Chapter 28: Nuclei

Introduction

Key Points

• Positive charge and mass concentrated in nucleus at atom's center.
• Nucleus volume ≈ 10⁻¹⁵ times atom's volume; atom is mostly empty space.
• Study of nucleus properties: size, mass, stability, radioactivity, fission, fusion.

Solved Examples

Ex 1

• Calculate ratio of nucleus volume to atom volume if nucleus radius is 10⁻¹⁵ m and atom radius is 10⁻¹⁰ m.
• Use volume ratio = (r_nucleus / r_atom)^3 = (10⁻¹⁵ / 10⁻¹⁰)^3 = (10⁻⁵)^3 = 10⁻¹⁵.

Practice Set

• Q1: Where is most of the atom's mass located?
  • A) Electron cloud
  • B) Nucleus
  • C) Entire atom evenly
  • D) Outer shells
• Q2: The volume of the nucleus is approximately how many times smaller than the atom?
  • A) 10⁻⁵
  • B) 10⁻¹⁰
  • C) 10⁻¹⁵
  • D) 10⁻²⁰
• Q3: Which phenomenon is NOT related to the nucleus?
  • A) Radioactivity
  • B) Fission
  • C) Chemical bonding
  • D) Fusion

Answer Key

• Q1: B - Nucleus contains most of the atom's mass.
• Q2: C - Nucleus volume is about 10⁻¹⁵ times the atom's volume.
• Q3: C - Chemical bonding involves electrons, not the nucleus.

Memory Aids

• "Nucleus = dense core, mass center."
• Volume ratio = (radius ratio)^3 = 10⁻¹⁵.
• Radioactivity, fission, fusion = nuclear phenomena.

Atomic Masses and Composition of Nucleus

Key Points

• Atomic mass = weighted average of isotopes' masses.
• 1 amu = \( \frac{1}{12} \times \) mass of carbon-12 atom = 1.660539 × 10⁻²⁷ kg = 931 MeV/c².
• Nucleons = protons + neutrons in nucleus.
• Atomic number (Z) = number of protons.
• Mass number (A) = protons (Z) + neutrons (N).
• Nuclear mass = total mass of protons and neutrons.
• Nuclide notation: \( ^A_Z X \), where X = chemical symbol.

Solved Examples

Ex 1

• Find mass of 1 amu in kg.
  • Given: 1 amu = 1/12 mass of carbon-12 = 1.660539 × 10⁻²⁷ kg.
• Ex 2
• Write nuclide notation for carbon-14.
  • Protons (Z) = 6, Mass number (A) = 14.
  • Notation: \( ^{14}_6 C \).

Practice Set

• Q1: 1 amu equals approximately:
  • A) 1.66 × 10⁻²⁷ kg
  • B) 1.66 × 10⁻²⁶ kg
  • C) 931 MeV/c²
  • D) Both A and C
• Q2: Atomic number (Z) represents:
  • A) Number of neutrons
  • B) Number of protons
  • C) Mass number
  • D) Number of electrons
• Q3: Mass number (A) is:
  • A) Number of protons
  • B) Number of neutrons
  • C) Sum of protons and neutrons
  • D) Number of electrons

Answer Key

• Q1: D - 1 amu = 1.66 × 10⁻²⁷ kg and 931 MeV/c².
• Q2: B - Atomic number = number of protons.
• Q3: C - Mass number = protons + neutrons.

Memory Aids

• 1 amu ≈ mass of 1 nucleon.
• Z = protons; A = protons + neutrons.
• Nuclide: \( ^A_Z X \).

Properties of Nucleons

Key Points

• Proton (p): charge +e, mass ≈ 1 amu, rest energy ≈ 938 MeV.
• Neutron (n): charge 0, mass ≈ 1 amu, rest energy ≈ 940 MeV.
• Masses of proton and neutron nearly equal.

Solved Examples

Ex 1

• Compare rest energies of proton and neutron.
  • Proton: 938 MeV, Neutron: 940 MeV; difference ≈ 2 MeV.

Practice Set

• Q1: Proton charge is:
  • A) 0
  • B) +e
  • C) -e
  • D) +2e
• Q2: Neutron mass compared to proton is:
  • A) Much less
  • B) Much more
  • C) Nearly equal
  • D) Zero
• Q3: Rest energy of proton is approximately:
  • A) 1 MeV
  • B) 938 MeV
  • C) 931 MeV
  • D) 940 MeV

Answer Key

• Q1: B - Proton charge is +e.
• Q2: C - Neutron mass nearly equals proton mass.
• Q3: B - Proton rest energy ≈ 938 MeV.

Memory Aids

• Proton: +e charge, neutron: neutral.
• Mass ~1 amu for both nucleons.
• Rest energy ~938 MeV proton, ~940 MeV neutron.

Isotopes, Isobars, Isotones and Isomers

Key Points

• Isotopes: same Z, different A (different neutrons).
• Isobars: same A, different Z.
• Isotones: same neutron number, different Z.
• Isomers: same Z and A, different energy states.

Solved Examples

Ex 1

• Identify isotopes of hydrogen:
  • ¹H (protium), ²H (deuterium), ³H (tritium) differ in neutrons.
• Ex 2
• Example of isobars: ³₁H and ³₂He (same A=3, different Z).

Practice Set

• Q1: Isotopes have:
  • A) Same Z, different A
  • B) Same A, different Z
  • C) Same neutron number, different Z
  • D) Same Z and A, different energy
• Q2: Which pair are isobars?
  • A) ³₁H and ³₂He
  • B) ¹H and ²H
  • C) ³⁷₁₇Cl and ³⁹₁₉K
  • D) ⁶Li and ⁷Li
• Q3: Isotones have:
  • A) Same Z, different A
  • B) Same neutron number, different Z
  • C) Same A, different Z
  • D) Same Z and A, different energy

Answer Key

• Q1: A - Isotopes have same atomic number, different mass number.
• Q2: A - ³₁H and ³₂He have same mass number (3), different atomic number.
• Q3: B - Isotones have same neutron number, different atomic number.

Memory Aids

• Isotopes: "Same Z, different N."
• Isobars: "Same A, different Z."
• Isotones: "Same N, different Z."
• Isomers: "Same Z & A, different energy."

Size of the Nucleus

Nuclear Size

• Volume of nucleus ∝ number of nucleons (mass number A)
• Formula: \( \frac{4}{3} \pi R^3 \propto A \) or \( R \propto A^{1/3} \)
• Radius: \( R = R_0 A^{1/3} \), where \( R_0 = 1.2 \times 10^{-15} \) m

Solved Examples

• Ex 1: Find radius of nucleus with \( A=27 \).
  • Use \( R = 1.2 \times 10^{-15} \times 27^{1/3} = 1.2 \times 10^{-15} \times 3 = 3.6 \times 10^{-15} \) m

Practice Set

• Q1: Radius of nucleus with \( A=64 \) is?
  • A) \(4.8 \times 10^{-15} \) m
  • B) \(1.2 \times 10^{-15} \) m
  • C) \(3.6 \times 10^{-15} \) m
  • D) \(6.0 \times 10^{-15} \) m
• Q2: If radius of nucleus is \( 6 \times 10^{-15} \) m, find mass number A.
  • A) 125
  • B) 216
  • C) 343
  • D) 64
• Q3: Radius of nucleus doubles, mass number changes by factor?
  • A) 2
  • B) 4
  • C) 8
  • D) 16

Answer Key

• Q1: A; \( R = 1.2 \times 10^{-15} \times 64^{1/3} = 1.2 \times 4 = 4.8 \times 10^{-15} \) m
• Q2: B; \( A = \left( \frac{R}{R_0} \right)^3 = \left( \frac{6 \times 10^{-15}}{1.2 \times 10^{-15}} \right)^3 = 5^3 = 125 \)
• Q3: C; \( R \propto A^{1/3} \Rightarrow A \propto R^3 \), doubling radius means \( A \) increases by \( 2^3 = 8 \)

Memory Aids

• Radius \( R = R_0 A^{1/3} \) (cube root relation)
• \( R_0 = 1.2 \times 10^{-15} \) m (constant nuclear radius parameter)
• Volume ∝ mass number (A)

Nuclear Density

Density of Nucleus

• Density \( \rho = \frac{\text{mass}}{\text{volume}} = \frac{A \times 1.66 \times 10^{-27} \text{ kg}}{\frac{4}{3} \pi R^3} \)
• Using \( R = R_0 A^{1/3} \), density simplifies to \( \rho = \frac{1.66 \times 10^{-27}}{\frac{4}{3} \pi R_0^3} = 2.29 \times 10^{17} \text{ kg/m}^3 \)
• Density is constant for all nuclei, independent of mass number
• Density highest at center, decreases towards edges; nucleus has no sharp boundary
• High density (~\(10^{17} \text{ kg/m}^3\)) indicates nucleus is highly compact

Solved Examples

• Ex 1: Calculate nuclear density using \( R_0 = 1.2 \times 10^{-15} \) m.
  • \( \rho = \frac{1.66 \times 10^{-27}}{\frac{4}{3} \pi (1.2 \times 10^{-15})^3} = 2.29 \times 10^{17} \text{ kg/m}^3 \)

Practice Set

• Q1: Nuclear density is approximately?
  • A) \(10^{3} \text{ kg/m}^3\)
  • B) \(10^{10} \text{ kg/m}^3\)
  • C) \(10^{17} \text{ kg/m}^3\)
  • D) \(10^{27} \text{ kg/m}^3\)
• Q2: If nuclear radius doubles, density changes by?
  • A) Doubles
  • B) Halves
  • C) Remains same
  • D) Quadruples
• Q3: Which statement is true about nuclear density?
  • A) Varies with mass number
  • B) Maximum at edges
  • C) Independent of mass number
  • D) Much less than atomic density

Answer Key

• Q1: C; Nuclear density is about \( 2.29 \times 10^{17} \text{ kg/m}^3 \)
• Q2: C; Density is constant, independent of radius or mass number
• Q3: C; Density does not depend on mass number, uniform for all nuclei

Memory Aids

• Density formula simplifies to constant value
• \( \rho \approx 2.3 \times 10^{17} \text{ kg/m}^3 \)
• Density highest at center, no sharp boundary

Mass-Energy and Nuclear Binding Energy

Key Concepts

• Mass-energy equivalence: \( E = mc^2 \) (Einstein's relation)
• Binding energy (\( E_b \)) is the energy holding nucleons (protons and neutrons) together in a nucleus.
• Mass defect (\( \Delta M \)) = \( [Z m_p + (A - Z) m_n] - M \), where \(Z\) = number of protons, \(A\) = mass number, \(m_p\) = proton mass, \(m_n\) = neutron mass, \(M\) = nucleus mass.
• Binding energy formula: \( E_b = \Delta M c^2 = \left([Z m_p + (A - Z) m_n] - M\right) c^2 \)
• Binding energy per nucleon: \( \Delta E_{bn} = \frac{E_b}{A} \) (average energy to remove one nucleon)

Solved Examples

Ex 1: Calculate binding energy if mass defect \( \Delta M = 0.0015 \) u. (1 u = 931.5 MeV/c²)

• \( E_b = \Delta M c^2 = 0.0015 \times 931.5 = 1.397 \text{ MeV} \)

Practice Set

• Q1: What is the mass defect for a nucleus with \( Z=2 \), \( A=4 \), proton mass = 1.0073 u, neutron mass = 1.0087 u, nucleus mass = 4.0026 u?
  • A) 0.0301 u
  • B) 0.0251 u
  • C) 0.0201 u
  • D) 0.0151 u
• Q2: Binding energy per nucleon for a nucleus with total binding energy 28 MeV and mass number 7 is:
  • A) 4 MeV
  • B) 3 MeV
  • C) 5 MeV
  • D) 6 MeV
• Q3: If mass defect is 0.002 u, what is the binding energy? (1 u = 931.5 MeV/c²)
  • A) 1.863 MeV
  • B) 0.931 MeV
  • C) 2.863 MeV
  • D) 3.863 MeV

Answer Key

• Q1: A) 0.0301 u — Calculated as \( (2 \times 1.0073 + 2 \times 1.0087) - 4.0026 = 0.0301 \) u
• Q2: A) 4 MeV — Binding energy per nucleon = \( 28 / 7 = 4 \) MeV
• Q3: A) 1.863 MeV — \( 0.002 \times 931.5 = 1.863 \) MeV

Memory Aids

• Mass defect = sum of nucleon masses - nucleus mass
• Binding energy \( E_b = \Delta M c^2 \)
• Binding energy per nucleon = total binding energy / mass number
• 1 atomic mass unit (u) = 931.5 MeV/c²

Binding Energy Curve and Nuclear Stability

Key Concepts

• Binding energy per nucleon vs mass number (A) graph shows nuclear stability.
• Peak at \( A \approx 56 \) (Iron-56) with max binding energy per nucleon ~8.75 MeV.
• Nuclei with low or very high mass numbers have lower binding energy per nucleon and are less stable.
• Higher binding energy per nucleon means more stable nucleus.
• Energy release occurs via:
  • Nuclear fission: Heavy nucleus splits into lighter nuclei with higher binding energy per nucleon.
  • Nuclear fusion: Light nuclei combine to form heavier nucleus with higher binding energy per nucleon.

Solved Examples

Ex 1: Why does fusion of light nuclei release energy?

• Because final nucleus has higher binding energy per nucleon than initial nuclei.

Ex 2: Why does fission of heavy nuclei release energy?

• Because resulting lighter nuclei have higher binding energy per nucleon than original heavy nucleus.

Practice Set

• Q1: Which nucleus is most stable based on binding energy per nucleon?
  • A) Hydrogen (A=1)
  • B) Iron (A=56)
  • C) Uranium (A=238)
  • D) Helium (A=4)
• Q2: Energy is released in nuclear fusion because:
  • A) Binding energy per nucleon decreases
  • B) Binding energy per nucleon increases
  • C) Mass number decreases
  • D) Mass number increases
• Q3: Nuclear fission occurs when:
  • A) Light nuclei combine
  • B) Heavy nucleus splits into lighter nuclei with higher binding energy per nucleon
  • C) Binding energy per nucleon decreases
  • D) Mass defect is zero

Answer Key

• Q1: B) Iron (A=56) — It has the highest binding energy per nucleon, hence most stable.
• Q2: B) Binding energy per nucleon increases — Fusion forms a nucleus with higher binding energy per nucleon, releasing energy.
• Q3: B) Heavy nucleus splits into lighter nuclei with higher binding energy per nucleon — This releases energy in fission.

Memory Aids

• Peak of binding energy curve = most stable nucleus (Iron-56)
• Fusion: light nuclei → heavier nucleus + energy
• Fission: heavy nucleus → lighter nuclei + energy
• Higher binding energy per nucleon = greater stability

Nuclear Force

Key Properties

• Electrostatic repulsion between protons is much stronger than gravitational attraction: \( \frac{F_e}{F_g} = 10^{36} \).
• Nuclear force is a strong attractive force acting at short range (~4.2 × 10⁻¹⁵ m).
• It balances electrostatic repulsion, holding nucleons (protons and neutrons) together in the nucleus.
• Charge independent: same magnitude between proton-proton, proton-neutron, and neutron-neutron pairs.
• Attractive force arises from exchange of π mesons (\( \pi^0, \pi^+, \pi^- \)).
• Acts only between nearest neighbors (saturated force), not all nucleons.
• Strongest force in nature within its range: ~10² times stronger than electrostatic, ~10³⁸ times stronger than gravitational force.
• Potential energy vs. separation graph shows:
  • Strong repulsion at distances < \( r_0 \) (~1 fm).
  • Strong attraction at distances > \( r_0 \), with minimum potential energy at \( r_0 \).

Solved Examples

Ex 1: Calculate the ratio of electrostatic to gravitational force between two protons.

• Given: \( \frac{F_e}{F_g} = 10^{36} \) (from nuclear force context)
• Answer: \( 10^{36} \)

Ex 2: Describe the nature of nuclear force at \( r < r_0 \) and \( r > r_0 \).

• At \( r < r_0 \): Force is strongly repulsive.
• At \( r > r_0 \): Force is attractive, stabilizing the nucleus.

Practice Set

Q1: Which of the following is true about nuclear force?

• A. Acts over long distances
• B. Is charge dependent
• C. Is saturated force
• D. Weaker than gravitational force

Q2: The nuclear force is approximately how many times stronger than the electrostatic force?

• A. 10²
• B. 10⁴
• C. 10⁶
• D. 10⁸

Q3: At what approximate distance does the nuclear force become repulsive?

• A. Greater than 4.2 × 10⁻¹⁵ m
• B. Less than 1 femtometer
• C. At 10⁻¹⁰ m
• D. At 1 meter

Answer Key

Q1: C. Nuclear force is a saturated force acting only with nearest neighbors.

Q2: A. Nuclear force is about 10² times stronger than electrostatic force within its range.

Q3: B. Nuclear force becomes repulsive at distances less than about 1 femtometer (\( r_0 \)).

Memory Aids

• "Nuclear force: Short, Strong, Saturated, and Charge-independent."
• Remember: \( F_e / F_g = 10^{36} \) shows electrostatic repulsion dominates gravity.
• Potential energy minimum at \( r_0 \) (~1 fm) indicates stable nucleon separation.

Radioactivity

Overview

• Discovered in 1896 by Antoine Henri Becquerel.
• Spontaneous disintegration of certain nuclei (radionuclides) emitting alpha (\(\alpha\)), beta (\(\beta\)) particles, and gamma (\(\gamma\)) rays.
• Three natural decay types: alpha decay, beta decay, gamma decay.

Solved Examples

Ex 1: Identify the type of radiation emitted when a nucleus emits a helium nucleus.

• Helium nucleus emission = alpha decay.

Practice Set

Q1: Who discovered radioactivity?

• A) Marie Curie
• B) Antoine Henri Becquerel
• C) Ernest Rutherford
• D) Niels Bohr

Q2: Radioactivity is the spontaneous disintegration of nuclei with emission of:

• A) Only alpha particles
• B) Alpha, beta particles and gamma rays
• C) Only beta particles
• D) Only gamma rays

Q3: Which of the following is NOT a type of natural radioactive decay?

• A) Alpha decay
• B) Beta decay
• C) Gamma decay
• D) Positron decay

Answer Key

Q1: B) Antoine Henri Becquerel - He discovered radioactivity in 1896.

Q2: B) Alpha, beta particles and gamma rays - All three are emitted in natural radioactivity.

Q3: D) Positron decay - Positron emission is a type of beta decay, not a separate decay type.

Memory Aids

• Radioactivity = spontaneous nucleus decay emitting \(\alpha\), \(\beta\), \(\gamma\).
• Discovered by Becquerel in 1896.
• Three decay types: alpha (helium nucleus), beta (electron/positron), gamma (photon).

Alpha Decay

Key Points

• Emission of alpha particle \( ^4_2He \) from heavy nucleus.
• General reaction: \( ^A_ZX \rightarrow ^{A-4}_{Z-2}Y + ^4_2He + Q \)
• Energy released (Q) = \( (m_X - m_Y - m_{He})c^2 \)
• Energy shared between daughter nucleus and alpha particle.
• Example: Alpha decay of \( ^{238}_{92}U \) releases 4.25 MeV.
• Proton emission requires more energy (7.68 MeV for \( ^{238}_{92}U \)), so not spontaneous.


Solved Examples

Ex 1: Calculate Q value if \( m_X = 238.05u \), \( m_Y = 234.04u \), \( m_{He} = 4.00u \), \( c^2 = 931.5 MeV/u \).

• \( Q = (238.05 - 234.04 - 4.00) \times 931.5 = 0.01 \times 931.5 = 9.315 \text{ MeV} \)

Practice Set

Q1: In alpha decay, the atomic number decreases by:

• A) 1
• B) 2
• C) 4
• D) 0

Q2: The energy released in alpha decay is given by:

• A) \( (m_X + m_Y + m_{He})c^2 \)
• B) \( (m_X - m_Y - m_{He})c^2 \)
• C) \( (m_Y - m_X - m_{He})c^2 \)
• D) \( (m_{He} - m_X - m_Y)c^2 \)

Q3: Which particle is emitted in alpha decay?

• A) Electron
• B) Positron
• C) Helium nucleus
• D) Photon

Answer Key

Q1: B) 2 - Alpha particle has 2 protons, so atomic number decreases by 2.

Q2: B) \( (m_X - m_Y - m_{He})c^2 \) - Energy released equals mass difference times \( c^2 \).

Q3: C) Helium nucleus - Alpha particle is a helium nucleus.

Memory Aids

• Alpha decay: loses 2 protons and 2 neutrons.
• Q value formula: \( Q = (m_X - m_Y - m_{He})c^2 \).
• Energy shared between alpha particle and daughter nucleus.

Beta Decay

Key Points

• Spontaneous emission of electron (\( \beta^- \)) or positron (\( \beta^+ \)) from nucleus.
• \( \beta^- \) decay: neutron \( \to \) proton + electron + antineutrino.
• Atomic number increases by 1; mass number \( A \) unchanged.
• \( \beta^+ \) decay: proton \( \to \) neutron + positron + neutrino.
• Atomic number decreases by 1; mass number \( A \) unchanged.


Solved Examples

Ex 1: In \( \beta^- \) decay, if original element has atomic number Z, what is atomic number after decay?

• New atomic number = \( Z + 1 \)

Practice Set

Q1: In \( \beta^- \) decay, a neutron converts into:

• A) Proton + electron + neutrino
• B) Proton + electron + antineutrino
• C) Neutron + positron + neutrino
• D) Proton + positron + neutrino

Q2: During beta decay, the mass number changes by:

• A) +1
• B) -1
• C) 0
• D) +2

Q3: In \( \beta^+ \) decay, the atomic number changes by:

• A) +1
• B) -1
• C) 0
• D) +2

Answer Key

Q1: B) Proton + electron + antineutrino - This is the \( \beta^- \) decay process.

Q2: C) 0 - Mass number remains unchanged in beta decay.

Q3: B) -1 - Atomic number decreases by 1 in \( \beta^+ \) decay.

Memory Aids

• \( \beta^- \): neutron \( \to \) proton + electron + antineutrino; Z increases by 1.
• \( \beta^+ \): proton \( \to \) neutron + positron + neutrino; Z decreases by 1.
• Mass number \( A \) unchanged.

Gamma Decay

Key Points

• Emission of high-energy photons (\( \gamma \)-rays) from excited nucleus to ground state.
• Occurs after alpha or beta decay leaves nucleus in excited state.
• Example: \( ^{60}_{27}Co \) undergoes beta decay to excited \( ^{60}_{28}Ni^* \), which emits two gamma rays (1.17 MeV and 1.33 MeV) to reach ground state.
• Gamma emission does not change atomic or mass number.


Solved Examples

Ex 1: What happens to the atomic number during gamma decay?

• Atomic number remains the same.

Practice Set

Q1: Gamma decay involves emission of:

• A) Electron
• B) Positron
• C) Photon
• D) Alpha particle

Q2: After beta decay, the nucleus is often:

• A) Stable
• B) Excited
• C) Ionized
• D) Neutral

Q3: The energies of gamma rays emitted by \( ^{60}Ni^* \) are approximately:

• A) 0.5 MeV and 0.7 MeV
• B) 1.17 MeV and 1.33 MeV
• C) 2 MeV and 3 MeV
• D) 4 MeV and 5 MeV

Answer Key

Q1: C) Photon - Gamma decay emits high-energy photons.

Q2: B) Excited - Beta decay leaves nucleus in excited state.

Q3: B) 1.17 MeV and 1.33 MeV - These are the gamma energies emitted by \( ^{60}Ni^* \).

Memory Aids

• Gamma decay = photon emission, no change in Z or A.
• Follows alpha or beta decay to release excess energy.
• Example: \( ^{60}Co \to ^{60}Ni^* \to ^{60}Ni + \gamma \).

Nuclear Energy

Key Concepts

• Nuclear reaction general form: A + a → B + b + Q (Q = energy released)
• Q-value = (mass of reactants - mass of products) × c²; positive Q = exothermic, negative Q = endothermic
• Conservation laws in nuclear reactions:
  • Number of nucleons conserved: A + a = B + b
  • Atomic number conserved: Z(A + a) = Z(B + b)
  • Energy change equals mass difference energy equivalent
  • Linear momentum conserved: total momentum before = total momentum after

Solved Examples

Ex 1

Calculate Q-value if mass of reactants = 10.005 u and products = 9.995 u. (1 u = 931.5 MeV/c²)

Q = (10.005 - 9.995) × 931.5 = 0.01 × 931.5 = 9.315 MeV (exothermic)

Practice Set

• Q1: What is conserved in a nuclear reaction?
  • A) Only mass
  • B) Only energy
  • C) Number of nucleons and atomic number
  • D) Neither mass nor energy
• Q2: If Q-value is negative, the reaction is:
  • A) Exothermic
  • B) Endothermic
  • C) Spontaneous
  • D) Impossible
• Q3: In a reaction, if mass of reactants is 12 u and products 11.995 u, energy released is:
  • A) 4.66 MeV
  • B) 0.005 MeV
  • C) 466 MeV
  • D) 0.5 MeV

Answer Key

• Q1: C - Number of nucleons and atomic number are conserved in nuclear reactions.
• Q2: B - Negative Q-value means energy is absorbed; reaction is endothermic.
• Q3: A - Energy = mass difference × 931.5 MeV = 0.005 × 931.5 = 4.66 MeV.

Memory Aids

• Q-value positive = energy released (exo)
• Q-value negative = energy absorbed (endo)
• Conservation: nucleons and charge always conserved

Nuclear Fission

Key Concepts

• Fission: splitting heavy nucleus into smaller nuclei + energy
• Example: \( _{92}^{235}U + _{0}^{1}n \to _{56}^{144}Ba + _{36}^{89}Kr + 3 _{0}^{1}n + Q \)
• Energy released per fission ≈ 200 MeV
• Applications:
  • Controlled energy in nuclear reactors
  • Sudden energy release in atomic bombs


Solved Examples

Ex 1

Calculate total energy released if 1 mole of \( _{92}^{235}U \) undergoes fission releasing 200 MeV per atom.

Energy = 200 MeV × Avogadro's number = 200 × 1.6×10⁻¹³ J × 6.022×10²³ ≈ 1.93×10¹³ J

Practice Set

• Q1: Nuclear fission involves:
  • A) Fusion of light nuclei
  • B) Splitting of heavy nucleus
  • C) Emission of alpha particles only
  • D) Electron capture
• Q2: Energy released per fission of uranium-235 is approximately:
  • A) 2 MeV
  • B) 20 MeV
  • C) 200 MeV
  • D) 2000 MeV
• Q3: Number of neutrons released in uranium-235 fission is:
  • A) 1
  • B) 2
  • C) 3
  • D) 4

Answer Key

• Q1: B - Fission is splitting of heavy nucleus into smaller nuclei.
• Q2: C - About 200 MeV energy is released per uranium-235 fission.
• Q3: C - Typically 3 neutrons are emitted per fission.

Memory Aids

• Fission = "F" for "Fragmentation" (splitting)
• U-235 + neutron → Ba + Kr + 3 neutrons + energy
• Energy ~ 200 MeV per fission event

Nuclear Fusion

Key Concepts

• Fusion: combining light nuclei to form heavier nucleus + energy
• Energy released due to higher binding energy per nucleon in heavier nucleus
• Examples:
  • \( _1^1H + _1^1H \to _1^2H + e^+ + \nu + 0.42 \text{ MeV} \)
  • \( _1^2H + _1^2H \to _2^3He + _0^1n + 3.27 \text{ MeV} \)
  • \( _1^2H + _1^2H \to _1^3H + _1^1H + 4.03 \text{ MeV} \)
• Thermonuclear fusion: fusion at very high temperatures (~10⁷–10⁸ K) to overcome Coulomb barrier
• Fusion powers stars; proton-proton cycle in sun converts hydrogen to helium releasing 26.7 MeV

Solved Examples

Ex 1

Calculate total energy released when 4 hydrogen nuclei fuse to form 1 helium nucleus in the sun.

Energy released = 26.7 MeV per 4 protons fused

Practice Set

• Q1: Fusion reaction releases energy because:
  • A) Heavier nucleus has lower binding energy
  • B) Heavier nucleus has higher binding energy per nucleon
  • C) Mass is created
  • D) Neutrons are lost
• Q2: The main fuel in the sun's core is:
  • A) Helium
  • B) Hydrogen
  • C) Carbon
  • D) Oxygen
• Q3: Energy released in proton-proton fusion cycle per 4 protons is approximately:
  • A) 2.67 MeV
  • B) 26.7 MeV
  • C) 267 MeV
  • D) 0.267 MeV

Answer Key

• Q1: B - Fusion releases energy as heavier nucleus has higher binding energy per nucleon.
• Q2: B - Hydrogen is the main fuel in the sun's core.
• Q3: B - About 26.7 MeV energy released per 4 protons fused.

Memory Aids

• Fusion = "F" for "Fusing" light nuclei
• Sun's core temperature ~1.5×10⁷ K for fusion
• Proton-proton cycle: 4H → He + energy (26.7 MeV)

Controlled Thermonuclear Fusion

Key Concepts

• Goal: steady power generation by heating fuel to ~10⁸ K
• Fuel is plasma: mixture of ions and electrons
• Challenge: confining plasma at high temperature without material container
• Global research ongoing; success could provide unlimited clean energy

Solved Examples

Ex 1

Explain why plasma confinement is necessary in fusion reactors.

Because no material can withstand ~10⁸ K, plasma must be confined by magnetic or inertial methods.

Practice Set

• Q1: Controlled fusion reactors aim to heat fuel to about:
  • A) 10³ K
  • B) 10⁵ K
  • C) 10⁸ K
  • D) 10¹¹ K
• Q2: Plasma in fusion reactors consists of:
  • A) Neutral atoms
  • B) Ions and electrons
  • C) Only electrons
  • D) Only ions
• Q3: Main challenge in controlled fusion is:
  • A) Fuel availability
  • B) Plasma confinement
  • C) Energy measurement
  • D) Cooling the reactor

Answer Key

• Q1: C - Fusion requires ~10⁸ K to overcome Coulomb barrier.
• Q2: B - Plasma is ionized gas of ions and electrons.
• Q3: B - Confining plasma at high temperature is the main challenge.

Memory Aids

• Controlled fusion needs "C" for "Confinement" and "Centillion" K (~10⁸)
• Plasma = charged particles, not neutral gas
• Magnetic confinement is key technology