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#nuclear physics

74 public questions tagged with this topic.

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.

Ref: NCERT > Physics Book > Atoms and Nuclei > Nuclear Reactions - Fission, Fusion and Energy Release

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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

Ref: NCERT > Physics Book > Atoms and Nuclei > Radioactive Decay, Nuclear Forces and Stability

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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

Ref: NCERT > Physics Book > Atoms and Nuclei > Radioactive Decay, Nuclear Forces and Stability

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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.

Ref: NCERT > Physics Book > Atoms and Nuclei > Radioactive Decay, Nuclear Forces and Stability

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Radioactive Decay, Nuclear Forces and Stability

A nucleus has a binding energy of \( 127.5 \, \text{MeV} \) and mass number 16. What is its binding energy per nucleon?

**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. Ebₙ = (E_b/A) . E_b = 127.5 MeV , A = 16 . Ebₙ = (127.5/16) ≈ 7.97 MeV ≈ 8 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.

Ref: NCERT > Physics Book > Atoms and Nuclei > Radioactive Decay, Nuclear Forces and Stability

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Nuclear Reactions - Fission, Fusion and Energy Release

What is the role of high temperature in nuclear fusion?

**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. High temperatures provide nuclei with sufficient kinetic energy to overcome the Coulomb barrier (electrostatic repulsion), allowing them to come close enough for the nuclear force to cause fusion. Using E_n = -13.6/n² eV, r_n = n² a₀, L = n h/2π, R = R₀

Ref: NCERT > Physics Book > Atoms and Nuclei > Nuclear Reactions - Fission, Fusion and Energy Release

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Radioactive Decay, Nuclear Forces and Stability

A nucleus with mass number 12 has a binding energy of \( 96 \, \text{MeV} \). 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 = 96 MeV , A = 12 . Ebₙ = (96/12) = 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

Ref: NCERT > Physics Book > Atoms and Nuclei > Radioactive Decay, Nuclear Forces and Stability

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Radioactive Decay, Nuclear Forces and Stability

Why is the nuclear force considered saturated in large nuclei?

**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. The nuclear force is short-ranged, affecting only a fixed number of neighboring nucleons, so adding more nucleons in large nuclei does not proportionally increase the binding energy, leading to saturation. 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 Limited range of interaction, consistent with Bohr model and nuclear binding

Ref: NCERT > Physics Book > Atoms and Nuclei > Radioactive Decay, Nuclear Forces and Stability

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Nuclear Reactions - Fission, Fusion and Energy Release

Which statement is true about the energy release in nuclear processes?

**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. Nuclear processes (fission and fusion) release energy when less tightly bound nuclei transform into more tightly bound ones, increasing the 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 c² and 1 u = 9

Ref: NCERT > Physics Book > Atoms and Nuclei > Nuclear Reactions - Fission, Fusion and Energy Release

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Radioactive Decay, Nuclear Forces and Stability

A nucleus with mass number 36 has a binding energy of \( 288 \, \text{MeV} \). 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 = 288 MeV , A = 36 . Ebₙ = (288/36) = 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

Ref: NCERT > Physics Book > Atoms and Nuclei > Radioactive Decay, Nuclear Forces and Stability

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Radioactive Decay, Nuclear Forces and Stability

A nucleus with mass number 100 has a binding energy of \( 850 \, \text{MeV} \). 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 = 850 MeV , A = 100 . Ebₙ = (850/100) = 8.5 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 8.5 MeV, consistent with Bohr model and nuclear binding energy systematics.

Ref: NCERT > Physics Book > Atoms and Nuclei > Radioactive Decay, Nuclear Forces and Stability

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Radioactive Decay, Nuclear Forces and Stability

What is the volume of a nucleus with radius \( 2.7 \times 10^{-15} \, \text{m} \)? (Use \( \pi = 3.14 \))

**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. Volume = (4/3) π R³ . R³ = (2.7 × 10⁻¹⁵)³ = 1.9683 × 10⁻⁴⁴ m³ . Volume = (4/3) × 3.14 × 1.9683 × 10⁻⁴⁴ ≈ 8.25 × 10⁻⁴⁴ m³ . 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 8.25 × 10⁻⁴⁴ m³, consistent with Bohr model and nuclear binding energy systematics.

Ref: NCERT > Physics Book > Atoms and Nuclei > Radioactive Decay, Nuclear Forces and Stability

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