13. Nuclei

Nuclear changes reveal a direct connection between mass, binding and energy. Begin with particle counts, then balance the complete system. Numerical constants and tabulated masses in worked examples are rounded classroom values.

The neutron’s identification by James Chadwick established a neutral nuclear constituent with a mass close to the proton’s. Nuclear mass measurements and reaction studies then showed how binding-energy differences account for large releases of energy.

A nucleus contains nearly all of an atom’s mass in a very small volume, yet its physics differs from ordinary chemistry.

Chemical reactions mainly rearrange electrons while leaving nuclear identities unchanged. Nuclear changes can alter the number of protons or neutrons and can release energies much larger per event. Their explanation needs nuclear composition, strong binding and mass–energy accounting.

Protons and neutrons are nucleons. The proton number identifies the element; the total nucleon number identifies an isotope together with the element. The same element may therefore have several isotopes with similar electron chemistry but different nuclear stability.

A nuclear model must explain why positive protons can remain close together despite electrostatic repulsion, why some nuclides are stable and others decay, and why both splitting some heavy nuclei and joining some light nuclei can release energy. Binding energy per nucleon provides a useful organising idea.

Two samples have identical proton number but different neutron number. Are they the same element, and must they have the same stability?

Element identity follows proton number. Neutrons change mass number and nuclear binding without changing that identity.

Result: They are isotopes of the same element; their stability can differ.

1. What particles are counted as nucleons?
2. Which number fixes the chemical element?

Answers

1. Protons and neutrons.
2. The proton number Z.

A nuclide symbol is a compact particle count; a measured atomic mass is not simply the sum of free nucleon masses.

Write a nuclide as $\!{}^{A}_{Z}X$, where Z is proton number, A is mass number, and neutron number is $N=A-Z$. The nucleus has charge $+Zq_{\rm e}$. A neutral atom has Z electrons outside the nucleus. An ion differs in electron count, not nuclear proton count.

Isotopes share Z and differ in neutron number. Isobars share A but have different Z. Isotones share neutron number. These categories compare counts; they do not imply that different nuclides have the same mass or stability.

The atomic mass unit u is defined from the mass of a neutral carbon-12 atom: one u is one twelfth of that atom’s mass. A mass number is an integer count, while an atomic mass in u is a measured quantity. They are related in scale but are not interchangeable.

Chadwick’s experiments showed that a penetrating neutral radiation could eject protons from hydrogen-rich material. Energy and momentum accounting supported a neutral particle with mass comparable to a proton, rather than an explanation entirely in terms of high-energy gamma rays. The neutron resolved major difficulties in nuclear mass bookkeeping.

A neutral nuclide has A=31 and Z=15. Find proton, neutron and electron counts. It then loses two electrons. What changes?

Protons equal Z. Neutrons equal A−Z. Neutrality initially requires Z electrons. Electron removal changes charge but not A or Z.

Result: Initially 15 protons, 16 neutrons and 15 electrons. The ion has 13 electrons and charge $+2q_{\rm e}$.

1. Are A and the atomic mass in u exactly the same quantity?
2. What is neutron number for A=27 and Z=13?

Answers

1. No. A counts nucleons; atomic mass is a measured mass.
2. 14.

Nuclear radii grow slowly with nucleon count, suggesting an approximately constant nuclear density.

A useful empirical estimate is $R=R_0A^{1/3}$, with $R_0$ about $1.2\,\mathrm{fm}$ and $1\,\mathrm{fm}=10^{-15}\,\mathrm m$. The nucleus has a diffuse quantum surface, so “radius” depends somewhat on how it is measured. The relation is an average model, not an exact hard-sphere boundary for every nuclide.

For a spherical estimate, volume is $4\pi R^3/3$. Substituting $R^3=R_0^3A$ makes volume proportional to A. Nuclear mass is approximately proportional to A as well, so their ratio is approximately independent of A. This is evidence of saturation: adding nucleons adds volume rather than compressing the entire nucleus without limit.

The radius of an atom is much larger than its nuclear radius. A drawing that shows both on the same convenient scale greatly enlarges the nucleus. Scattering probes suitable to nuclear length scales can reveal nuclear size; the Coulomb turning distance of a projectile is not automatically the same thing.

Estimate the radius for A=125 using $R_0=1.2\,\mathrm{fm}$. Compare its radius with A=1 in the same formula.

The cube root of 125 is 5. The radius ratio is therefore 5, whereas the volume ratio is 125.

Result: $R=6\,\mathrm{fm}$ and $R_{125}/R_1=5$.

1. If A is multiplied by eight, how does estimated radius change?
2. What assumption is used when deriving volume from the radius relation?

Answers

1. It doubles.
2. A roughly spherical average shape.

A bound nucleus has less mass than its separated constituents; the difference is the energy needed to separate it.

Rest energy is $E_0=mc^2$. If a complete isolated system loses energy to its surroundings, its mass decreases by $\Delta E/c^2$. Mass and energy are not two separately conserved stores that can be counted twice; total energy, including rest energy, must be balanced.

The useful conversion is $1\,\mathrm u\,c^2\approx931.5\,\mathrm{MeV}$. A tiny mass difference can therefore correspond to a large energy per nuclear event. Use rest masses for the complete initial and final sets of particles when computing a reaction Q-value.

Using bare nuclear mass $M_{\rm nuc}$, the mass defect is $\Delta m=Zm_p+(A-Z)m_n-M_{\rm nuc}$. Binding energy is $B=\Delta mc^2$. This is energy that must be supplied to separate the nucleus into free, stationary nucleons. Formation of the bound nucleus can release the same energy when the overall energy and momentum balance permits.

With neutral atomic masses, a convenient approximation is $\Delta m=Zm_{\rm H}+(A-Z)m_n-M_{\rm atom}$. The electron rest masses cancel between Z neutral hydrogen atoms and the neutral atom. Differences in electronic binding energy are negligible for the classroom nuclear-energy scale, but should be included for precision work.

Use $m_{\rm H}=1.007825\,\mathrm u$, $m_n=1.008665\,\mathrm u$, and helium-4 atomic mass $4.002603\,\mathrm u$. Estimate its binding energy with $1\,\mathrm u\,c^2=931.5\,\mathrm{MeV}$.

There are two protons and two neutrons. Use the atomic-mass formula so electron masses cancel.

$\Delta m=2m_{\rm H}+2m_n-M_{\rm He}$.

Multiply the mass defect in u by the energy conversion, then divide by four for energy per nucleon.

Result: $\Delta m=0.030377\,\mathrm u$; $B=28.2962\,\mathrm{MeV}$; $B/A=7.074\,\mathrm{MeV}$ per nucleon.

Binding energy per nucleon generally rises rapidly among light nuclei, reaches a broad maximum in the iron–nickel region, then decreases gradually towards very heavy nuclei. The detailed curve is not smooth: pairing and nuclear shell structure produce exceptions. A higher binding energy per nucleon usually means more energy is required per nucleon to separate that nucleus, but it alone does not determine every possible decay or reaction.

Both fusion of suitable light nuclei and fission of suitable heavy nuclei can lead to products with greater total binding energy. Compare the sum of all product binding energies with the initial sum; do not compare only one product or rely on mass number alone.

1. Does a bound nucleus weigh more or less than its free constituent nucleons?
2. Is total binding energy or binding energy per nucleon used in a complete reaction-energy balance?

Answers

1. Less, by its binding energy divided by c squared.
2. Total binding energy of all reactants and products; per-nucleon values help compare trends.

Electrostatic repulsion alone cannot hold a nucleus together; a short-range attractive interaction is required.

The effective nuclear interaction between nucleons is strongly attractive over a short range, while a very short-distance repulsive component prevents unlimited collapse. Its range is comparable to nuclear dimensions. A nucleon interacts strongly mainly with nearby nucleons, which helps explain approximately constant nuclear density and saturation of binding.

The strong interaction is not an electrostatic attraction between opposite charges: neutrons are electrically neutral, and proton–proton nuclear attraction must compete with Coulomb repulsion. To a useful approximation the nuclear part shows similar behaviour for proton–proton, neutron–neutron and proton–neutron pairs when compared in appropriate quantum states, but spin and other quantum details matter.

Coulomb repulsion between protons extends across the nucleus and grows in importance for heavy nuclei. Extra neutrons can contribute nuclear attraction without adding electric repulsion, but an arbitrarily large neutron excess is not stable. The observed pattern of stable nuclides reflects several competing energy contributions.

Why can nuclear matter have roughly constant density even as A grows, while very heavy nuclei become harder to stabilise?

Short-range saturation limits each nucleon’s nearby strong interactions. Long-range Coulomb repulsion acts among the protons across the nucleus.

Result: Adding nucleons mainly adds volume, but accumulating many protons increases the competing Coulomb energy.

1. Why do neutrons help bind a nucleus without adding proton–proton repulsion?
2. What does short-range saturation mean?

Answers

1. They participate in nuclear interactions but carry no electric charge.
2. Each nucleon’s strong binding interactions mainly involve nearby neighbours, rather than every nucleon equally.

A single unstable nucleus decays unpredictably, but a large population follows a predictable statistical law.

Let $\lambda_{\rm d}$ be the decay constant, the probability per unit time in the exponential-decay model. The expected decrease satisfies $dN/dt=-\lambda_{\rm d}N$. Integrating gives $N=N_0e^{-\lambda_{\rm d}t}$. Activity is the expected decay rate $\mathcal A=\lambda_{\rm d}N$, measured in becquerels, with $1\,\mathrm{Bq}=1\,\mathrm{s^{-1}}$.

Set $N=N_0/2$ to find half-life: $T_{1/2}=\ln2/\lambda_{\rm d}$. The mean lifetime is $\tau=1/\lambda_{\rm d}$. Thus $N/N_0=2^{-t/T_{1/2}}$. Half-life is not the scheduled waiting time of an individual nucleus, and the exponential law does not predict when the very last nucleus will decay.

In beta decay, the emitted electron or positron is created in the weak-interaction process; it is not an electron that was stored in a classical nuclear orbit. Neutrinos carry energy, momentum and angular momentum as required. A daughter nucleus may remain excited and then emit gamma radiation.

A parent population initially contains 6400 nuclei and has half-life 6 hours. Find the expected parent count after 18 hours.

The elapsed time is three half-lives, so multiply the original count by $(1/2)^3$.

Result: $N=800$ nuclei expected. This is a statistical expectation, not a guarantee for an individual sample.

A sample contains $2.0\times10^6$ parent nuclei and has $\lambda_{\rm d}=0.002\,\mathrm{s^{-1}}$. Find its initial activity.

Activity is decay constant times population; use seconds throughout.

Result: $\mathcal A=4000\,\mathrm{Bq}$.

1. After two half-lives, what fraction of the initial parent nuclei remains?
2. Does gamma emission alone change the element?

Answers

1. $0.25$.
2. No. It changes the nuclear energy state without changing A or Z.

Nuclear energy release is determined by the mass of the complete initial and final systems.

For a reaction with consistent particle masses, $Q=(M_{\rm initial}-M_{\rm final})c^2$. Positive Q means energy can be released as product kinetic energy and radiation; negative Q means energy must be supplied. Momentum must also be conserved, so reaction thresholds can exceed a simple rest-energy deficit when a target is initially at rest.

In fission, a heavy nucleus separates into lighter fragments, often with neutrons and gamma radiation. Many fragment combinations are possible, so a single fission equation is one channel rather than a unique outcome. An incoming neutron can excite a suitable nucleus and make fission more likely.

A chain reaction is possible if emitted neutrons induce further fissions. Its growth depends on neutron production, absorption and escape. A controlled reactor maintains a useful steady rate through engineered neutron balance and heat removal. A moderator slows neutrons in reactor designs that need thermal neutrons; it is distinct from an absorber used for control.

One possible balanced channel is $\!{}^{235}_{92}\mathrm U+{}^{1}_{0}\mathrm n\rightarrow{}^{141}_{56}\mathrm{Ba}+{}^{92}_{36}\mathrm{Kr}+3{}^{1}_{0}\mathrm n$. Check both totals: $235+1=141+92+3$ and $92=56+36$. These conservation checks do not by themselves predict the channel probability or energy.

In fusion, light nuclei combine into a more tightly bound product system. Positive nuclei must approach close enough for nuclear attraction to matter. Coulomb repulsion opposes that approach. High temperature gives a distribution of kinetic energies, and quantum tunnelling helps fusion occur at energies below a classical barrier estimate.

Stellar hydrogen burning proceeds through reaction sequences rather than one collision that directly joins four isolated protons. The net conversion to helium releases energy because the complete products have lower mass. Some energy leaves as neutrinos. Temperature, density and confinement determine whether energy production can outpace losses in a laboratory plasma.

A hypothetical nuclear reaction has an initial total mass exceeding the final total mass by $0.018\,\mathrm u$. Find Q using $931.5\,\mathrm{MeV/u}$.

The initial-minus-final difference is positive, so the reaction is exoergic in rest-energy accounting.

Multiply the mass difference by the mass-energy conversion.

Result: $Q=16.767\,\mathrm{MeV}$.

Energy conservation never means that missing mass disappears without accounting. Lower final rest energy appears in other energy forms. The masses and binding energies used must refer to the same complete set of particles, including any emitted leptons when relevant.

1. Why can fission of a heavy nucleus release energy?
2. What two conservation totals must be checked in an elementary nuclear reaction equation?

Answers

1. Its products can have greater total binding energy and lower total rest mass.
2. Total nucleon number and electric charge, with all emitted particles included.

• Nuclide notation gives proton number Z, nucleon number A and neutron number $A-Z$.
• A useful radius estimate is $R=R_0A^{1/3}$.
• Binding energy is the separated-constituent mass excess multiplied by $c^2$.
• A simple parent population obeys $N=N_0e^{-\lambda_{\rm d}t}$, $T_{1/2}=\ln2/\lambda_{\rm d}$ and activity $\lambda_{\rm d}N$.
• Reaction energy uses the complete initial-minus-final rest-mass balance.

• Conservation of A and charge is necessary for a simple reaction equation but does not establish its probability.
• Binding energy per nucleon is a trend guide; reaction energy needs the complete total binding or mass balance.
• A decay curve predicts averages and statistical fluctuations remain in finite populations.

Try these newly written questions before reading the answers. Show a diagram, the governing equation, substitutions and units where appropriate.

13.1 A nuclide has Z=26 and A=58. Find neutron number.

13.2 Distinguish isotopes and isobars.

13.3 Why does removing electrons not change isotope identity?

13.4 Estimate nuclear radius for A=64 with $R_0=1.2\,\mathrm{fm}$.

13.5 If mass number rises by a factor of 27, how does the simple radius estimate change?

13.6 Use R∝A^(1/3) to explain roughly constant nuclear density.

13.7 A nucleus has mass defect $0.025\,\mathrm u$. Find binding energy using $931.5\,\mathrm{MeV/u}$.

13.8 A nucleus with A=20 has binding energy $156\,\mathrm{MeV}$. Find binding energy per nucleon.

13.9 Why is atomic hydrogen mass used with neutral atomic masses in a convenient binding formula?

13.10 Does a larger total binding energy always mean greater binding per nucleon?

13.11 State two ways the effective nuclear force differs from electrostatic force.

13.12 Why do very heavy nuclei face increasing Coulomb competition?

13.13 A population of 9600 parent nuclei undergoes four half-lives. Find expected remaining count.

13.14 A sample has half-life $8\,\mathrm s$. Find its decay constant.

13.15 For $\lambda_{\rm d}=0.05\,\mathrm{s^{-1}}$, find mean lifetime.

13.16 A sample activity initially is $1200\,\mathrm{Bq}$. What is it after three half-lives if only that parent contributes?

13.17 A nuclide with A=226 and Z=88 emits one alpha particle. Find daughter A and Z.

13.18 How do A and Z change in beta-minus decay?

13.19 Why does gamma decay not change the element?

13.20 What is wrong with saying that every nucleus decays exactly one half-life after being made?

13.21 A reaction loses $0.012\,\mathrm u$ of total rest mass. Find released energy using $931.5\,\mathrm{MeV/u}$.

13.22 Explain why total energy is conserved even when final rest mass is smaller.

13.23 What distinguishes a moderator from a neutron absorber?

13.24 Why is high temperature relevant to fusion but not proof of net useful energy production?

13.25 Why can both suitable light-nucleus fusion and heavy-nucleus fission release energy?

13.26 Check nucleon and charge totals for $\!{}^{2}_{1}\mathrm H+{}^{3}_{1}\mathrm H\rightarrow{}^{4}_{2}\mathrm{He}+{}^{1}_{0}\mathrm n$.

13.27 Why is an arbitrary cluster of labelled nucleon balls not a literal measured nuclear structure?

Answers

13.1 32 neutrons.

13.2 Isotopes share Z and differ in A; isobars share A and differ in Z.

13.3 Isotope identity depends on the nuclear proton and neutron counts.

13.4 $R=4.8\,\mathrm{fm}$.

13.5 Radius rises by a factor of three.

13.6 Volume is proportional to R³ and therefore to A; mass is also approximately proportional to A, so density is approximately constant.

13.7 $B=23.2875\,\mathrm{MeV}$.

13.8 $7.8\,\mathrm{MeV}$ per nucleon.

13.9 It lets the electron rest masses cancel on both sides, neglecting small electronic binding-energy differences.

13.10 No. Divide by A before making a per-nucleon comparison.

13.11 It is short-ranged and depends on quantum state; it can bind neutral neutrons and includes a short-distance repulsive component.

13.12 Many protons contribute repulsive interactions across the nucleus, while strong binding mainly saturates among near neighbours.

13.13 600 nuclei.

13.14 $\lambda_{\rm d}=\ln2/8=0.086643\,\mathrm{s^{-1}}$.

13.15 $\tau=20\,\mathrm s$.

13.16 $150\,\mathrm{Bq}$.

13.17 $A=222$, $Z=86$.

13.18 A is unchanged; Z increases by one, and an electron and electron antineutrino are emitted.

13.19 The excited nucleus loses energy without changing proton number.

13.20 Half-life describes a statistical population decrease; individual decay times are unpredictable.

13.21 $Q=11.178\,\mathrm{MeV}$.

13.22 The rest-energy decrease appears as kinetic energy, radiation or other accounted energy forms.

13.23 A moderator slows neutrons mainly by scattering; an absorber captures them and removes them from the neutron population.

13.24 It increases energetic collision probabilities; losses, density and confinement must also be controlled.

13.25 Both can increase the total binding energy of the complete product system.

13.26 $A:2+3=5=4+1$; $Z:1+1=2=2+0$.

13.27 Nucleons occupy quantum states, and a fixed ball arrangement omits their dynamics and probability distributions.