#mass defect

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

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

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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.

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

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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.

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 binding energy of a nucleus if its mass defect is \( 0.15 \, \text{u} \)? (Given \( 1 \, \text{u} = 931.5 \,

**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. E_b = Δ M · c² . Δ M = 0.15 u . E_b = 0.15 × 931.5 = 139.725 MeV ≈ 139.73 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 139.73 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 mass defect of \( 0.136 \, \text{u} \). What is its binding energy in MeV? (Given \( 1 \, \text{u} = 931

**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. Binding energy = Δ M · c² . Δ M = 0.136 u . Energy = 0.136 × 931.5 ≈ 126.7 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

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

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Nuclear Size, Density and Structure

What is the binding energy of a nucleus with mass defect \( 0.12 \, \text{u} \)? (Given \( 1 \, \text{u} = 931.5 \, \tex

**Nuclear size** order 10⁻¹⁵ m femtometer, atomic size 10⁻¹⁰ m, ratio 10⁵, nucleus contains protons and neutrons bound by strong force, density independent of A indicates incompressibility, R₀ determined from electron scattering experiments. E_b = Δ M · c² . Δ M = 0.12 u . E_b = 0.12 × 931.5 = 111.78 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 111.78 MeV, consistent with Bohr model and nuclear binding energy systematics.

Ref: NCERT > Physics Book > Atoms and Nuclei > Nuclear Size, Density and Structure

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Nuclear Size, Density and Structure

What is the binding energy of a nucleus with a mass defect of \( 0.09 \, \text{u} \)? (Given \( 1 \, \text{u} = 931.5 \,

**Nuclear radius** R = R₀ A^{1/3}, R₀=1.2×10⁻¹⁵ m, A mass number, for A=16 R=1.2×10⁻¹⁵×2.52=3.02×10⁻¹⁵ m, for A=36 R=1.2×10⁻¹⁵×3.30=3.96×10⁻¹⁵ m, nuclear density ρ = mass/volume = (A×1.66×10⁻²⁷ kg)/(4/3 π R³) ≈2.3×10¹⁷ kg/m³ independent of A, extremely high, mass 3.67×10⁻²⁷ kg radius 2.0×10⁻¹⁵ m gives density =3.67×10⁻²⁷/(4/3 π×8×10⁻⁴⁵)=1.09×10¹⁷ kg/m³. Binding energy = Δ M · c² . Δ M = 0.09 u . E_b = 0.09 × 931.5 = 83.835 MeV ≈ 83.84 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 83.84 MeV, consistent

Ref: NCERT > Physics Book > Atoms and Nuclei > Nuclear Size, Density and Structure

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Mass Defect, Binding Energy and Binding Energy per Nucleon

A nucleus has a mass defect of \( 0.05 \, \text{u} \). What is its binding energy in MeV? (Given \( 1 \, \text{u} = 931.

**Energy equivalent** E= m c², 0.005 kg matter E=0.005×9×10¹⁶=4.5×10¹⁴ J, 0.01 kg 9×10¹⁴ J, mass defect 0.1 u => BE=0.1×931.5=93.15 MeV, mass defect from BE 149.04 MeV => Δm=149.04/931.5=0.16 u, BE per nucleon 8.5 MeV A=20 total BE=170 MeV. Binding energy = Δ M · c² . Δ M = 0.05 u . E_b = 0.05 × 931.5 = 46.575 MeV ≈ 46.58 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 46.58 MeV, consistent with Bohr model and nuclear binding energy systematics.