Nuclei | ISC Class 12 Physics Notes

This note covers nuclear composition, atomic masses, isotopes, isobars, isotones, nuclear size and density, mass-energy equivalence, mass defect, binding energy, nuclear reactions, pair production and annihilation, fission, nuclear reactors, and fusion in the Sun and stars.

What particles make up a nucleus?

The nucleus is the small central part of an atom containing its positive charge and most, more than 99.9%, of its mass. Its radius is smaller than the atomic radius by a factor of about 10⁴. The atom is therefore almost empty.

A proton is a positively charged nuclear particle. A neutron is electrically neutral and has almost the same mass as a proton. Either particle is called a nucleon. The electrons of an atom lie outside its nucleus.

How are nuclear particles counted?

The atomic number Z counts protons, the neutron number N counts neutrons, and the mass number A counts all nucleons. These are dimensionless whole-number counts, not masses measured in a unit.

A = Z + N

N = A − Z

A nuclide is a nuclear species specified by its proton and neutron numbers. In notation such as ¹⁹⁷₇₉Au, the upper number is A, the lower number is Z, and Au is the chemical symbol for gold. This nucleus contains 79 protons and 118 neutrons.

Let e denote the magnitude of the electron's charge, 1.6 × 10⁻¹⁹ coulomb; the coulomb has symbol C. A proton has charge +e and an electron has charge −e. The nuclear charge is +Ze. A neutral atom has Z electrons, whose combined charge is −Ze.

Note: Mass number counts protons plus neutrons. Atomic number counts protons. Electrons contribute to atomic mass but are not included in either the nuclear mass number or the neutron number.

For deuterium, a hydrogen isotope written ²₁H, Z = 1 and A = 2, so its nucleus contains one proton and one neutron. Tritium, ³₁H, contains one proton and two neutrons. Adding a neutron changes the isotope without changing the element.

How do atomic masses, isotopes, isobars and isotones differ?

The unified atomic mass unit, symbol u, is one-twelfth of the mass of one carbon-12 atom. Its approximate value is 1 u = 1.66 × 10⁻²⁷ kg, where kg means kilogram. The SI unit of mass is the kilogram; SI means the International System of Units.

Atomic mass includes the nucleus and its surrounding electrons. Nuclear mass refers to the nucleus itself. A mass expressed in u need not be a whole number, even though the corresponding mass number is an integer.

Which count stays the same in each relationship?

Isotopes have identical electronic structures as neutral atoms and hence identical chemical behaviour. Their nuclear compositions differ. Isobars and isotones instead compare nuclei of different elements, because their atomic numbers are different.

Most elements are mixtures of two or more isotopes. The atomic mass of an element is a weighted average: each isotope's mass contributes according to its relative abundance, meaning its proportion in the sample. This helps explain why an element's quoted atomic mass may be non-integral.

To classify two nuclides, first read A and Z, then calculate N = A − Z for each. Compare Z for isotopes, A for isobars and N for isotones. Comparing the printed masses alone cannot establish all three relationships.

How are nuclear radius and density related to mass number?

The nuclear radius R represents the size of a nucleus treated as spherical. Scattering experiments using fast electrons allow nuclear sizes to be measured. The radius is related to mass number by the empirical relation below, where R₀ is a constant equal to 1.2 × 10⁻¹⁵ m.

R = R₀∛A

Here ∛A means the cube root of A. The SI unit of length is the metre, symbol m. One femtometre, symbol fm, equals 10⁻¹⁵ m. Thus R₀ = 1.2 fm. Radius increases with the cube root of mass number, rather than directly with mass number.

The volume V of a spherical nucleus is V = 4πR³/3, where π is the circle constant. The density ρ is mass divided by volume. The SI unit of density is kilogram per cubic metre, kg m⁻³.

Derivation: Why is nuclear density nearly independent of A?

1. Let mₙ be the neutron mass, used here as an approximate mass for each nucleon. The nuclear mass is approximately A mₙ because proton and neutron masses are almost equal.
2. Cubing R = R₀∛A gives R³ = R₀³A. Therefore V = 4πR₀³A/3.
3. Divide the approximate mass by this volume: ρ ≈ A mₙ/(4πR₀³A/3). The factor A cancels.

ρ ≈ 3mₙ/(4πR₀³), so nuclear density is nearly independent of mass number. Its approximate value is 2.3 × 10¹⁷ kg m⁻³. The approximation treats the mass per nucleon as constant.

Worked example 1. An iron nucleus has A = 56 and a given mass of 55.85 u, equivalent to 9.27 × 10⁻²⁶ kg. Calculate its density using R₀ = 1.2 × 10⁻¹⁵ m and π ≈ 3.1416.

Formula: V = 4πR₀³A/3; ρ = mass/V.

Substitute: V = (4/3) × 3.1416 × (1.2 × 10⁻¹⁵)³ × 56 m³; ρ = (9.27 × 10⁻²⁶)/V.

Answer: ρ ≈ 2.29 × 10¹⁷ kg m⁻³, or 229,000,000,000,000,000 kg m⁻³.

This near constancy concerns nuclear matter. It does not imply equal densities for ordinary materials. Ordinary matter contains atoms with much empty space, whereas almost all atomic mass is concentrated within the very small nuclear volume.

What does mass-energy equivalence mean?

Mass-energy equivalence means that mass has an associated energy. In Einstein's relation, E is the energy equivalent of mass m, while c is the speed of light in vacuum, approximately 3 × 10⁸ m s⁻¹. Here s denotes the second.

E = mc²

The SI unit of energy is the joule, symbol J. In base units, 1 J = 1 kg m² s⁻². With mass in kilograms and speed in metres per second, the equation gives energy in joules. The SI unit of speed is metre per second.

Mass and energy cannot be treated as separately conserved in these processes. The conservation of mass-energy requires initial and final total energies to be equal, provided the energy associated with mass is included.

Which energy units are useful for nuclei?

An electron volt, symbol eV, is the energy gained by a charge of magnitude e through a potential difference of one volt. A volt, symbol V in this unit context, is one joule per coulomb. Thus 1 eV = 1.602 × 10⁻¹⁹ J.

A megaelectron volt, symbol MeV, equals 10⁶ eV. Nuclear energies are conveniently expressed in MeV. Mass can instead be expressed in MeV/c², because dividing energy by c² gives a mass. These two units describe different quantities.

Worked example 2. Find the energy equivalent of 1 g of substance. Use 1 g = 10⁻³ kg and c = 3 × 10⁸ m s⁻¹, where g denotes gram.

Formula: E = mc².

Substitute: E = 10⁻³ × (3 × 10⁸)² J.

Answer: E = 9 × 10¹³ J, or 90,000,000,000,000 J. This is the energy equivalent of the entire mass, not the energy necessarily released in a particular reaction.

Worked example 3. Calculate the energy equivalent of 1 u using 1 u = 1.6605 × 10⁻²⁷ kg, c = 2.9979 × 10⁸ m s⁻¹ and 1 eV = 1.602 × 10⁻¹⁹ J.

Formula: E = mc².

Substitute: E = 1.6605 × 10⁻²⁷ × (2.9979 × 10⁸)² J.

Answer: E ≈ 1.4924 × 10⁻¹⁰ J ≈ 931.5 MeV. In decimal form, the energy is 0.00000000014924 J. Consequently, 1 u = 931.5 MeV/c².

For dimensional checks, square brackets indicate dimensions: mass has [M], length [L] and time [T]. Energy has [ML²T⁻²], speed [LT⁻¹], volume [L³] and density [ML⁻³]. The M inside dimensional brackets is a base dimension, distinct from a mass symbol used in a calculation.

How are mass defect and binding energy calculated?

A bound nucleus has less mass than its constituent protons and neutrons would have separately. The difference is the mass defect Δm, where Δ indicates a difference. Let mₚ be proton mass and M be the mass of the nucleus.

Δm = Zmₚ + (A − Z)mₙ − M

The binding energy B is the energy needed to separate a nucleus into its individual nucleons. The same energy is released when those free nucleons form the bound nucleus. Binding energy is an energy, while mass defect is a mass.

B = Δmc²

The binding energy per nucleon b is total binding energy divided by the number of nucleons. It is the average energy per nucleon needed for complete separation, rather than the energy required to remove a particular nucleon.

b = B/A

Derivation: How can atomic masses give nuclear binding energy?

1. Let Mₐ denote the mass of a neutral atom and mₑ the electron mass. Neglecting electron binding energies, M ≈ Mₐ − Zmₑ.
2. Substitute this expression in the nuclear mass-defect formula: Δm ≈ Zmₚ + Nmₙ − Mₐ + Zmₑ.
3. Let mₕ denote the mass of a neutral hydrogen atom. Using mₕ ≈ mₚ + mₑ combines the proton and electron terms.

Δm ≈ Zmₕ + Nmₙ − Mₐ. Multiplying by c² gives the nuclear binding energy to this approximation. Use atomic hydrogen mass with atomic masses, or proton mass with nuclear masses, consistently.

Worked example 4. Oxygen-16, ¹⁶₈O, has atomic mass 15.99493 u. Given mₚ = 1.00727 u, mₙ = 1.00866 u, mₑ = 0.00055 u and 1 u c² = 931.5 MeV, find its binding energy. Use 1 MeV = 1.6 × 10⁻¹³ J.

Formula: M = Mₐ − Zmₑ; Δm = Zmₚ + Nmₙ − M; B = Δmc².

Substitute: M = 15.99493 − 8 × 0.00055 = 15.99053 u. The separate nucleons have mass 8 × (1.00727 + 1.00866) = 16.12744 u. Thus Δm = 0.13691 u.

Answer: B ≈ 127.5 MeV ≈ 2.04 × 10⁻¹¹ J, or 0.0000000000204 J.

Worked example 5. Find the binding energy of nitrogen-14, ¹⁴₇N, from atomic mass 14.00307 u. Use mₕ = 1.007825 u, mₙ = 1.008665 u, 1 u c² = 931.5 MeV and 1 MeV = 1.6 × 10⁻¹³ J.

Formula: Δm = Zmₕ + Nmₙ − Mₐ; B = Δmc².

Substitute: Δm = 7 × 1.007825 + 7 × 1.008665 − 14.00307 = 0.11236 u. Then B = 0.11236 × 931.5 MeV.

Answer: B ≈ 104.66 MeV ≈ 1.675 × 10⁻¹¹ J, or 0.00000000001675 J.

In ¹⁴₇N, the letter N is the chemical symbol for nitrogen; in formulas, N denotes neutron number. The notation and context distinguish them. In both examples, the positive mass defect corresponds to energy that must be supplied to separate the bound nucleus.

What does the binding-energy curve reveal about stability?

The binding-energy curve plots b against mass number A. It compares how tightly nucleons are bound on average. A higher binding energy per nucleon indicates tighter binding, so this quantity is more useful for such comparisons than total binding energy alone.

The horizontal axis is mass number A and the vertical axis is binding energy per nucleon in MeV. The plotted curve rises irregularly among light nuclei, reaches a high region near iron-56, then declines gradually towards uranium-238.

See Fig. 13.1 in your NCERT textbook

Which features should be read from the graph?

Fusion joins light nuclei; fission splits a heavy nucleus. Either can release energy when the products have greater total binding energy than the original system. The graph therefore explains two apparently opposite routes to nuclear energy.

Why is the middle region nearly flat?

The nuclear force binds protons and neutrons. It is much stronger than electrical repulsion at relevant nuclear separations, but falls rapidly to zero beyond a few femtometres. A nucleon mainly interacts with neighbours within this short range.

This limited range leads to saturation: adding more distant nucleons does not greatly increase an interior nucleon's binding. Since most nucleons in a large nucleus are inside it rather than on its surface, binding energy per nucleon changes little through the middle region.

The vertical axis is potential energy, meaning interaction energy, in MeV; the horizontal axis is nucleon separation r in fm. The curve has a minimum at r₀, about 0.8 fm, rises steeply to the left, and approaches zero to the right.

See Fig. 13.2 in your NCERT textbook

Here r₀ is the separation at minimum potential energy. The force is attractive for separations greater than r₀ within its effective range, and strongly repulsive below r₀. The nuclear force between proton-proton, neutron-neutron and proton-neutron pairs is approximately the same.

How do annihilation and pair production illustrate mass-energy conversion?

A positron, written e⁺, is the electron's antiparticle: it has the same mass and an equal but opposite electric charge. An electron is written e⁻. A photon is a quantum, or discrete packet, of electromagnetic radiation.

Mutual annihilation occurs when an electron and a positron disappear as a particle pair and their energy appears as radiation. The symbol γ denotes a gamma-ray photon, a high-energy photon. A two-photon annihilation process is represented by e⁻ + e⁺ → γ + γ.

How is the energy accounted for?

Rest energy is the energy equivalent of a particle's mass when it is stationary. For an electron and positron initially at rest, their combined rest energy is 2mₑc². The two photons carry this energy. Each electron or positron has rest energy approximately 0.511 MeV, so the pair contributes approximately 1.022 MeV.

If the pair has kinetic energy, meaning energy of motion, that energy must also be included. Mass-energy conservation does not permit the moving pair's initial kinetic energy to disappear from the final energy balance.

Pair production is the creation of an electron and a positron from photon energy. In the electric field around a nucleus, the nucleus recoils: it moves to carry some momentum. Momentum is the conserved quantity associated with directed motion. This recoil allows both energy and momentum to be conserved.

E₀ = 2mₑc² gives the approximate minimum photon energy when recoil energy is neglected. Here E₀ denotes this approximate threshold energy. It is about 1.022 MeV. Energy above the particles' rest-energy requirement appears as kinetic energy, including recoil.

Note: Pair production is not the separation of pre-existing electrons and positrons from a nucleus. It creates a particle-antiparticle pair. Annihilation and pair production illustrate conversion between particle rest energy and radiation while conserving total energy and charge.

The net charge of the electron-positron pair is zero, matching the photon's zero charge. Accounting for charge does not replace the energy condition: the photon must also supply the rest energies of both particles.

How are nuclear reactions balanced and their energy changes found?

A nuclear reaction changes nuclear composition or produces new nuclei through an interaction. It may change one element into another, a process called transmutation. This differs from a chemical reaction, which rearranges combinations of atoms.

When writing nuclear equations, account for electric charge and nucleon number. In a reaction involving electrons or positrons, include their charges explicitly. Do not assume that proton and neutron counts must separately remain unchanged in every kind of nuclear transformation.

What is the Q-value?

The Q-value Q is the energy released by a nuclear process. Let M₁ denote the sum of initial rest masses and M₂ the sum of final rest masses. Rest mass means the mass associated with a particle or system at rest.

Q = (M₁ − M₂)c²

When the products are material particles, Q is the final total kinetic energy minus the initial total kinetic energy. If photons are emitted, their energy must also be included in the energy balance.

For fission with the same total proton and neutron counts before and after, the free-nucleon mass sums cancel. The increase in total binding energy therefore equals the energy release. Multiply binding energy per nucleon by the appropriate nucleon number before comparing the totals.

Worked example 6. A nucleus with A = 240 and binding energy per nucleon about 7.6 MeV splits into two equal nuclei, each with A = 120 and binding energy per nucleon about 8.5 MeV. Estimate Q; use 1 MeV = 1.6 × 10⁻¹³ J.

Formula: B₁ = 240 × 7.6 MeV; B₂ = 2 × 120 × 8.5 MeV; Q = B₂ − B₁. Here B₁ and B₂ are the initial and final total binding energies.

Substitute: Q = 2040 − 1824 MeV = 216 MeV.

Answer: Q ≈ 216 MeV ≈ 3.46 × 10⁻¹¹ J, or 0.0000000000346 J, an energy release of the order of 200 MeV per fission.

Mass-energy conversion also applies to chemical reactions. Their binding-energy differences and associated mass defects are much smaller than in nuclear processes. It is incorrect to infer that mass-energy equivalence operates exclusively in nuclei.

How do fission and a nuclear reactor release useful energy?

Neutron-induced fission occurs when a neutron initiates the splitting of a heavy nucleus into intermediate-mass fragments. One possible reaction uses uranium-235, written ²³⁵₉₂U. The notation ¹₀n denotes a neutron, with mass number one and charge zero.

¹₀n + ²³⁵₉₂U → ²³⁶₉₂U → ¹⁴⁴₅₆Ba + ⁸⁹₃₆Kr + 3 ¹₀n

Here U, Ba and Kr denote uranium, barium and krypton. The mass-number sums are 236 on each side: 1 + 235 = 144 + 89 + 3. The charge-number sums are also equal: 92 = 56 + 36.

The same initial nucleus can produce other pairs of fragments. Thus this equation is one possible fission channel, not a claim that every uranium-235 fission produces the same nuclei. The fragment products are radioactive, meaning their unstable nuclei can undergo further transformations.

What makes the reaction self-sustaining?

A chain reaction occurs when neutrons released in one fission cause further fissions. Some neutrons escape or are absorbed without causing fission. A continuing chain therefore depends on how many neutrons remain available to cause the next generation of fissions.

Criticality is the condition for a steady, self-sustaining chain reaction: on average, one neutron from each fission causes a subsequent fission. Fewer continuing neutrons make the chain die away; more make the fission rate increase.

A controlled chain reaction maintains a manageable energy output in a reactor. An uncontrolled chain reaction produces a rapidly increasing release of energy, as in a nuclear fission bomb. These descriptions concern the reaction's behaviour, not different definitions of fission.

What are the main reactor parts and their functions?

Draw a casing around fuel elements, with moderator around the fuel and control rods entering the core. Add arrows for coolant entering and leaving. Label a heat-transfer stage leading to steam, a turbine and a generator.

The fission energy first appears as kinetic energy of fragments and neutrons, then becomes heat in surrounding matter. The coolant transports heat for steam production. Steam drives a turbine, a rotating machine, which drives a generator that converts mechanical energy into electrical energy.

Keep moderation, neutron absorption and heat removal distinct. The moderator changes neutron speeds, control rods regulate the neutron population, and coolant transports thermal energy. Confusing these functions obscures how the reactor both sustains and controls its energy output.

Why does fusion require high temperature and power the stars?

In nuclear fusion, light nuclei combine to form a heavier nucleus. Energy is released when the resulting system is more tightly bound. The binding-energy curve provides the qualitative explanation: suitable light nuclei move towards greater binding energy per nucleon on combining.

Both reacting nuclei carry positive charge and repel electrically. The Coulomb barrier is the energy barrier associated with this repulsion. The nuclei must approach closely enough for the short-range attractive nuclear force to become effective.

What is thermonuclear fusion?

Thermonuclear fusion uses very high temperature to provide energetic nuclear motion. Temperature is measured in kelvin, symbol K. Temperatures of the order of millions of kelvin are involved; reaching and maintaining suitable conditions is difficult.

The Sun's interior has a temperature of about 1.5 × 10⁷ K. Fusion there involves protons whose energies are much above the average energy. It is therefore incorrect to treat the average particle energy as the energy of every proton.

How does hydrogen become helium?

Hydrogen in the Sun's core is the fuel. A sequence of nuclear reactions converts it into helium. Including electron-positron annihilation, the overall energy accounting can be written as follows:

4 ¹₁H + 2 e⁻ → ⁴₂He + 2 ν + 6 γ + 26.7 MeV

Here ν denotes a neutrino, an electrically neutral elementary particle produced in the reaction sequence. The hydrogen and helium symbols in this equation represent nuclei. This is the net result of several steps, rather than a simultaneous collision of four protons.

Four nucleons occur on each side. The initial charge is +4e − 2e = +2e, equal to that of the helium nucleus. The neutrinos and photons carry no electric charge. The energy release is associated with the smaller mass of the final system.

A hydrogen bomb involves uncontrolled thermonuclear fusion. Producing steady useful power by controlled fusion requires maintaining and confining extremely hot fuel. At such temperatures the fuel is plasma, a mixture of positive ions and electrons; an ion is an electrically charged atom or nuclear species.

No ordinary container can withstand direct contact with fuel at the temperatures envisaged for controlled fusion. Confinement is therefore a central difficulty. Fission and fusion both exploit increased binding, but their initiating conditions and practical methods of control differ substantially.

Glossary

• Nucleon — A proton or neutron belonging to the particles that constitute an atomic nucleus.
• Nuclide — A nuclear species identified by its particular numbers of protons and neutrons.
• Isotopes — Nuclides with the same proton number but different neutron numbers and mass numbers.
• Isobars — Nuclides with equal mass numbers but different numbers of protons in their nuclei.
• Isotones — Nuclides with equal neutron numbers but different numbers of protons in their nuclei.
• Unified atomic mass unit — A mass unit defined as one-twelfth of the mass of a carbon-12 atom.
• Mass defect — The excess of the combined masses of free constituent nucleons over the mass of their bound nucleus.
• Binding energy — The energy required to separate a nucleus completely into its individual constituent protons and neutrons.
• Binding energy per nucleon — Total nuclear binding energy divided by mass number, expressing the average binding energy associated with each nucleon.
• Q-value — The energy release of a nuclear process, determined by the difference between initial and final total rest energies.
• Criticality — The condition in which a chain reaction sustains itself steadily through one subsequent fission per fission on average.
• Thermonuclear fusion — Fusion produced under very high temperature conditions that supply energetic motion to the reacting light nuclei.

Common errors and misconceptions

• Misconception: Atomic number and mass number both count all nuclear particles. Correct: Atomic number counts protons; mass number counts protons plus neutrons. Subtracting the former from the latter gives neutron number.
• Misconception: Atomic mass and nuclear mass can be substituted interchangeably. Correct: Atomic mass includes electrons. Use atomic hydrogen masses with atomic masses, or proton masses with nuclear masses, consistently.
• Misconception: Nuclear radius is directly proportional to mass number. Correct: Radius is proportional to its cube root. Nuclear volume is proportional to mass number, giving nearly constant nuclear density.
• Misconception: Larger total binding energy necessarily means tighter average binding. Correct: Compare binding energy per nucleon when assessing average binding in nuclei containing different numbers of nucleons.
• Misconception: MeV and MeV/c² both measure energy. Correct: MeV measures energy, whereas MeV/c² measures mass. Multiplying a mass defect by c² gives its binding-energy equivalent.
• Misconception: Moderator and control rods perform the same job. Correct: The moderator slows neutrons, whereas control rods absorb them. The coolant has the separate function of carrying away heat.
• Misconception: Fusion happens easily because it releases energy. Correct: Positively charged nuclei first have to approach against electrical repulsion. High temperature supplies energetic motion, and maintaining suitable conditions is difficult.

Exam-style questions with model answers

Q1. Define isotopes and explain why ²₁H and ³₁H are isotopes. The upper number is mass number and the lower number is atomic number. [2 marks]

1. Isotopes are nuclides with the same atomic number but different neutron numbers, and therefore different mass numbers.
2. Both given nuclei contain one proton. Their neutron numbers are 2 − 1 = 1 and 3 − 1 = 2, so they are isotopes of hydrogen.

Q2. Show that nuclear density is nearly independent of mass number A. Use R = R₀∛A, spherical volume V = 4πR³/3 and approximate nuclear mass A mₙ, where R is radius, R₀ is constant and mₙ is the approximate mass per nucleon. [3 marks]

1. Cubing the radius relation gives R³ = R₀³A. Hence the nuclear volume is V = 4πR₀³A/3 and is proportional to mass number.
2. Density ρ is mass divided by volume, so ρ ≈ A mₙ/(4πR₀³A/3). Both mass and volume contain the same factor A.
3. Cancelling A gives ρ ≈ 3mₙ/(4πR₀³). With R₀ and the approximate nucleon mass fixed, density is nearly independent of A.

Q3. Calculate the energy equivalent of 1 g of matter, using 1 g = 10⁻³ kg and speed of light c = 3 × 10⁸ m s⁻¹. State what the result represents. [3 marks]

1. Einstein's relation is E = mc², where E is energy equivalent and m is mass. Convert the given mass into SI units: m = 10⁻³ kg.
2. Substitution gives E = 10⁻³ × (3 × 10⁸)² = 9 × 10¹³ J, using the speed of light given in the question.
3. This represents the rest-energy equivalent of the entire gram of matter. A reaction releases the energy corresponding to its mass decrease, not necessarily this entire amount.

Q4. Calculate the nuclear mass, mass defect and binding energy of ¹⁶₈O. Its atomic mass is 15.99493 u. Proton, neutron and electron masses are 1.00727 u, 1.00866 u and 0.00055 u respectively. Use 1 u c² = 931.5 MeV and neglect electron binding energies. [4 marks]

1. The nucleus has eight protons and eight neutrons. Subtract the eight electron masses from the atomic mass: M = 15.99493 − 8 × 0.00055 = 15.99053 u.
2. The mass of the separate constituent nucleons is 8 × 1.00727 + 8 × 1.00866 = 16.12744 u.
3. The mass defect is the separate-nucleon mass minus nuclear mass: Δm = 16.12744 − 15.99053 = 0.13691 u.
4. The binding energy is B = Δmc² = 0.13691 × 931.5 ≈ 127.5 MeV. This energy must be supplied for complete separation into individual nucleons.

Q5. A nucleus of mass number 240 has binding energy per nucleon about 7.6 MeV. It splits into two nuclei of mass number 120 each, with binding energy per nucleon about 8.5 MeV. Calculate the energy release Q and explain its sign using the binding-energy curve. [5 marks]

1. The initial total binding energy is the mass number multiplied by binding energy per nucleon: 240 × 7.6 = 1824 MeV.
2. Each product has binding energy 120 × 8.5 = 1020 MeV. For the two product nuclei together, the total is 2040 MeV.
3. The gain in total binding energy is 2040 − 1824 = 216 MeV. Thus the estimated energy release Q is 216 MeV.
4. Q is positive because the final nucleons are more tightly bound. The decrease in the system's rest mass supplies the released energy.
5. The curve places intermediate-mass nuclei at greater binding energy per nucleon than very heavy nuclei. The result is therefore an exothermic fission estimate of the order of 200 MeV.

Q6. State the functions of five reactor components: fuel elements, moderator, control rods, coolant, and casing with shielding. [5 marks]

1. Fuel elements contain material that undergoes nuclear fission. Fission releases energy and neutrons, supplying the heat source and particles needed to continue the chain reaction.
2. The moderator slows neutrons in a reactor designed to use slow neutrons. Its purpose is to change neutron speeds for effective continuation of fission.
3. Control rods absorb neutrons. Adjusting their insertion regulates the neutron population and therefore the rate of the chain reaction.
4. The coolant carries thermal energy away from the reactor core. This heat can be used to produce steam for driving a turbine and generator.
5. The casing encloses the reactor system, while shielding reduces radiation reaching the surroundings. Together they provide containment and protection around the operating reactor.

Q7. An electron and a positron each have rest energy 0.511 MeV. Find the total photon energy when they annihilate at rest, and the approximate threshold for photon production of such a pair near a nucleus. Neglect nuclear recoil energy in the threshold calculation. [3 marks]

1. For annihilation at rest, the combined initial rest energy is 0.511 + 0.511 = 1.022 MeV. This becomes the total energy of the emitted photons.
2. Pair production must supply both particle rest energies, so the approximate threshold under the stated assumption is also 1.022 MeV.
3. The nearby nucleus enables momentum conservation by taking recoil. Neglecting its recoil energy gives the quoted approximate threshold; energy above the rest-energy requirement contributes to motion.

Q8. Explain why hydrogen fusion in the Sun can release energy yet requires very high temperature. Use the net result 4 ¹₁H + 2 e⁻ → ⁴₂He + 2 ν + 6 γ + 26.7 MeV, where e⁻ denotes an electron, ν a neutrino and γ a photon. [4 marks]

1. Four hydrogen nuclei are converted into one helium nucleus through a sequence of reactions. The equation gives the overall result, including annihilation, rather than one simultaneous collision.
2. The final system is more tightly bound and has smaller total rest mass. The associated energy release is 26.7 MeV for the net process given.
3. The reacting hydrogen nuclei are positively charged and repel electrically. They must approach closely enough for the attractive, short-range nuclear force to become effective.
4. Very high temperature supplies energetic nuclear motion. Fusion in the Sun involves protons with energies much above the average, helping them overcome the electrical barrier.

Key takeaways

• Atomic number counts protons, mass number counts all nucleons, and their difference gives the neutron number of a nucleus.
• Isotopes share proton number, isobars share mass number, and isotones share neutron number; calculate each count before classifying nuclides.
• Nuclear radius increases as the cube root of mass number, so nuclear volume and mass grow together and density stays nearly constant.
• Mass defect becomes binding energy through multiplication by c²; maintain consistent use of atomic or nuclear masses throughout calculations.
• Binding energy per nucleon compares average binding, with middle-mass nuclei more tightly bound than light or very heavy nuclei.
• A positive Q-value represents energy release and a decrease in total rest mass; increased total binding explains favourable fission reactions.
• Reactor fuel supplies fission energy, the moderator slows neutrons, control rods absorb neutrons, and the coolant transports heat.
• Fusion joins light nuclei under demanding conditions; stellar hydrogen conversion releases energy because the final nuclear system is more tightly bound.

Test yourself

How many neutrons are in ¹⁹⁷₇₉Au?

Subtract atomic number from mass number: N = 197 − 79 = 118 neutrons.

Why are ³₁H and ³₂He isobars?

Both have mass number three, but their atomic numbers are one and two respectively.

What distinguishes one unified atomic mass unit from a mass number?

One u is a defined unit of mass based on carbon-12; mass number is a dimensionless count of nucleons.

Why does nuclear density not increase substantially with mass number?

Nuclear mass and nuclear volume are both approximately proportional to mass number, so their ratio remains nearly constant.

What is the difference between binding energy and binding energy per nucleon?

Binding energy is the total energy for complete nuclear separation; dividing it by mass number gives binding energy per nucleon.

What sign of Q indicates an energy-releasing reaction?

A positive Q-value indicates that initial total rest mass exceeds final total rest mass and energy is released.

What does criticality mean for a steady fission chain?

On average, one neutron from each fission causes another fission, keeping the chain reaction self-sustaining at a steady rate.

Why does a favourable fusion energy balance not remove the need for high temperature?

Positive nuclei still repel and must approach within the short range of nuclear attraction. High temperature supplies the required energetic motion.