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Structure of Atom · TOPIC LESSON

Nuclear Reactions

Keep element identity, isotope identity and ionic charge separate before balancing nuclear change.

JEE Advanced · NEET UG43 min study guide9 indexed sections

BY THE END, YOU CAN

  • Count protons, neutrons and electrons for isotopic ions.
  • Distinguish isotopes, isobars, isotones and average atomic mass.
  • Balance alpha, beta, positron, electron-capture and gamma transformations.
  • Relate mass defect and reaction mass change to binding energy and Q value.
  • Use decay constants, half-lives and activities with explicit closed-sample assumptions.

BEFORE YOU START

  • Atomic number and ionic charge.
  • Moles, weighted averages and logarithms.
  • Energy units and exponential first-order decay.

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.

No source-verified exam questions have been published for this lesson yet. The worked examples and practice below are original teaching material, not past-exam questions.

01 / UNDERSTAND & APPLY

1. Three numbers describe three different features

Atomic number Z is the number of protons and defines the element. Mass number A is the integer number of nucleons, protons plus neutrons, in a particular nucleus. Neutron number is N=A−Z. Ionic charge describes the imbalance between protons and electrons and does not ordinarily change the nucleus.

For an ion with charge q in units of the positive elementary charge, electron count is Z−q. A +2 ion has two fewer electrons than protons, while a −1 ion has one extra electron. Treating the sign algebraically avoids memorizing separate rules for cations and anions.

A nuclide symbol can be written with A at upper left and Z at lower left of the element symbol, with ionic charge at upper right if needed. The element symbol already implies Z, but showing both makes conservation bookkeeping easier. The mass number is not the same as the measured atomic mass in unified atomic mass units.

Protons and neutrons contain nearly all the atomic mass and occupy a tiny nucleus compared with the spatial extent of the electron cloud. Electrons dominate ordinary chemical bonding and ion formation. A chemical reaction generally rearranges electrons and nuclei among molecules without changing the identity of each nucleus; a nuclear reaction can change Z and therefore the element.

protons=Z; neutrons=A−Z; electrons=Z−q

q is signed ionic charge in units of +e.

ORIGINAL WORKED EXAMPLE

Count an isotopic ion

For ²⁷Al³⁺ with Z=13, count protons, neutrons and electrons.

  1. The element has Z=13 protons.
  2. Neutrons are A−Z=27−13=14.
  3. The +3 charge means three electrons have been removed, so electrons=13−3=10.
  4. Removing electrons does not change A or the aluminum identity.

Answer 13 protons, 14 neutrons and 10 electrons.

02 / UNDERSTAND & APPLY

2. Isotopes share an element but differ in neutron number

Isotopes have the same Z and different A, so they are nuclei of the same element with different neutron counts. Isobars have the same A but different Z. Isotones have the same neutron number but different Z. These classifications compare nuclei; isoelectronic species instead share an electron count and can involve different nuclei and charges.

Isotopes of an element have closely related ordinary chemical behavior because their neutral atoms have the same electron number and similar electronic structure. Their different masses can nevertheless affect vibrational frequencies, diffusion and reaction rates through isotope effects. Saying isotopes have exactly identical physical and chemical behavior is too strong.

The unified atomic mass unit u is defined through carbon-12. A specific isotope’s atomic mass is usually near its integer mass number but is not generally exactly equal to it. Nuclear binding and the masses of the constituent particles explain the difference. A mass number counts particles; an atomic mass measures mass.

The average atomic mass of a sample is Σf_i m_i, where f_i is the fraction of atoms of isotope i and the fractions sum to one. Use atom fractions, not mass fractions, unless a conversion is made. The sample’s isotopic composition determines the weighted mean, which need not equal the mass of any individual atom.

A reported periodic-table atomic weight characterizes an element’s isotopic composition under the stated convention and may vary among natural samples for some elements. In a numerical exercise, use the provided isotope masses and abundances instead of assuming rounded mass numbers are exact measured masses.

average atomic mass=Σf_i m_i; Σf_i=1

f_i is an atom-number fraction of isotope i.

ORIGINAL WORKED EXAMPLE

Infer an isotope abundance

A hypothetical element has two isotopes of masses 62.0 u and 64.0 u. A sample’s average mass is 62.6 u. Find the atom fraction of the heavier isotope.

  1. Let the heavier-isotope fraction be x and the lighter fraction 1−x.
  2. 62.0(1−x)+64.0x=62.6.
  3. Simplifying gives 62.0+2.0x=62.6, so x=0.300.
  4. The mean lies between the two masses, as a valid weighted average must.

Answer 30.0% heavier isotope and 70.0% lighter isotope in this hypothetical model.

03 / UNDERSTAND & APPLY

3. Nuclear equations conserve more than the element symbol

In an introductory nuclear equation, the sum of mass numbers and the sum of charge numbers match on both sides. A neutron is written ¹₀n, a proton ¹₁p, an alpha particle ⁴₂He, an electron ⁰₋₁e and a positron ⁰₊₁e. The zero mass number of an electron means zero nucleon count, not zero physical rest mass.

Balancing A and charge identifies missing species in many exercises, but it is not the full set of physical conservation laws. Energy, momentum, angular momentum and lepton number also constrain actual processes. Neutrinos or antineutrinos are essential in beta processes even when a simplified classroom notation omits them.

A balanced candidate equation does not prove that the reaction is energetically possible or likely. The nuclear masses determine its Q value, and dynamics determine probabilities and rates. Balancing is the first consistency check, not a universal prediction engine for nuclear stability.

Ion charge and nuclear charge require care. An electron-capture equation refers to a nuclear transformation involving an orbital electron; the atom’s subsequent electron-shell rearrangement is a separate detail. When balancing nuclide notation, Z tracks proton number, whereas an ion’s superscript charge tracks its electron deficit or excess.

ORIGINAL WORKED EXAMPLE

Find an alpha-decay daughter

Determine the daughter nuclide when ²³⁸₉₂U emits one alpha particle.

  1. Mass-number balance gives A_daughter=238−4=234.
  2. Charge-number balance gives Z_daughter=92−2=90.
  3. Atomic number 90 identifies thorium.
  4. The equation is ²³⁸₉₂U→²³⁴₉₀Th+⁴₂He, with energy and possible subsequent emissions treated separately.

Answer ²³⁴₉₀Th.

04 / UNDERSTAND & APPLY

4. Track what changes inside the nucleus

In beta-minus decay, a neutron changes into a proton while an electron and an electron antineutrino are emitted: n→p+e⁻+ν̄_e. The daughter’s mass number stays the same and Z increases by one. The emitted electron is produced in the transformation; it is not an electron that was previously orbiting inside the nucleus.

In beta-plus decay, a proton changes into a neutron with a positron and electron neutrino emitted: p→n+e⁺+ν_e within the nucleus when energetically allowed. A stays fixed and Z decreases by one. Electron capture also lowers Z by one: p+e⁻→n+ν_e, consuming an atomic electron rather than emitting a positron.

Gamma emission lowers a nucleus from an excited state to a lower nuclear-energy state without changing A or Z. A star on the nuclide label can indicate excitation. The emitted photon carries energy and angular momentum, but no nucleon number or electric charge.

Using neutral atomic masses, a simple beta-minus Q value is approximately the parent-minus-daughter atomic mass difference times c², while positron emission requires accounting for two electron rest masses in that atomic-mass convention. Electron capture uses a different electron bookkeeping. Rather than mixing memorized formulas, identify whether supplied masses are nuclear or neutral-atomic masses and conserve the actual particles consistently.

Beta particles have a continuous kinetic-energy spectrum because decay energy is shared with a neutrino or antineutrino and nuclear recoil. A two-body alpha decay has a much more sharply determined alpha energy for a specified daughter state, because the energy sharing is then constrained mainly by the two-body kinematics.

ORIGINAL WORKED EXAMPLE

Compare two beta directions

State the changes in A and Z for ¹⁴₆C beta-minus decay and for a generic positron decay of ᴬ_ZX.

  1. In carbon beta-minus decay, A remains 14 and Z rises from 6 to 7, giving ¹⁴₇N.
  2. The full lepton bookkeeping includes e⁻ and ν̄_e among the products.
  3. For positron decay, A remains A and Z becomes Z−1, with e⁺ and ν_e emitted.
  4. The positron process still requires a favorable energy balance; charge balance alone is insufficient.

Answer ¹⁴₆C→¹⁴₇N+e⁻+ν̄_e; generic β⁺ daughter has A unchanged and Z−1.

05 / UNDERSTAND & APPLY

5. Binding energy measures the mass difference of a bound nucleus

The mass of a bound nucleus is less than the sum of the masses of its separated free protons and neutrons. The positive difference Δm=Zm_p+Nm_n−m_nucleus is the mass defect, and binding energy is B=Δmc². Energy must be supplied to separate the nucleus completely into those free nucleons.

If neutral atomic masses are used, a convenient equivalent expression is Δm=Zm_H+Nm_n−m_atom, using the neutral hydrogen-atom mass rather than the bare proton mass. Electron masses then cancel to the intended precision, with small electronic binding corrections neglected. Combining a bare-nucleus formula with a neutral-atom mass without adjustment produces an avoidable error.

Binding energy per nucleon B/A provides a useful measure for comparing average nuclear binding. It is not by itself a complete prediction of every radioactive decay channel, because allowed daughter combinations and conservation constraints matter. Broadly, moving toward more strongly bound products can release nuclear energy.

A handy conversion is 1 u c²≈931.5 MeV. This relates a mass difference to energy, not a mass number difference to energy. Mass numbers can balance exactly while measured rest mass decreases, with the difference appearing in kinetic energy and radiation.

B=[Zm_p+Nm_n−m_nucleus]c²

Use nuclear mass with bare nucleon masses.

B≈[Zm_H+Nm_n−m_atom]c²; 1 u c²≈931.5 MeV

Neutral atomic-mass convention, neglecting small electronic binding corrections.

ORIGINAL WORKED EXAMPLE

Binding energy from a supplied defect

A hypothetical nucleus with A=20 has mass defect 0.160 u. Find binding energy and binding energy per nucleon.

  1. B=0.160×931.5=149.04 MeV.
  2. Divide by 20 nucleons: B/A=149.04/20=7.452 MeV per nucleon.
  3. The binding energy is positive because energy is needed to separate the bound system.
  4. The defect is a measured-mass difference; it is not a missing fraction of a nucleon count.

Answer B≈149 MeV; B/A≈7.45 MeV per nucleon.

06 / UNDERSTAND & APPLY

6. A Q value compares all initial and final masses

For a specified nuclear reaction, Q=(Σm_initial−Σm_final)c² using a consistent mass convention. Q>0 means rest-mass energy is released into kinetic energy and radiation; Q<0 means energy must be supplied. For a reaction initiated by a moving projectile, the actual threshold energy can exceed |Q| because momentum conservation requires final kinetic energy.

Nuclear stability reflects a balance among the strong interaction, electrostatic repulsion, neutron–proton composition and quantum shell effects. Light stable nuclei often have neutron number near proton number; heavier stable nuclei generally need more neutrons than protons. A simple N/Z rule is a trend, not an exact test for every nuclide.

Fission divides a heavy nucleus into lighter nuclei and other emitted particles, while fusion combines light nuclei into heavier products. Some such processes release energy because the products are more strongly bound overall. Their practical initiation and control involve physics beyond equation balancing; this lesson uses only basic conservation and energy accounting.

A nuclear transformation can change the element, unlike ordinary electron-transfer redox chemistry. Oxidation states describe how electrons are assigned in chemical bonding, whereas nuclear charge number counts actual protons. A change in oxidation state is not a beta decay, and chemical oxidation does not normally transmute one element into another.

ORIGINAL WORKED EXAMPLE

Calculate a reaction Q value

A hypothetical reaction has total initial rest mass 25.0100 u and total final rest mass 25.0060 u in a consistent convention. Find Q.

  1. The mass difference is 25.0100−25.0060=0.0040 u.
  2. Q=0.0040×931.5=3.726 MeV.
  3. The positive sign means energy is released.
  4. The total final kinetic and radiation energy exceeds the initial kinetic energy by this Q value, with total energy conserved.

Answer Q≈+3.7 MeV at the precision of the supplied mass difference.

07 / UNDERSTAND & APPLY

7. Half-life describes a population, not a scheduled lifetime for one nucleus

For a single radioactive nuclide with constant decay probability per unit time λ, the expected parent population obeys dN/dt=−λN and N=N₀e^(−λt). The half-life is t_1/2=ln2/λ. Individual decay times are random; a nucleus does not carry a clock forcing it to decay exactly one half-life after preparation.

Activity A_act is the expected decay rate, λN, and has SI unit becquerel, one decay per second. Do not confuse activity with mass number A. For one nuclide with constant λ, activity decreases with the same exponential fraction as the parent population.

After k half-lives the remaining parent fraction is 2^(−k). For an arbitrary measured fraction f, elapsed time is t=−ln(f)/λ. The ideal exponential never becomes exactly zero at finite time, although an actual small sample eventually contains no surviving parent nuclei because nuclei are discrete.

These equations assume no new parent production, no physical loss or gain of sample, and a constant decay environment for the mode being considered. Most nuclear decay rates are insensitive to ordinary chemical conditions, but electron-capture rates can show electronic-environment dependence. Avoid an absolute claim that no decay rate can ever be influenced by its surroundings.

N=N₀e^(−λt); t_1/2=ln2/λ; A_act=λN

Single parent nuclide, constant λ, no production or physical sample loss.

ORIGINAL WORKED EXAMPLE

Population and activity after several half-lives

A nuclide has half-life 6.00 h. A closed sample initially has activity 800 Bq. Find activity after 18.0 h.

  1. Elapsed time is 18.0/6.00=3.00 half-lives.
  2. The parent and activity fractions are (1/2)³=1/8.
  3. Activity is 800/8=100 Bq.
  4. This assumes the measured activity is from the parent alone, without an additional radioactive daughter contribution.

Answer 100 Bq parent activity.

08 / UNDERSTAND & APPLY

8. State what is actually measured

If a parent decays directly into a stable daughter with one daughter per decay and no initial daughter, daughter number is N_D=N₀−N_P. The daughter-to-parent ratio is therefore N_D/N_P=e^(λt)−1. This relation needs a closed system and the stated initial daughter condition; inherited daughter material changes the inference.

If the daughter is itself radioactive, its number depends on both production from the parent and its own decay. The simple stable-daughter subtraction does not predict its activity. A detector that counts radiation from several chain members may therefore not follow the parent’s single exponential over the entire interval.

Detector count rate is not automatically equal to activity. Only a fraction of emitted radiation may reach and register in the detector, and environmental background contributes counts. With known constant efficiency ε and an appropriate single-channel model, net count rate can be related to activity by rate_net=εA_act.

Activity alone is not a measure of biological harm or absorbed dose. Radiation type, energy, exposure geometry and material interactions matter. The equations here describe decay bookkeeping, not a safety clearance for handling radioactive material or a medical interpretation of exposure.

  • Mass number is nucleon count, not exact atomic mass.
  • Ionic charge changes electron count; atomic number changes element identity.
  • Beta particles have zero mass number but nonzero rest mass.
  • Include neutrino bookkeeping when explaining beta processes.
  • Keep nuclear and neutral-atomic masses in consistent formulas.
  • A detector count rate can include efficiency losses, background and daughter emissions.

09 / UNDERSTAND & APPLY

9. Eight solved practice checks

Use the stated models and avoid assuming that an algebraically balanced transformation must be an observed spontaneous decay.

ORIGINAL PRACTICE · 1

Count particles in ³⁵Cl⁻ with Z=17.

Reveal answer & reasoning

17 protons, 18 neutrons and 18 electrons.

Neutrons are 35−17=18; charge −1 means electrons=17−(−1)=18.

ORIGINAL PRACTICE · 2

Are ¹⁴C and ¹⁴N isotopes or isobars?

Reveal answer & reasoning

Isobars.

They share mass number 14 but have different atomic numbers, so they are different elements rather than isotopes of one element.

ORIGINAL PRACTICE · 3

A sample has 40% isotope mass 10.0 u and 60% isotope mass 11.0 u by atom number. Find its average atomic mass.

Reveal answer & reasoning

10.6 u.

Weighted mean=0.40×10.0+0.60×11.0=10.6 u.

ORIGINAL PRACTICE · 4

What changes in A and Z accompany gamma emission?

Reveal answer & reasoning

Neither changes.

The nucleus loses excitation energy by a photon without changing proton or neutron count.

ORIGINAL PRACTICE · 5

A nuclide undergoes one alpha decay followed by two beta-minus decays. Give net changes in A and Z.

Reveal answer & reasoning

A decreases by four; Z has no net change.

Alpha gives ΔA=−4 and ΔZ=−2. Each beta-minus gives ΔA=0 and ΔZ=+1, so two restore Z.

ORIGINAL PRACTICE · 6

A closed parent sample has 1/16 of its initial parent nuclei remaining. How many half-lives elapsed?

Reveal answer & reasoning

Four.

(1/2)⁴=1/16. This is an expected population relation for the stated single-nuclide model.

ORIGINAL PRACTICE · 7

A nuclide has λ=0.0100 s⁻¹ and N=2.00×10⁶ parent nuclei. Find activity and half-life.

Reveal answer & reasoning

Activity 2.00×10⁴ Bq; half-life 69.3 s.

A_act=λN=0.0100×2.00×10⁶. The half-life is 0.693147/0.0100=69.3 s.

ORIGINAL PRACTICE · 8

A parent decays directly to a stable daughter in a closed sample with no initial daughter. If daughter/parent=3, how many half-lives have passed?

Reveal answer & reasoning

Two.

N_D=3N_P means N₀=N_P+N_D=4N_P, so the parent fraction is 1/4=(1/2)².

KEEP THE BIG IDEAS

Quick revision

  • Z fixes element identity, A fixes nucleon count and ion charge fixes the electron imbalance.
  • Isotope averages require atom fractions and measured isotope masses.
  • Nuclear equations conserve charge and nucleon number while energy and lepton accounting add further constraints.
  • Binding energy and reaction Q values use consistent rest-mass conventions.
  • Half-life and activity describe statistical populations with explicit production and loss assumptions.
  • Count rate, activity and dose are different quantities.
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The teaching notes, worked explanations and original practice are written for this lesson. Separately labeled selected exam questions identify their exam, year, source and whether wording is paraphrased or an exact excerpt. Educational references do not certify current exam coverage.

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