#mass-energy equivalence

Nuclear Size, Density and Structure

What is the energy equivalent of \( 0.001 \, \text{kg} \) of matter in Joules? (Given \( c = 3 \times 10^8 \, \text{m/s}

**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 = m c² . m = 0.001 kg , c² = (3 × 10⁸)² = 9 × 10¹⁶ m²/s² . E = 0.001 × 9 × 10¹⁶ = 9 × 10¹³ 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 yields 9 × 10¹³ J, consistent with Bohr model and nuclear binding energy systematics.

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Mass Defect, Binding Energy and Binding Energy per Nucleon

What is the energy equivalent of \( 0.1 \, \text{g} \) of matter in Joules? (Given \( c = 3 \times 10^8 \, \text{m/s} \)

**Mass defect** Δm = Z m_p + N m_n - M_nucleus, binding energy BE = Δm c², 1 u =931.5 MeV/c², BE per nucleon = BE/A, measures stability, peak ~8.8 MeV at Fe-56, for A=36 BE=288 MeV BE/A=8 MeV, for A=12 BE=96 MeV BE/A=8 MeV, for A=16 BE=127.5 MeV BE/A≈7.97 MeV, higher BE/A more stable. E = m c² . m = 0.1 × 10⁻³ kg = 10⁻⁴ kg , c² = 9 × 10¹⁶ m²/s² . E = 10⁻⁴ × 9 × 10¹⁶ = 9 × 10¹² J . Using E_n = -13.6/n² eV, r_n = n² a₀, L = n h/2π, R =

Ref: NCERT > Physics Book > Atoms and Nuclei > Mass Defect, Binding Energy and Binding Energy per Nucleon

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Mass Defect, Binding Energy and Binding Energy per Nucleon

Why does nuclear fission release more energy than chemical reactions?

**Mass defect** Δm = Z m_p + N m_n - M_nucleus, binding energy BE = Δm c², 1 u =931.5 MeV/c², BE per nucleon = BE/A, measures stability, peak ~8.8 MeV at Fe-56, for A=36 BE=288 MeV BE/A=8 MeV, for A=12 BE=96 MeV BE/A=8 MeV, for A=16 BE=127.5 MeV BE/A≈7.97 MeV, higher BE/A more stable. Nuclear fission involves changes in nuclear binding energy (order of MeV), which is about a million times larger than the energy from chemical bond changes (order of eV). Using E_n = -13.6/n² eV, r_n = n² a₀, L = n h/2π, R = R₀ A^¹/³, BE = Δm c² and

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De Broglie Hypothesis and Quantization in Bohr Model

How much energy is released when \( 5 \, \text{g} \) of matter is converted into energy? (Given \( c = 3 \times 10^8 \,

**Quantization basis** de Broglie standing wave requires constructive interference, integer wavelengths in orbit, otherwise destructive, so only certain radii allowed r_n = n² a₀, a₀=0.53 Å, angular momentum L = n h/2π, de Broglie explains why orbits are stationary - electron wave closed on itself. E = m c² . m = 5 × 10⁻³ kg , c² = 9 × 10¹⁶ m²/s² . E = 5 × 10⁻³ × 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 u = 931.5

Ref: NCERT > Physics Book > Atoms and Nuclei > De Broglie Hypothesis and Quantization in Bohr Model

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Mass Defect, Binding Energy and Binding Energy per Nucleon

What is the significance of the mass-energy equivalence in nuclear physics?

**Binding energy calculation** from mass defect, for nucleus mass number 28 BE 224 MeV BE/A=8 MeV, A=18 BE 144 MeV BE/A=8 MeV, BE/A indicates stability, fusion of light nuclei and fission of heavy release energy because product has higher BE/A, difference released. Einstein’s E = m c² explains how mass defect converts into binding energy, unifying mass and energy conservation and enabling the understanding of energy release in nuclear reactions. 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 Explains binding energy, consistent

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Mass Defect, Binding Energy and Binding Energy per Nucleon

What is the energy equivalent of a neutron with mass \( 1.6749 \times 10^{-27} \, \text{kg} \) in Joules? (Given \( c =

**Binding energy calculation** from mass defect, for nucleus mass number 28 BE 224 MeV BE/A=8 MeV, A=18 BE 144 MeV BE/A=8 MeV, BE/A indicates stability, fusion of light nuclei and fission of heavy release energy because product has higher BE/A, difference released. E = m c² . m = 1.6749 × 10⁻²⁷ kg , c² = (3 × 10⁸)² = 9 × 10¹⁶ m²/s² . E = 1.6749 × 10⁻²⁷ × 9 × 10¹⁶ ≈ 1.507 × 10⁻¹⁰ 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 =

Ref: NCERT > Physics Book > Atoms and Nuclei > Mass Defect, Binding Energy and Binding Energy per Nucleon

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