☒️ Class 12 Physics · Chapter 13 · AAI ATC Exam

βš›οΈ Nuclei

Nuclear Size, Mass Defect, Binding Energy, Radioactivity, Fission & Fusion β€” Complete AAI ATC Written Exam Coverage

πŸ“š NCERT Class 12 Physics
✈️ AAI ATC Relevant
🎯 18 Numerical MCQs
⏱️ 6 Subtopics Covered

πŸ“Œ Chapter Overview

βš›οΈ Nuclear Composition & Terminology
πŸ“ Size & Density of Nucleus
πŸ’₯ Mass-Energy & Binding Energy
πŸ”— Nuclear Forces
☒️ Radioactivity (α, β, γ)
🌟 Nuclear Fission & Fusion
1
Nuclear Composition & Atomic Mass

Atomic Mass Unit (u)

Defined as 1/12th of the mass of ΒΉΒ²C atom.

1 u = 1.660539 Γ— 10⁻²⁷ kg = 931.5 MeV/cΒ²

Nuclear Terminology

SymbolNameDefinition
ZAtomic numberNumber of protons in nucleus
NNeutron numberNumber of neutrons
AMass numberA = Z + N (total nucleons)

Particle Masses

mβ‚š = 1.00727 u = 1.67262 Γ— 10⁻²⁷ kg (proton)
mβ‚™ = 1.00866 u = 1.67493 Γ— 10⁻²⁷ kg (neutron)

Key Definitions

πŸ”‘ Isotopes: Same Z, different N (same element, different mass) β€” e.g., ΒΉH, Β²H, Β³H
πŸ”‘ Isobars: Same A, different Z β€” e.g., ³₁H and Β³β‚‚He
πŸ”‘ Isotones: Same N, different Z β€” e.g., ΒΉβΉβΈβ‚ˆβ‚€Hg and ¹⁹⁷₇₉Au

A free neutron is unstable (mean life ~1000 s) and decays into a proton, electron, and antineutrino. Inside the nucleus, neutrons are stable.

🎯 Practice MCQs β€” Nuclear Composition

Q1 Chlorine has two isotopes with masses 34.98 u (75.4% abundance) and 36.98 u (24.6% abundance). The average atomic mass of chlorine is:
A 35.98 u
B 35.47 u
C 35.00 u
D 36.00 u
βœ… Avg mass = (75.4 Γ— 34.98 + 24.6 Γ— 36.98) / 100
= (2637.49 + 909.71) / 100 = 3547.2 / 100 = 35.47 u
Q2 The gold nucleus ¹⁹⁷₇₉Au has how many neutrons?
A 79
B 197
C 118
D 276
βœ… N = A βˆ’ Z = 197 βˆ’ 79 = 118 neutrons
Q3 The energy equivalent of 1 gram of matter (using E = mcΒ²) is:
A 9 Γ— 10¹⁰ J
B 3 Γ— 10⁸ J
C 9 Γ— 10ΒΉΒ³ J
D 9 Γ— 10¹⁢ J
βœ… E = mcΒ² = 10⁻³ Γ— (3Γ—10⁸)Β² = 10⁻³ Γ— 9Γ—10¹⁢ = 9 Γ— 10ΒΉΒ³ J
2
Size & Density of the Nucleus

Nuclear size was determined by Geiger-Marsden scattering experiments. The radius of the nucleus follows an empirical relation:

R = Rβ‚€ A^(1/3) where Rβ‚€ = 1.2 Γ— 10⁻¹⁡ m = 1.2 fm

Since Volume ∝ R³ ∝ A, the nuclear density is constant for all nuclei, independent of mass number A.

Nuclear density β‰ˆ 2.3 Γ— 10¹⁷ kg m⁻³
πŸ’‘ Nuclear density is ~10¹⁴ times greater than water (10Β³ kg/mΒ³). This is because atoms are mostly empty space β€” the nucleus occupies only ~10⁻¹² of the atomic volume.
🌟 Neutron stars have density comparable to nuclear density (~2Γ—10¹⁷ kg/mΒ³) β€” they are essentially giant nuclei!
πŸ–ΌοΈ
ADD IMAGE HERE
Nuclear size comparison & R = Rβ‚€A^(1/3) concept diagram

🎯 Practice MCQs β€” Nuclear Size & Density

Q4 The ratio of nuclear radii of ¹⁹⁷₇₉Au to ¹⁰⁷₄₇Ag is approximately: (R = Rβ‚€A^(1/3))
A 1.23
B 1.225
C 1.40
D 1.09
βœ… R(Au)/R(Ag) = (A_Au/A_Ag)^(1/3) = (197/107)^(1/3) = (1.841)^(1/3) β‰ˆ 1.225
Q5 The nuclear radius of ⁢⁴₂₉Cu is 4.8 fm. What is the radius of ²⁷₁₃Al? (Rβ‚€ = 1.2 fm)
A 3.6 fm
B 2.4 fm
C 4.8 fm
D 6.0 fm
βœ… R(Al)/R(Cu) = (27/64)^(1/3) = (0.422)^(1/3) = 0.75
R(Al) = 0.75 Γ— 4.8 = 3.6 fm
Q6 Given mass of iron nucleus = 55.85 u and A = 56. The nuclear density is approximately: (Rβ‚€ = 1.2 fm, 1 u = 1.66 Γ— 10⁻²⁷ kg)
A 1.5 Γ— 10¹⁢ kg/mΒ³
B 5 Γ— 10¹⁴ kg/mΒ³
C 2.29 Γ— 10¹⁷ kg/mΒ³
D 9 Γ— 10ΒΉΒ³ kg/mΒ³
βœ… m = 55.85 Γ— 1.66Γ—10⁻²⁷ = 9.27Γ—10⁻²⁢ kg
R = 1.2Γ—10⁻¹⁡ Γ— 56^(1/3) = 1.2Γ—10⁻¹⁡ Γ— 3.826 = 4.59Γ—10⁻¹⁡ m
ρ = m / (4Ο€RΒ³/3) = 9.27Γ—10⁻²⁢ / (4Ο€/3 Γ— (4.59Γ—10⁻¹⁡)Β³) β‰ˆ 2.29 Γ— 10¹⁷ kg/mΒ³
3
Mass Defect & Nuclear Binding Energy

Mass Defect (Ξ”M)

The actual nuclear mass is always less than the sum of masses of its constituent protons and neutrons. This difference is the mass defect:

Ξ”M = [ZΒ·mβ‚š + (Aβˆ’Z)Β·mβ‚™] βˆ’ M_nucleus

Binding Energy (Eᡦ)

The energy equivalent of the mass defect β€” the energy needed to completely disassemble the nucleus into free protons and neutrons:

Eᡦ = Ξ”M Γ— cΒ² | 1 u = 931.5 MeV/cΒ²

Binding Energy per Nucleon (Eᡦₙ)

Eᡦₙ = Eᡦ / A (average energy per nucleon)

Key Features of the B.E. per Nucleon Curve (Fig. 13.1)

πŸ“Œ Middle mass nuclei (30 < A < 170): Eᡦₙ β‰ˆ 8 MeV (nearly constant) β€” most stable
πŸ“Œ Peak at A = 56 (⁡⁢Fe): Eᡦₙ β‰ˆ 8.75 MeV β€” most tightly bound nucleus
πŸ“Œ Light nuclei (A < 30): Lower Eᡦₙ β†’ energy released in fusion
πŸ“Œ Heavy nuclei (A > 170): Lower Eᡦₙ β†’ energy released in fission
🎯 1 u = 931.5 MeV/cΒ² β€” this conversion is extremely important for all nuclear energy calculations in the AAI ATC exam.
πŸ–ΌοΈ
ADD IMAGE HERE
Binding energy per nucleon vs mass number curve (Fig. 13.1)

🎯 Practice MCQs β€” Mass Defect & Binding Energy

Q7 The mass defect of ΒΉβΆβ‚ˆO nucleus is 0.13691 u. Its binding energy in MeV is: (1 u = 931.5 MeV/cΒ²)
A 0.13691 MeV
B 100 MeV
C 127.5 MeV
D 200 MeV
βœ… Eᡦ = Ξ”M Γ— 931.5 = 0.13691 Γ— 931.5 = 127.5 MeV
Q8 The binding energy of ¹⁴₇N nucleus is 104.66 MeV. Its binding energy per nucleon is: (A = 14)
A 104.66 MeV
B 8.75 MeV
C 7.476 MeV
D 14.95 MeV
βœ… Eᡦₙ = Eᡦ/A = 104.66/14 = 7.476 MeV per nucleon
Q9 For a nucleus with A = 240 (Eᡦₙ = 7.6 MeV) splitting into two A = 120 fragments (Eᡦₙ = 8.5 MeV), the total energy released is:
A 0.9 MeV
B 7.6 MeV
C 216 MeV
D 8.5 MeV
βœ… Gain per nucleon = 8.5 βˆ’ 7.6 = 0.9 MeV
Total gain = 240 Γ— 0.9 = 216 MeV (approximately 200 MeV is the standard stated value for uranium fission)
4
Nuclear Forces

To bind nucleons in the tiny nuclear volume, a force far stronger than Coulomb repulsion between protons must exist β€” the Strong Nuclear Force.

Key Properties of Nuclear Force

β‘  Strongest force in nature β€” much stronger than Coulomb force or gravity
β‘‘ Short-range β€” effective only up to ~2–3 fm. Rapidly falls to zero beyond that
β‘’ Charge-independent β€” same between n-n, p-n, and p-p pairs
β‘£ Saturating β€” each nucleon interacts only with its nearest neighbours
β‘€ At r < 0.8 fm: strongly repulsive (prevents nuclear collapse)
β‘₯ At r > 0.8 fm: attractive (holds nucleus together)

The potential energy between two nucleons has a minimum at rβ‚€ β‰ˆ 0.8 fm. Unlike Coulomb or gravitational forces, there is no simple mathematical formula for the nuclear force.

πŸ–ΌοΈ
ADD IMAGE HERE
Potential energy vs distance graph for two nucleons (Fig. 13.2)

🎯 Practice MCQs β€” Nuclear Forces

Q10 The nuclear force between two nucleons is repulsive when the distance between them is:
A Greater than 2 fm
B Between 0.8 fm and 2 fm
C Less than 0.8 fm
D Exactly 0.8 fm
βœ… The nuclear potential energy minimum is at rβ‚€ = 0.8 fm. For r < 0.8 fm, the nuclear force is strongly repulsive. For r > 0.8 fm, it is attractive.
Q11 The binding energy per nucleon for a nucleus with A = 56 (iron) is approximately 8.75 MeV. Its total binding energy is:
A 8.75 MeV
B 100 MeV
C 490 MeV
D 8750 MeV
βœ… Eᡦ = Eᡦₙ Γ— A = 8.75 Γ— 56 = 490 MeV
Q12 The constancy of binding energy per nucleon in the range 30 < A < 170 is due to:
A Long-range nature of nuclear force
B Charge independence of nuclear force
C Short-range (saturation) property of nuclear force
D Repulsive nature of nuclear force at all distances
βœ… Each nucleon interacts only with its nearest neighbours (short-range, saturating property). Adding more nucleons doesn't change the binding of inner nucleons. Hence Eᡦₙ stays approximately constant β€” this is the saturation property.
5
Radioactivity

Discovered by A.H. Becquerel in 1896. Radioactivity is a nuclear phenomenon where an unstable nucleus spontaneously emits radiation to become more stable.

Three Types of Radioactive Decay

TypeParticle EmittedChange in ZChange in ANature
Ξ±-decay⁴₂He nucleus (Ξ±-particle)Zβ†’Zβˆ’2Aβ†’Aβˆ’4Ionising, low penetration
Ξ²-decayElectron (β⁻) or Positron (β⁺)Zβ†’ZΒ±1A unchangedMedium penetration
Ξ³-decayHigh-energy photon (Ξ³-ray)UnchangedUnchangedHighly penetrating, EM radiation
☒️ Radioactive Decay Law: N(t) = Nβ‚€ e^(βˆ’Ξ»t)
Half-life: T₁/β‚‚ = 0.693/Ξ»
Mean life: Ο„ = 1/Ξ» = T₁/β‚‚/0.693
Activity: R = βˆ’dN/dt = Ξ»N (unit: Becquerel = 1 decay/s)
🎯 Ξ³-rays are shortest wavelength EM radiation (shorter than X-rays). They are emitted when a nucleus transitions between energy states after Ξ± or Ξ² decay β€” nucleus de-excitation.

🎯 Practice MCQs β€” Radioactivity

Q13 A radioactive sample has half-life T₁/β‚‚ = 20 days. After 60 days, what fraction of the original sample remains?
A 1/2
B 1/4
C 1/8
D 1/16
βœ… Number of half-lives = 60/20 = 3
Fraction remaining = (1/2)Β³ = 1/8
Q14 The decay constant of a radioactive element is 4.33 Γ— 10⁻⁴ s⁻¹. Its half-life in seconds is:
A 2310 s
B 4330 s
C 1600 s
D 6930 s
βœ… T₁/β‚‚ = 0.693/Ξ» = 0.693 / (4.33Γ—10⁻⁴) = 1600 s
Q15 In β⁻ decay, ²³⁸₉₂U β†’ ? The daughter nucleus has:
A Z = 91, A = 234
B Z = 93, A = 238
C Z = 90, A = 234
D Z = 92, A = 238
βœ… In β⁻ decay: a neutron converts to a proton β†’ Z increases by 1, A unchanged
Daughter: Z = 92+1 = 93 (Np), A = 238 β†’ ²³⁸₉₃Np
6
Nuclear Fission & Fusion

Nuclear Fission

A heavy nucleus splits into two intermediate-mass fragments when bombarded by a neutron. Example β€” Uranium fission:

ΒΉβ‚€n + ²³⁡₉₂U β†’ ²³⁢₉₂U β†’ ¹⁴⁴₅₆Ba + ⁸⁹₃₆Kr + 3ΒΉβ‚€n + ~200 MeV

Energy released per fission β‰ˆ 200 MeV. Source of energy in nuclear reactors and atom bombs (uncontrolled fission).

Fission of 1 kg uranium generates ~10¹⁴ J vs burning 1 kg coal = 10⁷ J β†’ nuclear is ~10 million times more energetic!

Nuclear Fusion

Two light nuclei combine to form a heavier, more tightly bound nucleus β€” releasing energy. Requires extremely high temperature (~10⁸ K) to overcome Coulomb barrier.

²₁H + ²₁H β†’ Β³β‚‚He + n + 3.27 MeV
4¹₁H β†’ ⁴₂He + 2e⁺ + 2Ξ½ + 6Ξ³ + 26.7 MeV (p-p cycle in sun)
β˜€οΈ Sun's energy source: Proton-proton (p-p) cycle β€” 4 hydrogen nuclei fuse to form helium-4 with release of 26.7 MeV.
🌑️ Thermonuclear fusion β€” fusion achieved by high temperature (particles get enough KE to overcome Coulomb barrier).
⚑ Controlled fusion β€” aim of fusion reactors (temperature needed ~10⁸ K, fuel = plasma).
πŸ”‘ Fission: Heavy (A>170) β†’ two medium nuclei β†’ ↑ Eᡦₙ β†’ energy released.
Fusion: Two light (A<30) β†’ one medium β†’ ↑ Eᡦₙ β†’ energy released.
Both exploit the binding energy curve!
πŸ–ΌοΈ
ADD IMAGE HERE
Fission chain reaction diagram & p-p fusion cycle in the Sun

🎯 Practice MCQs β€” Fission & Fusion

Q16 The fission properties of ²³⁹₉₄Pu are similar to ²³⁡U with average energy released = 180 MeV per fission. How much energy (in MeV) is released when all atoms in 1 kg of Pu undergo fission? (N_A = 6.023Γ—10Β²Β³, A = 239)
A 1.08 Γ— 10²⁡ MeV
B 4.54 Γ— 10²⁢ MeV
C 2.25 Γ— 10Β²Β³ MeV
D 1.8 Γ— 10Β² MeV
βœ… Number of atoms in 1 kg = (1000/239) Γ— 6.023Γ—10Β²Β³ = 2.52 Γ— 10²⁴
Total energy = 2.52Γ—10²⁴ Γ— 180 MeV = 4.54 Γ— 10²⁢ MeV
Q17 In the fusion reaction Β²H + Β²H β†’ Β³He + n + 3.27 MeV, how long can a 100 W electric lamp glow using 2 kg of deuterium? (Molar mass of Β²H = 2 g/mol)
A 1.4 Γ— 10⁸ s
B 2.4 Γ— 10¹⁰ s
C 4.9 Γ— 10¹⁰ s
D 1.0 Γ— 10⁡ s
βœ… Moles of Β²H in 2 kg = 2000/2 = 1000 mol β†’ N = 1000 Γ— 6.023Γ—10Β²Β³ = 6.023Γ—10²⁢ atoms
Pairs = 3.01Γ—10²⁢; Energy = 3.01Γ—10²⁢ Γ— 3.27 MeV = 9.84Γ—10²⁢ Γ— 1.6Γ—10⁻¹³ J = 1.57Γ—10¹⁴ J
Time = E/P = 1.57Γ—10¹⁴/100 β‰ˆ ~4.9 Γ— 10¹⁰ s (~1550 years!)
Q18 The Coulomb barrier height for head-on collision of two deuterons (radius = 2.0 fm each) is: (e = 1.6Γ—10⁻¹⁹ C, k = 9Γ—10⁹ NΒ·mΒ²/CΒ²)
A 400 eV
B 72 keV
C 360 keV
D 1.44 MeV
βœ… When two deuterons just touch, r = 2+2 = 4 fm = 4Γ—10⁻¹⁡ m
V = keΒ²/r = (9Γ—10⁹ Γ— (1.6Γ—10⁻¹⁹)Β²) / (4Γ—10⁻¹⁡)
V = 9Γ—10⁹ Γ— 2.56Γ—10⁻³⁸ / 4Γ—10⁻¹⁡ = 5.76Γ—10⁻¹⁴ J
V = 5.76Γ—10⁻¹⁴ / 1.6Γ—10⁻¹⁹ eV = 3.6Γ—10⁡ eV = 360 keV

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