Atoms and Nuclei

Nuclear Reactions - Fission, Fusion and Energy Release

What is the mass number of a nucleus with radius \( 5.4 \times 10^{-15} \, \text{m} \)? (Given \( R_0 = 1.2 \times 10^{-

**Nuclear fusion** source of energy in Sun, proton-proton cycle 4p→He+2e⁺+2ν+26.7 MeV, high temperature ~10⁷ K needed to give kinetic energy to overcome repulsion, thermal motion at high T allows tunneling, energy release because He BE/A higher than H. Fission releases energy because heavy nucleus BE/A ~7.6 MeV splits to intermediate ~8.5 MeV. R = R₀ A¹/³ . 5.4 × 10⁻¹⁵ = 1.2 × 10⁻¹⁵ × A¹/³ . A¹/³ = (5.4/1.2) = 4.5 . A = (4.5)³ = 91.125 ≈ 91 . Using E_n = -13.6/n² eV, r_n = n² a₀, L = n h/2π, R = R₀ A^¹/³, BE = Δm c² and 1 u

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Nuclear Reactions - Fission, Fusion and Energy Release

What is the mass defect of a nucleus with binding energy \( 186.3 \, \text{MeV} \)? (Given \( 1 \, \text{u} = 931.5 \, \

**Nuclear fission** splitting heavy nucleus like U-235 into intermediate mass fragments Ba and Kr plus neutrons, releases ~200 MeV per fission because product BE/A higher, mass defect converted to energy, controlled in reactors, uncontrolled in bombs. Fusion combining light nuclei D+T→He+n releases ~17.6 MeV, requires high temperature to overcome Coulomb barrier to bring nuclei close for strong force to act. Δ M = (E_b/c²) . Δ M = (186.3/931.5) ≈ 0.2 u . Using E_n = -13.6/n² eV, r_n = n² a₀, L = n h/2π, R = R₀ A^¹/³, BE = Δm c² and 1 u = 931.5 MeV, evaluation yields 0.2 u, consistent

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Nuclear Reactions - Fission, Fusion and Energy Release

What is the energy equivalent of \( 0.01 \, \text{kg} \) of matter in Joules? (Given \( c = 3 \times 10^8 \, \text{m/s}

**Energy release in nuclear processes** always because final BE/A higher than initial, mass defect difference appears as kinetic energy of fragments and radiation, 1 u =931.5 MeV, high temperature in fusion provides kinetic energy to overcome Coulomb barrier, confinement needed, Sun's core temperature ~1.5×10⁷ K enables fusion. E = m c² . m = 0.01 kg , c² = 9 × 10¹⁶ m²/s² . E = 0.01 × 9 × 10¹⁶ = 9 × 10¹⁴ J . Using E_n = -13.6/n² eV, r_n = n² a₀, L = n h/2π, R = R₀ A^¹/³, BE = Δm c² and 1 u = 931.5 MeV, evaluation

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Nuclear Reactions - Fission, Fusion and Energy Release

What is the energy equivalent of \( 0.005 \, \text{kg} \) of matter in Joules? (Given \( c = 3 \times 10^8 \, \text{m/s}

**Nuclear fusion** source of energy in Sun, proton-proton cycle 4p→He+2e⁺+2ν+26.7 MeV, high temperature ~10⁷ K needed to give kinetic energy to overcome repulsion, thermal motion at high T allows tunneling, energy release because He BE/A higher than H. Fission releases energy because heavy nucleus BE/A ~7.6 MeV splits to intermediate ~8.5 MeV. E = m c² . m = 0.005 kg , c² = 9 × 10¹⁶ m²/s² . E = 0.005 × 9 × 10¹⁶ = 4.5 × 10¹⁴ J . Using E_n = -13.6/n² eV, r_n = n² a₀, L = n h/2π, R = R₀ A^¹/³, BE = Δm c² and 1

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Nuclear Reactions - Fission, Fusion and Energy Release

Which process involves the splitting of a heavy nucleus into two intermediate mass fragments?

**Nuclear fission** splitting heavy nucleus like U-235 into intermediate mass fragments Ba and Kr plus neutrons, releases ~200 MeV per fission because product BE/A higher, mass defect converted to energy, controlled in reactors, uncontrolled in bombs. Fusion combining light nuclei D+T→He+n releases ~17.6 MeV, requires high temperature to overcome Coulomb barrier to bring nuclei close for strong force to act. Nuclear fission is the process where a heavy nucleus splits into two intermediate mass fragments, releasing energy due to the higher binding energy per nucleon in the resulting nuclei. Using E_n = -13.6/n² eV, r_n = n² a₀, L = n h/2π, R = R₀

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Nuclear Reactions - Fission, Fusion and Energy Release

What is the mass defect of a nucleus with binding energy \( 149.04 \, \text{MeV} \)? (Given \( 1 \, \text{u} = 931.5 \,

**Energy release in nuclear processes** always because final BE/A higher than initial, mass defect difference appears as kinetic energy of fragments and radiation, 1 u =931.5 MeV, high temperature in fusion provides kinetic energy to overcome Coulomb barrier, confinement needed, Sun's core temperature ~1.5×10⁷ K enables fusion. Δ M = (E_b/c²) . Δ M = (149.04/931.5) ≈ 0.16 u . Using E_n = -13.6/n² eV, r_n = n² a₀, L = n h/2π, R = R₀ A^¹/³, BE = Δm c² and 1 u = 931.5 MeV, evaluation yields 0.16 u, consistent with Bohr model and nuclear binding energy systematics.

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Nuclear Reactions - Fission, Fusion and Energy Release

What is the primary source of energy in stars like the Sun?

**Nuclear fusion** source of energy in Sun, proton-proton cycle 4p→He+2e⁺+2ν+26.7 MeV, high temperature ~10⁷ K needed to give kinetic energy to overcome repulsion, thermal motion at high T allows tunneling, energy release because He BE/A higher than H. Fission releases energy because heavy nucleus BE/A ~7.6 MeV splits to intermediate ~8.5 MeV. The energy in stars like the Sun is generated through nuclear fusion, where light nuclei (e.g., hydrogen) combine to form heavier nuclei (e.g., helium), releasing energy due to increased binding energy per nucleon. Using E_n = -13.6/n² eV, r_n = n² a₀, L = n h/2π, R = R₀ A^¹/³, BE = Δm

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Radioactive Decay, Nuclear Forces and Stability

The binding energy per nucleon of a nucleus is \( 8.5 \, \text{MeV} \). What is the total binding energy for a nucleus w

**Nuclear force** strong, short-range ~1 fm, attractive, charge independent, saturated in large nuclei because each nucleon interacts only with neighbors, not all others, so BE/A saturates ~8 MeV, primary factor limiting stable nuclei size is Coulomb repulsion between protons growing as Z² vs strong force saturating, beyond Z≈83 no stable nuclei, competition between Coulomb and strong. Total binding energy = Ebₙ × A . Ebₙ = 8.5 MeV , A = 20 . E_b = 8.5 × 20 = 170 MeV . Using E_n = -13.6/n² eV, r_n = n² a₀, L = n h/2π, R = R₀ A^¹/³, BE = Δm c² and 1

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Radioactive Decay, Nuclear Forces and Stability

How much energy is equivalent to a mass defect of \( 0.1 \, \text{u} \)? (Given \( 1 \, \text{u} = 931.5 \, \text{MeV/c}

**Stability** belt of stability N≈Z for light, N>Z for heavy due to Coulomb, beyond leads to alpha decay, fission, stability requires balance, nuclear force saturated explains constant density and BE/A. Energy = Δ M · c² . Δ M = 0.1 u . Energy = 0.1 × 931.5 = 93.15 MeV . Using E_n = -13.6/n² eV, r_n = n² a₀, L = n h/2π, R = R₀ A^¹/³, BE = Δm c² and 1 u = 931.5 MeV, evaluation yields 93.15 MeV, consistent with Bohr model and nuclear binding energy systematics.

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Radioactive Decay, Nuclear Forces and Stability

What is the primary factor limiting the size of stable nuclei?

**Radioactive decay** occurs when nucleus unstable, alpha decay emits He-4, beta decay neutron→proton+electron+antineutrino, gamma decay photon emission, decay law N=N₀ e^{-λt}, half-life T½=ln2/λ, nuclear density ~10¹⁷ kg/m³, nuclear force saturated means BE/A constant for A>20. In large nuclei, the Coulomb repulsion between protons increases with atomic number, counteracting the nuclear force and reducing stability, limiting the size of stable nuclei. Using E_n = -13.6/n² eV, r_n = n² a₀, L = n h/2π, R = R₀ A^¹/³, BE = Δm c² and 1 u = 931.5 MeV, evaluation yields Coulomb repulsion, consistent with Bohr model and nuclear binding energy systematics.

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Radioactive Decay, Nuclear Forces and Stability

A nucleus with mass number 18 has a binding energy of \( 144 \, \text{MeV} \). What is its binding energy per nucleon?

**Nuclear force** strong, short-range ~1 fm, attractive, charge independent, saturated in large nuclei because each nucleon interacts only with neighbors, not all others, so BE/A saturates ~8 MeV, primary factor limiting stable nuclei size is Coulomb repulsion between protons growing as Z² vs strong force saturating, beyond Z≈83 no stable nuclei, competition between Coulomb and strong. Ebₙ = (E_b/A) . E_b = 144 MeV , A = 18 . Ebₙ = (144/18) = 8.0 MeV . Using E_n = -13.6/n² eV, r_n = n² a₀, L = n h/2π, R = R₀ A^¹/³, BE = Δm c² and 1 u = 931.5 MeV, evaluation yields

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Radioactive Decay, Nuclear Forces and Stability

A nucleus with mass number 28 has a binding energy of \( 224 \, \text{MeV} \). What is its binding energy per nucleon?

**Stability** belt of stability N≈Z for light, N>Z for heavy due to Coulomb, beyond leads to alpha decay, fission, stability requires balance, nuclear force saturated explains constant density and BE/A. Ebₙ = (E_b/A) . E_b = 224 MeV , A = 28 . Ebₙ = (224/28) = 8.0 MeV . Using E_n = -13.6/n² eV, r_n = n² a₀, L = n h/2π, R = R₀ A^¹/³, BE = Δm c² and 1 u = 931.5 MeV, evaluation yields 8.0 MeV, consistent with Bohr model and nuclear binding energy systematics.

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