not the time for every nucleus to decay and not the lifetime of one particular nucleus. For measured count rate, subtract constant background before halving.

Decay constant and half-life are reciprocal through ln 2

λ=ln2/t½=0.693/t½t½=0.693/λλ=ln2/t½=0.693/t½ t½=0.693/λ

A larger λ means a greater decay probability per unit time and therefore a shorter half-life. The unit of λ is the reciprocal of the time unit used for t½.

For t½=110 min=(110)(60)=6600 s, λ=0.693/6600=1.05×10⁻⁴ s⁻¹.

If λ=0.048 min⁻¹, t½=0.693/0.048=14 min.

Do not mix minutes with an s⁻¹ answer. The numerator is dimensionless, so λ and t½ must have reciprocal units.

A constant decay probability produces exponential decay

dx/dt=−λx⇒x=x0e(−λt)xmaybeN,activityAorbackground−correctedreceivedcountrate.dx/dt=−λx ⇒ x=x₀e^(−λt) x may be N, activity A or background-corrected received count rate.

Because A=λN, the rate of decrease is proportional to the amount still undecayed. As x becomes smaller, the magnitude of the negative gradient also becomes smaller.

Sketch: start at (0,x₀), decrease continuously with a steepest negative gradient initially, curve upward as the slope becomes less negative, and approach zero asymptotically without crossing it.

For x₀=180 Bq, λ=0.048 min⁻¹ and t=8.4 min: x=180e^(−0.048×8.4)=120 Bq.

lnx=lnx0−λtAgraphoflnxagainsttisastraightlineofgradient−λandinterceptlnx0.ln x=ln x₀−λt A graph of ln x against t is a straight line of gradient −λ and intercept ln x₀.

If measured count C includes constant background B, apply the decay law to C−B, not C. Add B back only if the question asks for the measured count.