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Nuclei · TOPIC LESSON

Nuclear Reactions and Binding Energy

Connect mass deficits, reaction energy and statistical decay with explicit model assumptions.

JEE Main · JEE Advanced · NEET UG80 min study guide9 indexed sections

BY THE END, YOU CAN

  • Calculate nuclear composition, radius scaling and binding energies with consistent masses.
  • Use Q-values and recoil to account for reaction energy.
  • Track alpha, beta and gamma changes and distinguish their spectra.
  • Solve decay, half-life, activity and background-counting problems.

BEFORE YOU START

  • Mass–energy equivalence and conservation of energy and momentum.
  • Exponentials, logarithms, scientific notation and unit conversion.

Original educational lesson. Work through the examples before revealing the practice answers. Check the current official syllabus separately; historical question frequency does not establish eligibility or predict future questions.

CONNECT THE CONCEPT TO THE EXAM

Selected exam questions & worked solutions

These selected examples are separate from the original practice below. Each identifies its exam, year and wording. They do not represent every paper or every selected year.

Available in this lesson: JEE Advanced 2018. This is a limited lesson sample, not a complete exam-question bank.

1 sourced sample question available across the listed exams and years.

Matched topics: Radioactive Decay.

JEE Advanced · 2018 · Paper 2 · Physics · Q5

Paraphrased exam question

Independently derived and checked solution. Archived key finality is unverified; this is not presented as an official final answer.

A decay chain takes ²³²₉₀Th to ²¹²₈₂Pb using α and β⁻ emissions. Determine the number of each emitted particle.

Reveal answer & worked solution

Answer: Five α and two β⁻ particles (independently derived; source A,C).

  1. Each α emission reduces mass number by4; β⁻ leaves it unchanged. Therefore Nα=(232−212)/4=5.
  2. Each α reduces atomic number by2, while β⁻ raises it by1. Hence 82=90−2(5)+Nβ.
  3. Solving gives Nβ=2. Check both mass and charge balance; balancing mass alone cannot determine β⁻ emissions.

Source: JEE (Advanced) 2018, Paper 2, Physics Q5; independently derived solution; key finality unverified. Solution written for this lesson.

01 / UNDERSTAND & APPLY

1. Count nucleons before balancing a nuclear change

A nuclide is specified by proton number Z and mass number A. It contains Z protons and N = A − Z neutrons. The chemical element is fixed by Z, while isotopes share Z but differ in neutron count. Isobars share A; isotones share N. The number of atomic electrons depends on ionisation and must not be confused with the nuclear proton count.

Mass number is an integer count, not a measured mass in atomic mass units. A nucleus with A nucleons does not generally have mass exactly A u. The atomic mass unit u is defined using neutral carbon-12 atoms; nuclear reaction calculations must use the actual supplied masses consistently rather than replace each mass by its mass number.

An empirical nuclear-radius estimate is R = r₀A^(1/3), with r₀ about 1.2 fm and 1 fm = 10⁻¹⁵ m. Volume then scales approximately as A, giving roughly similar nuclear mass density across many nuclides. The nucleus has a diffuse quantum boundary, so this radius law is a useful scale estimate rather than a rigid-ball geometry valid to unlimited precision.

The residual strong interaction between nucleons is short-ranged and approximately saturating: a nucleon interacts strongly mainly with nearby nucleons. Proton–proton electric repulsion is long-ranged. Their competition helps explain why heavy stable nuclei generally need more neutrons than protons and why there is no unlimited progression of stable heavier nuclei.

ORIGINAL WORKED EXAMPLE

Nuclear size and composition

An illustrative nuclide has A = 125 and Z = 52. Find its neutron count and approximate radius using r₀ = 1.2 fm. Compare its radius with a nuclide having A = 27.

  1. Neutron count is A − Z = 125 − 52 = 73.
  2. Since 125^(1/3) = 5, R = 1.2 × 5 = 6.0 fm.
  3. The A = 27 radius is 1.2 × 3 = 3.6 fm, so the radius ratio is 5/3, while the volume ratio is 125/27.

Answer 73 neutrons; approximately 6.0 fm; radius ratio 5/3.

02 / UNDERSTAND & APPLY

2. A bound nucleus has less rest energy than separated nucleons

The binding energy B is the positive energy required to separate a nucleus completely into free protons and neutrons at rest, with all particles far apart. A bound system has lower total rest energy than those separated constituents. Thus the mass defect is Δm = Zmₚ + Nmₙ − M_nucleus and B = Δmc². Mass has not disappeared from conservation of total energy; energy released during formation leaves the system.

For mass differences expressed in u, use 1 u c² ≈ 931.5 MeV. If a mass is already given in MeV/c², multiplying by c² leaves MeV. Multiplying an energy expressed in MeV by c² a second time is a dimensional mistake. Nuclear energies are commonly millions of electron volts per event, but each event still involves a very small number of joules.

Tables often provide neutral atomic masses rather than bare nuclear masses. For binding-energy work, the convenient approximate atomic-mass expression is B = [Z m_H + N mₙ − M_atom]c², where m_H is the neutral hydrogen-atom mass. Electron rest masses then cancel. Small electron binding-energy corrections are usually negligible at the stated nuclear precision but matter in high-precision work.

Total binding energy and binding energy per nucleon answer different questions. B/A is an average measure of binding; the energy to remove one specified neutron is a separation energy obtained from a difference of actual neighbouring masses. It is not generally equal to B/A. A larger nucleus can have greater total B yet a lower average binding per nucleon.

B = [Zmₚ + (A − Z)mₙ − M_nucleus]c²

Use nuclear masses throughout. B is positive for a bound nucleus relative to complete separation.

B ≈ [Zm_H + (A − Z)mₙ − M_atom]c²

Neutral atomic masses; electron rest masses cancel, with small electronic binding corrections neglected.

ORIGINAL WORKED EXAMPLE

A supplied mass defect

A nucleus with A = 20 has a measured mass defect of 0.1720 u relative to its free nucleons. Find total binding energy and binding energy per nucleon using 931.5 MeV/u.

  1. Binding energy is 0.1720 × 931.5 = 160.218 MeV.
  2. Divide total energy by the 20 nucleons: 160.218/20 = 8.0109 MeV per nucleon.
  3. The supplied mass defect has four significant figures, so report about 160.2 MeV and 8.011 MeV per nucleon. This is an illustrative mass-defect calculation, not a claim identifying a particular measured isotope.

Answer 160.2 MeV total; 8.011 MeV per nucleon.

03 / UNDERSTAND & APPLY

3. Read the binding curve as an energy comparison, not a stability certificate

The broad binding-energy-per-nucleon curve rises rapidly among light nuclei, reaches a high region around iron and nickel, and decreases slowly toward very heavy nuclei. Individual nuclei show structure around this smooth trend. Avoid treating a rough classroom curve as exact isotope data or assuming one isotope must lead every possible measure of stability.

If a reaction conserves the same total numbers of constituent protons and neutrons and increases total binding energy, it releases energy. Fusion of suitable light nuclei and fission of suitable heavy nuclei can move products toward more tightly bound arrangements. The released energy is the increase in total binding, not merely a difference between two unweighted B/A values.

For example, when one A-nucleon parent splits into fragments with A₁ and A₂, compare A₁(B/A)₁ + A₂(B/A)₂ with A(B/A)_parent, including any other products. Averaging the two fragment B/A values without weighting by their nucleon counts is generally wrong. Full reaction masses are the more universal method when particle species change.

An energetically favourable rearrangement is not necessarily fast. Coulomb barriers, quantum tunnelling, angular-momentum constraints and available decay pathways affect rates. Binding per nucleon alone cannot determine half-life, and a positive Q-value is not a guarantee of an easily initiated reaction. Energy accounting and reaction probability are separate parts of the physics.

Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.

Fusion and fission can increase total binding energyAn illustrative broad binding-energy-per-nucleon curve rises from small mass number toward a maximum region near iron and nickel and decreases slowly for heavier nuclei. Arrows show suitable light-nucleus fusion and heavy-nucleus fission moving toward greater binding.060120180240048Light-nucleus fusionHeavy-nucleus fissionHigh-binding regionnear iron / nickelBinding energy per nucleon (MeV)Mass number A
Schematic trend, not isotope measurements. Positions on the curve do not determine half-lives or guarantee a reaction channel. Reaction energy uses total binding or complete masses, not an unweighted difference in plotted averages.

ORIGINAL WORKED EXAMPLE

Weighted binding-energy release

An illustrative A = 200 nucleus with B/A = 7.6 MeV splits into two A = 100 fragments, each with B/A = 8.4 MeV. Ignore other products for this idealised bookkeeping calculation. Find energy released.

  1. Initial total binding is 200 × 7.6 = 1520 MeV.
  2. Final total binding is 2 × 100 × 8.4 = 1680 MeV.
  3. The final state is more tightly bound by 160 MeV; that is the released energy, not just 8.4 − 7.6 = 0.8 MeV.

Answer 160 MeV in the stated idealised model.

04 / UNDERSTAND & APPLY

4. Q-values compare complete initial and final states

For a nuclear reaction, Q = (sum of initial rest masses − sum of final rest masses)c². Positive Q means rest energy can become kinetic energy or radiation; negative Q means energy must be supplied. In a closed reaction, total energy, momentum, electric charge and the relevant conserved quantum numbers must balance. Merely balancing mass numbers does not finish the problem.

For many reactions written using neutral atoms, electron rest masses cancel when the total atomic charge numbers balance and no explicit electron or positron is mishandled. Otherwise account for electrons separately. In particular, using neutral atomic masses gives Q for beta-minus decay approximately [M_parent − M_daughter]c², but beta-plus decay requires subtracting 2mₑc². Electron capture has different electron bookkeeping from positron emission.

An endothermic reaction on a target initially at rest generally needs projectile kinetic energy greater than |Q| because momentum conservation leaves kinetic energy in the products. For a nonrelativistic two-body threshold with negative Q small compared with rest energies, a common approximation is K_threshold ≈ |Q|(1 + m_projectile/M_target). This approximation is not a universal threshold formula for arbitrary relativistic or multiparticle events.

In a two-body decay of a parent initially at rest, daughter momenta are equal and opposite. For nonrelativistic products, kinetic energies are inversely proportional to their masses: K₁/K₂ = m₂/m₁. A light alpha particle therefore receives most, but not all, of the available decay energy. Neglecting the recoil may be acceptable only when the requested precision permits it.

ORIGINAL WORKED EXAMPLE

Energy sharing with recoil

A parent at rest decays into an alpha particle of mass approximately 4 u and a daughter of mass approximately 196 u. The available kinetic energy is 5.00 MeV. Find each kinetic energy nonrelativistically.

  1. Momentum conservation gives equal momentum magnitudes.
  2. K_alpha/K_daughter = 196/4 = 49.
  3. Thus K_alpha = (196/200) × 5.00 = 4.90 MeV and K_daughter = (4/200) × 5.00 = 0.100 MeV.

Answer Alpha: 4.90 MeV; daughter recoil: 0.100 MeV.

05 / UNDERSTAND & APPLY

5. Identify what changes in alpha, beta and gamma emission

Alpha emission ejects a helium-4 nucleus: the daughter has A − 4 and Z − 2. In beta-minus decay a neutron converts into a proton while an electron and electron antineutrino are emitted; A is unchanged and Z increases by one. The emitted electron is created in the weak process, rather than being a pre-existing nuclear constituent shaken loose.

In beta-plus decay a proton converts into a neutron with a positron and electron neutrino emitted; A is unchanged and Z decreases by one. Electron capture also reduces Z by one at fixed A by absorbing an atomic electron and emitting a neutrino. Whether a channel is allowed depends on actual masses and energy requirements, not just whether charge numbers can be balanced.

Gamma emission lowers the excitation energy of a nucleus without changing A or Z. The atom may also produce characteristic X-rays after electron-shell rearrangements, but X-rays and gamma rays are distinguished by their origin rather than by a perfectly nonoverlapping energy boundary. Recoil gives a small correction to the photon energy relative to the full level gap.

Alpha decay often has discrete characteristic energies because it is essentially a two-body transition to particular daughter levels. Beta electrons or positrons have a continuous energy distribution because energy and momentum are shared among the charged particle, neutrino and recoiling daughter. A listed beta endpoint is therefore not the energy carried by every emitted electron.

After a sequence of decays, balance changes in A first to determine the number of alpha events, then use changes in Z to infer beta events under the specified channels. Do not infer a unique physical chain from algebra alone: gamma steps and alternative branches can occur without changing the final A and Z labels.

06 / UNDERSTAND & APPLY

6. Constant decay probability produces an exponential population

For a single independently decaying nuclide with constant decay probability per unit time λ, the expected number of decays in a short interval is λN dt. Thus dN/dt = −λN and N(t) = N₀e^(−λt). λ is positive and has inverse-time units. The minus sign belongs to the decrease of the undecayed population; the activity is the positive rate A_activity = λN.

Individual decay times are unpredictable in this model. Exponential decay describes an expected population or an ensemble average, not a clock forcing every nucleus to decay at exactly one half-life. Radioactive nuclei do not grow deterministically old until they reach a fixed expiry time. For constant λ, the conditional probability of survival over a further interval is independent of how long a nucleus has already survived.

The half-life T₁/₂ satisfies e^(−λT₁/₂) = 1/2, giving T₁/₂ = ln 2/λ. The mean lifetime is τ = 1/λ, so τ = T₁/₂/ln 2 and is longer than the half-life. After n half-lives, the expected surviving fraction is 2^(−n); the decayed fraction is 1 − 2^(−n). These two fractions are often confused.

Activity is measured in becquerels, with 1 Bq = one decay per second. It depends both on how much material is present and on λ. Two samples with equal numbers of nuclei need not have equal activity if their half-lives differ. Equal masses do not imply equal numbers of nuclei when molar masses differ.

The simple law assumes no additional production or removal and a constant decay constant. Ordinary temperature and pressure changes usually have negligible effects on common nuclear decay rates, but blanket statements of perfect environmental independence are too strong: electron-capture rates can depend on electronic conditions. State the isolated constant-rate model used in a calculation.

N(t) = N₀e^(−λt); activity = λN(t)

A single nuclide with constant λ, no replenishment and independent decay.

T₁/₂ = ln 2/λ; τ = 1/λ

Use the same time units in λ and t; mean lifetime and half-life differ.

Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.

Every half-life halves the population still presentNormalized exponential decay passes through one,one half,one quarter,one eighth and one sixteenth at zero through four half-lives. The same normalized curve applies to activity for a single nuclide with constant decay constant.0123411/21/41/8N / N₀ = activity / activity₀= 2^(−t / T½)Expected surviving fractionElapsed half-lives, t / T½
Quantitative curve for the expected surviving population. Individual nuclei decay at random times. A detector background or a radioactive mixture must not be substituted directly into this normalized single-nuclide law.

ORIGINAL WORKED EXAMPLE

Population and activity after several half-lives

A sample initially contains 8.0 × 10¹² radioactive nuclei with a half-life of 2.0 h. Find expected surviving nuclei and activity after 6.0 h. Use ln 2 = 0.693.

  1. Six hours is three half-lives, so N = 8.0 × 10¹²/8 = 1.0 × 10¹².
  2. Convert half-life to seconds: 2.0 h = 7200 s, giving λ = 0.693/7200 = 9.625 × 10⁻⁵ s⁻¹.
  3. Activity is λN = 9.625 × 10⁷ Bq. To two significant figures this is 9.6 × 10⁷ Bq.

Answer 1.0 × 10¹² nuclei remain; activity approximately 9.6 × 10⁷ Bq.

07 / UNDERSTAND & APPLY

7. Separate physical decay from what a detector records

A detector records only some emitted radiation. In a simple model, measured count rate C = εA_activity + C_background, with constant detection efficiency ε. Subtract background before using count-rate ratios to infer a half-life. Counts are not automatically the total number of nuclear decays, especially when emission branching fractions and geometry matter.

For independent counts with Poisson statistics, the standard deviation of a large count n is approximately √n. Relative counting uncertainty is therefore about 1/√n. Counting for four times as long reduces relative statistical uncertainty by roughly a factor of two under constant-rate conditions. This is a statistical estimate, not a bound guaranteeing every observed count lies within one standard deviation.

If a detector sees a mixture of independent nuclides, activity is a sum Σ λᵢNᵢ₀e^(−λᵢt), generally not a single exponential. A parent producing a radioactive daughter requires production as well as decay: dN_d/dt = λ_pN_p − λ_dN_d when the branch is fully to that daughter. Daughter activity can initially rise even though both species are radioactive.

Dating from a surviving fraction uses t = ln(N₀/N)/λ under an appropriate closed-system model. The starting abundance, contamination, replenishment and environmental history can matter more than arithmetic. A numerical decay-age exercise provides these assumptions; an actual sample cannot be dated reliably from an unqualified activity ratio alone.

ORIGINAL WORKED EXAMPLE

Subtract background before finding half-life

A fixed detector records 520 counts per minute initially and 145 counts per minute 12 h later. Background is constant at 20 counts per minute, efficiency is unchanged and only one nuclide contributes. Find half-life.

  1. Net initial rate is 520 − 20 = 500 counts per minute.
  2. Net later rate is 145 − 20 = 125 counts per minute, a surviving ratio of 1/4.
  3. A factor of 1/4 is two half-lives, so T₁/₂ = 12/2 = 6 h. Dividing the raw rates would give the wrong ratio.

Answer 6 h under the stated single-nuclide model.

08 / UNDERSTAND & APPLY

8. Original practice with complete reasoning

These problems are original exercises in nuclear bookkeeping and exponential decay. Use supplied masses or energy differences; do not reconstruct precise isotope data from the smooth binding-energy trend.

ORIGINAL PRACTICE · 1

For A = 64 and Z = 29, how many neutrons are present?

Reveal answer & reasoning

35.

N = A − Z = 64 − 29.

ORIGINAL PRACTICE · 2

A mass number increases from 8 to 64. How does the empirical nuclear radius change?

Reveal answer & reasoning

It doubles.

R scales as A^(1/3); (64/8)^(1/3) = 2.

ORIGINAL PRACTICE · 3

A nuclear mass defect is 0.0100 u. Find binding energy using 931.5 MeV/u.

Reveal answer & reasoning

9.32 MeV.

Multiply 0.0100 by 931.5 to obtain 9.315 MeV, then round to three significant figures.

ORIGINAL PRACTICE · 4

Does B/A give the exact energy needed to remove the last neutron?

Reveal answer & reasoning

No.

Neutron separation energy is a difference of actual masses or total binding energies of neighbouring nuclei, not the average per nucleon.

ORIGINAL PRACTICE · 5

An alpha emitter has A = 220 and Z = 86. Give the daughter labels.

Reveal answer & reasoning

A = 216; Z = 84.

Alpha emission removes two protons and two neutrons.

ORIGINAL PRACTICE · 6

How do A and Z change during beta-minus emission?

Reveal answer & reasoning

A unchanged; Z increases by one.

A neutron becomes a proton, electron and antineutrino; nucleon count stays fixed.

ORIGINAL PRACTICE · 7

A nuclide emits two alpha particles and one beta-minus particle. What are its net label changes?

Reveal answer & reasoning

ΔA = −8; ΔZ = −3.

Two alpha emissions give −8 and −4; beta-minus adds one to Z.

ORIGINAL PRACTICE · 8

A reaction has initial rest mass 0.00300 u greater than final rest mass. Find Q.

Reveal answer & reasoning

+2.79 MeV.

Q = +0.00300 × 931.5 MeV. Positive Q is energy available to final kinetic energy or radiation.

ORIGINAL PRACTICE · 9

A sample passes through four half-lives. What fractions remain and have decayed?

Reveal answer & reasoning

1/16 remains; 15/16 has decayed.

The surviving fraction is 2^(−4); subtract it from one for the decayed fraction.

ORIGINAL PRACTICE · 10

The decay constant is 0.0200 s⁻¹. Find mean lifetime and half-life.

Reveal answer & reasoning

50.0 s and 34.7 s.

τ = 1/λ = 50.0 s; T₁/₂ = 0.693/0.0200 ≈ 34.7 s.

ORIGINAL PRACTICE · 11

Two samples have equal N but one has half the half-life. Compare activities.

Reveal answer & reasoning

The shorter-half-life sample has twice the activity.

λ = ln 2/T₁/₂ and activity = λN.

ORIGINAL PRACTICE · 12

A detector records 10,000 independent counts. Estimate relative Poisson standard deviation.

Reveal answer & reasoning

Approximately 1%.

√n = 100, so √n/n = 100/10,000 = 0.01.

ORIGINAL PRACTICE · 13

Why can daughter activity initially increase in a decay chain?

Reveal answer & reasoning

Production can exceed its own decay rate.

The derivative λ_pN_p − λ_dN_d can be positive, especially when initially little daughter is present.

ORIGINAL PRACTICE · 14

An endothermic reaction has Q = −2.0 MeV, projectile mass m and stationary target mass 4m. Estimate the nonrelativistic threshold under the stated two-body approximation.

Reveal answer & reasoning

Approximately 2.5 MeV.

K_threshold ≈ |Q|(1 + m/M) = 2.0(1 + 1/4). Momentum conservation prevents all projectile kinetic energy from becoming rest energy.

09 / UNDERSTAND & APPLY

9. Audit energy signs, mass conventions and decay assumptions

Begin by distinguishing an atom from its nucleus, a mass number from a measured mass, and a population from an activity. Then select the process: separating a nucleus calls for binding energy; transforming initial particles into final particles calls for Q; waiting for an isolated radioactive sample calls for exponential decay. These operations use related ideas but different inputs.

Check the physical conclusion after the arithmetic. Positive binding energy means energy must be supplied for complete separation. Positive Q means the chosen reaction releases rest energy, without guaranteeing a high reaction rate. A half-life divides the expected remaining population, not the original population anew each time. The same check prevents most apparent paradoxes in this chapter.

  • Balance charge, nucleon labels and all particles required by the decay channel.
  • Never mix nuclear and atomic masses without electron bookkeeping.
  • Use total binding energies, not unweighted averages, to compare reaction states.
  • Account for recoil whenever energy is shared between products.
  • Convert half-life to seconds before reporting activity in Bq.
  • Subtract background and state efficiency assumptions for detector ratios.
  • A mixture or replenished daughter need not follow one exponential.

KEEP THE BIG IDEAS

Quick revision

  • Binding lowers the rest energy of a composite nucleus relative to separated nucleons.
  • Q compares complete initial and final states, while a rate also depends on reaction dynamics.
  • Half-life describes a statistical population with a constant decay probability.
  • Activity, detected counts and remaining nuclei are related but distinct quantities.
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