CLASS 12 PHYSICS · CHAPTER 13 · STUDY NOTES

Nuclei

An exploration of the atomic nucleus, covering its composition, size, mass-energy equivalence, binding energy, and the powerful forces that hold it together, alongside nuclear phenomena like fission and fusion.

EXPLORE

Atomic Masses and Composition of Nucleus

Count particles before comparing masses.

Count the protons and neutrons separately

Isotopes share proton number Z but differ in neutron number N. Orange circles are protons; blue circles are neutrons. Circle counts are exact in these small examples; nucleon positions are schematic.

Read the model limits

Circles and time stages are enlarged teaching representations. Rest-mass changes correspond to energy exchange; matter is not simply disappearing, and total mass-energy is conserved. Source values are approximate.

Original teaching models · NCERT Physics Part II, chapter 13 · local exploration, no recorded completion

Mass number A: The integer count of protons plus neutrons in a nucleus: A = Z + N. It is not the measured atomic mass in u.

Term	Symbol	Definition
Atomic Number	Z	Number of protons in the nucleus
Neutron Number	N	Number of neutrons in the nucleus
Mass Number	A	Total number of protons and neutrons (Z + N)
Nucleon	-	A collective term for a proton or a neutron

Protons have charge +e, neutrons have zero electric charge; both are nucleons. Atomic number Z identifies the element. Neutron number N = A − Z identifies how many neutrons accompany those protons. A neutral atom also has Z electrons outside its nucleus.

Isotopes share Z; isobars share A; isotones share N. A mass spectrometer distinguishes isotopic masses. The atomic mass of an element can be a weighted mean of isotope masses, which need not equal an integer A. One atomic mass unit is one-twelfth the mass of a neutral carbon-12 atom, approximately 1.66054 × 10⁻²⁷ kg.

Let’s look a little closer.

Go deeper: Nuclear identity can change

A free neutron undergoes beta decay into a proton, electron and electron antineutrino, with mean lifetime about 880 s. Binding inside a nucleus does not guarantee stability: nuclear beta decays are possible when the complete energy balance allows them. An electron emitted in beta-minus decay is not an electron previously stored inside the nucleus.

Alpha decay changes the daughter’s A by −4 and Z by −2. Beta-minus changes Z by +1; beta-plus changes Z by −1, with A unchanged. Gamma emission lowers an excited nucleus’s energy without changing A or Z. Gamma and X-ray energy ranges can overlap; nuclear versus electronic origin distinguishes them.

Tritium is radioactive and rare, but occurs naturally through cosmic-ray reactions as well as being produced artificially. Do not interpret the chapter’s statement about its absence as an exact modern claim.

Take a moment. What do you think?

PAUSE & TRY · think first, then check with a free account

Question

A nucleus has Z = 14 and A = 30. Compare it with Z = 14, A = 29 and with Z = 15, A = 30. Which pair are isotopes, and which are isobars?

Write your own answer and compare it with ours. It’s free.

NCERT reference: chapter PDF pages 1, 2, 3, 9, 14, 15.

EXPLORE

Size and Density of the Nucleus

Compare circles on one scale, then compare mass with volume.

Larger nucleus, nearly the same matter density

Mass number A = 64

For A = 64, the approximate spherical radius is R = 1.2A^(1/3) = 4.80 fm. Volume scales with A, as does approximate mass A u; their ratio is about 2.29 × 10¹⁷ kg/m³. The circle is a projection, not a literal arrangement or proof that every nucleus is perfectly spherical.

Read the model limits

Original teaching models · NCERT Physics Part II, chapter 13 · local exploration, no recorded completion

Approximate nuclear radius: For the chapter’s spherical model, R ≈ R₀A^(1/3), with R₀ ≈ 1.2 fm and 1 fm = 10⁻¹⁵ m.

Scattering experiments reveal nuclear size on a femtometre scale. The empirical radius law means that multiplying A by eight doubles R, rather than multiplying R by eight. Since volume is 4πR³/3, this approximate volume grows in proportion to A.

With nuclear mass approximated by A u, density is 3u/(4πR₀³) ≈ 2.3 × 10¹⁷ kg/m³. Its weak dependence on A supports a nearly constant-density picture. The formula is an approximation, particularly for very small nuclei; real nuclei need not be perfectly spherical.

Go deeper: Why a nucleus is not a solid ball

The circle shows a projected radius, not the positions of literal touching beads. Nuclear charge distributions have diffuse edges and quantum structure. A single sharp radius and uniform density are useful estimates, not universal exact properties.

The estimated nuclear density is approximately 2.3 × 10¹⁴ times water’s density (about 10³ kg/m³). Carefully divide the densities; the inherited notes understated this ratio by a factor of 100.

PAUSE & TRY · think first, then check with a free account

Question

Using R₀ = 1.2 fm, compare A = 27 and A = 216. Find their radii, volume ratio and approximate density ratio.

Write your own answer and compare it with ours. It’s free.

NCERT reference: chapter PDF pages 4, 5.

EXPLORE

Mass-Energy Equivalence and Mass Defect

Track the energy of the whole system, not disappearing particles.

Binding changes the mass of the whole system

For ⁴He, the atomic-mass route gives Δm = 2mH + 2mn − mHe = 0.030377 u, and binding energy Eb ≈ 28.30 MeV. Formation releases energy and lowers the bound system’s rest mass. Relative energy zero here is the free-nucleon rest-energy sum; no proton or neutron is lost.

Read the model limits

Original teaching models · NCERT Physics Part II, chapter 13 · local exploration, no recorded completion

Binding energy: The energy needed to separate a bound nucleus completely into free protons and neutrons; Eb = Δm c² for the corresponding rest-mass difference.

A bound nucleus has less rest mass than its separated constituent nucleons. The mass defect is Δm = Zmp + Nmn − Mnucleus. Formation can release the corresponding binding energy; separation requires energy supplied. Total mass-energy is conserved when the emitted or supplied energy is included.

Use consistent mass data. With neutral-atom masses, Δm ≈ ZmH + Nmn − Matom: hydrogen-atom masses supply the same number of electron masses as the final neutral atom. Electronic binding corrections are small on the nuclear-energy scale. For mass in u, Eb ≈ Δm × 931.5 MeV.

Go deeper: Energy units and reference zero

1 u c² is approximately 931.5 MeV, while 1 MeV = 1.60218 × 10⁻¹³ J. Binding energy has units of energy, not MeV/c². The latter is a mass unit; a source extraction that ends an oxygen binding-energy calculation in MeV/c² must be corrected.

The energy-level drawing sets the separated constituent rest-energy sum to zero. The bound nucleus then sits below it by Eb. No protons or neutrons vanish on binding: interaction energy changes the total system’s rest mass.

PAUSE & TRY · think first, then check with a free account

Question

A hypothetical bound system has mass defect 0.0200 u. Find its binding energy in MeV and joules, and state the energy required for complete separation.

Write your own answer and compare it with ours. It’s free.

NCERT reference: chapter PDF pages 5, 6.

EXPLORE

Binding Energy per Nucleon

Read the trend, then calculate the full binding gain.

Energy release depends on the total binding gain

This schematic redraw uses rounded values read from the chapter’s binding curve. Middle-mass nuclei lie near 8–9 MeV per nucleon; light and very heavy examples lie lower. Joining points is illustrative, not a measured continuous function or a claim that every isotope has that binding.

Read the model limits

Original teaching models · NCERT Physics Part II, chapter 13 · local exploration, no recorded completion

Binding energy per nucleon: Total binding energy divided by A; it measures average binding, not the complete criterion for stability against every decay.

The chapter’s rounded curve rises steeply among light nuclei, has a broad middle-mass plateau near 8–9 MeV per nucleon, and falls towards very heavy nuclei. This is a teaching trend across selected nuclei; different isotopes at the same A need not have identical binding.

Short-range nuclear attraction mainly involves nearby nucleons and approximately saturates. Light nuclei have a larger fraction of surface nucleons with fewer neighbours. In very heavy nuclei, long-range proton–proton repulsion increasingly offsets attraction. Suitable fusion and fission channels can release energy by increasing total binding.

Go deeper: Per nucleon is not total energy

For a reaction, compare the full initial and final masses or total binding energies, counting all products and any free nucleons. A higher final average can indicate energy release only when the complete mass-number and composition balance is accounted for.

The graph redraw uses the chapter’s approximate plotted values. The broad iron-region peak should not be turned into an exact claim that A = 56 is the global maximum for all measured isotopes. Binding per nucleon alone also does not decide beta-decay stability or every possible reaction.

PAUSE & TRY · think first, then check with a free account

Question

Two light nuclei each contain A = 4 nucleons at an illustrative binding of 7.0 MeV per nucleon. They form one A = 8 product at 7.5 MeV per nucleon with unchanged total proton/neutron numbers. Estimate Q; is this a prediction of a real reaction?

Write your own answer and compare it with ours. It’s free.

NCERT reference: chapter PDF pages 7, 8, 9, 10.

EXPLORE

The Nuclear Force

Separate the sign of the potential from the sign of its slope.

Read the force from the slope of the potential

Separation 1.2 fm

At separation r = 1.2 fm, the illustrative potential has positive slope: force attracts the pair. Use F = −dU/dr. The minimum at 0.8 fm follows the source’s schematic; the curve shape and vertical magnitude are a teaching function, not a measured nuclear potential.

Read the model limits

The plotted potential is an original analytic teaching function with the source’s qualitative repulsion/attraction pattern. It is not a universal nuclear-force formula. Actual interactions depend on quantum state and spin as well as distance.

Original teaching models · NCERT Physics Part II, chapter 13 · local exploration, no recorded completion

Force from a potential: The radial force follows F = −dU/dr: a rising potential with separation means attraction; a falling potential means repulsion.

The residual strong interaction binds nucleons over short ranges of a few femtometres. In the chapter’s schematic it is repulsive below roughly 0.8 fm, attractive at somewhat larger separations, and weak at large separations. At a potential minimum the plotted force is zero, not maximally attractive.

The nuclear interaction is approximately charge independent; p–p, p–n and n–n pairs can all interact. Proton pairs additionally repel electrically. Short range and saturation mean that a selected nucleon does not attract every distant nucleon equally strongly.

Go deeper: The curve is a model

The interactive potential is an original analytic teaching function with the source’s qualitative pattern and minimum. Its energy scale and exact shape are not measured nuclear-potential data. Actual interactions depend on spin and quantum state as well as separation.

A negative U alone does not reveal whether the force points inward or outward: inspect the slope. Charge independence concerns the nuclear interaction, not the sum of all interactions, so p–p total behaviour need not equal n–n behaviour.

PAUSE & TRY · think first, then check with a free account

Question

At two separations U is negative. At the first, U decreases as r increases; at the second, U increases. State the force direction at each. What happens at a smooth minimum?

Write your own answer and compare it with ours. It’s free.

NCERT reference: chapter PDF pages 8, 9.

EXPLORE

Nuclear Fission

Follow one possible channel, then test conservation and continuation.

One reaction, conservation and continuation

An incoming neutron approaches ²³⁵U. Capture is one possible outcome; not every neutron causes fission.

Read the model limits

Selected products conserve mass number and charge. Other channels exist; branching is an expectation model, not a reactor simulation. Energy and momentum must also be conserved.

Original teaching models · NCERT Physics Part II, chapter 13 · local exploration, no recorded completion

Fission Q-value: The energy released by a specified split, obtained from the full initial minus final rest masses multiplied by c².

After neutron capture, an excited heavy nucleus may split into intermediate fragments and additional neutrons. One chapter channel is ²³⁵U + n → ¹⁴⁴Ba + ⁸⁹Kr + 3n. Other fragment channels exist; capture does not guarantee this particular result or even fission.

A typical uranium-235 fission releases about 200 MeV, distributed among fragment motion, neutrons, radiation and later decay products. Transfers to surrounding matter produce heat. In an electricity-producing system, heat drives conversion stages; electrical output is smaller than total released energy.

Go deeper: One event versus a chain

Some released neutrons can cause further fissions while others escape or are absorbed without fission. The toy factor k describes the expected number of successor events: k < 1 declines, k = 1 sustains on average, k > 1 grows. It is not a guarantee of integer counts or a reactor-design calculation.

For the selected channel, A balances at 236 = 144 + 89 + 3 and nuclear charge at 92 = 56 + 36. Conservation of energy and momentum also applies. Proton and neutron numbers are not separately conserved in all nuclear transformations: weak reactions can convert one into the other.

PAUSE & TRY · think first, then check with a free account

Question

An illustrative chain has k = 0.8 and initially 100 fission events. Estimate expected events in the next two generations. Does each event necessarily release 200 MeV as electrical energy?

Write your own answer and compare it with ours. It’s free.

NCERT reference: chapter PDF pages 9, 10, 11, 14, 15.

EXPLORE

Nuclear Fusion

Connect the products to binding and the barrier to tunnelling.

Close approach, new binding and stellar energy

The selected deuterium channel is ²H + ²H → ³He + n + about 3.27 MeV. Four initial nucleons become three in helium plus one free neutron, with total charge 2e conserved. This is one possible channel, not every deuterium collision.

Read the model limits

Original teaching models · NCERT Physics Part II, chapter 13 · local exploration, no recorded completion

Fusion energy balance: For a specified combination channel, Q = (sum of initial rest masses − sum of final rest masses)c²; energy release is not guaranteed merely by joining nuclei.

A selected deuterium channel is ²H + ²H → ³He + n + about 3.27 MeV. Four nucleons and total charge +2e balance. Positive nuclei repel before close approach; at short distances the nuclear interaction can bind suitable products.

High temperature increases relative collision energies and reaction rates in thermonuclear plasma. In the Sun, quantum tunnelling allows reactions through the Coulomb barrier even when collision energies are below its height. The proton–proton chain makes helium through intermediate reactions; it is not four protons colliding simultaneously.

Go deeper: Count the stellar intermediates

For the illustrated branch, p + p → ²H + e⁺ + νₑ occurs twice, and ²H + p → ³He + γ occurs twice. Then ³He + ³He → ⁴He + 2p returns two protons. Thus six protons enter these steps and two return: net consumption is four.

Including annihilation of the two positrons with two electrons, the net reaction is 4p + 2e⁻ → ⁴He + 2νₑ + energy, about 26.7 MeV. Some energy escapes with neutrinos, so this total is not all deposited heat. The first step involves weak conversion of a proton to a neutron.

Controlled fusion requires fuel conditions, confinement and energy balance. Magnetic fields influence charged-particle motion in plasma; the schematic torus is not an operating device or a claim of unlimited energy without engineering limits.

PAUSE & TRY · think first, then check with a free account

Question

In the illustrated pp-chain branch, the first two reactions each run twice. Count protons used and returned. Write the net charge-balanced reaction when positrons annihilate with electrons.

Write your own answer and compare it with ours. It’s free.

NCERT reference: chapter PDF pages 10, 11, 12, 13, 14, 15.

Nuclei Class 12 Notes

Chapter notes

Nuclei

Atomic Masses and Composition of Nucleus

Count the protons and neutrons separately

Size and Density of the Nucleus

Larger nucleus, nearly the same matter density

Mass-Energy Equivalence and Mass Defect

Binding changes the mass of the whole system

Binding Energy per Nucleon

Energy release depends on the total binding gain

The Nuclear Force

Read the force from the slope of the potential

Nuclear Fission

One reaction, conservation and continuation

Nuclear Fusion

Close approach, new binding and stellar energy

Now practise Nuclei, free