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Mass Defect, Binding Energy and Binding Energy per Nucleon

This category gathers questions that deal with the concepts of mass defect, binding energy and binding energy per nucleon. It helps learners understand how these quantities are defined, calculated and why they matter for nuclear reactions and stability.

30 questions

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.

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

What is the energy equivalent of \( 2 \, \text{kg} \) 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 = 2 kg , c² = 9 × 10¹⁶ m²/s² . E = 2 × 9 × 10¹⁶ = 1.8 × 10¹⁷ J . Using E_n = -13.6/n² eV, r_n = n² a₀, L = n h/2π, R = R₀ A^¹/³, BE = Δm

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

Which process is responsible for the energy release in an atomic bomb?

**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. The energy in an atomic bomb comes from uncontrolled nuclear fission, where a heavy nucleus splits into lighter fragments, 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 c² and 1 u = 931.5 MeV, evaluation yields Nuclear fission, consistent with Bohr model and nuclear bind

Ref: NCERT > Physics Book > Atoms and Nuclei > 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

What is the energy equivalent of \( 0.002 \, \text{kg} \) of matter in Joules? (Given \( c = 3 \times 10^8 \, \text{m/s}

**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. E = m c² . m = 0.002 kg , c² = 9 × 10¹⁶ m²/s² . E = 0.002 × 9 × 10¹⁶ = 1.8 × 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 yields 1.8 × 10¹⁴ J, consistent with Bohr

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

What is the binding energy per nucleon if a nucleus with mass number 56 has a total binding energy of \( 490 \, \text{Me

**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. Ebₙ = (E_b/A) . E_b = 490 MeV , A = 56 . Ebₙ = (490/56) ≈ 8.75 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 8.75

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

What is the energy equivalent of \( 0.5 \, \text{g} \) 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 = 0.5 × 10⁻³ kg , c² = 9 × 10¹⁶ m²/s² . E = 0.5 × 10⁻³ × 9 × 10¹⁶ = 4.5 × 10¹³ J . Using E_n = -13.6/n² eV, r_n = n² a₀, L = n h/2π, R = R₀

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

InEinstein's mass-energy equivalence, what does the term \( c^2 \) represent?

**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. In E = m c² , c² is the square of the speed of light in a vacuum, acting as the conversion factor between mass and energy, indicating the immense energy equivalent of a small mass. 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,

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

What is the mass defect of a nucleus if its binding energy is \( 186 \, \text{MeV} \)? (Given \( 1 \, \text{u} = 931.5 \

**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. E_b = Δ M · c² . Δ M = (E_b/c²) = (186/931.5) ≈ 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 with Bohr model and nuclear binding energy systematics.

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

What is the main source of energy in nuclear reactors?

**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. Nuclear reactors produce energy through nuclear fission, where heavy nuclei split into lighter fragments, releasing energy due to the increase in 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 = 931.5 MeV, evaluation yields Nuclear fission, consistent with Bohr model an

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

What is the mass defect of a nucleus with binding energy \( 93.15 \, \text{MeV} \)? (Given \( 1 \, \text{u} = 931.5 \, \

**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. Δ M = (E_b/c²) . Δ M = (93.15/931.5) = 0.1 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 0.1 u, consistent with Bohr model and nuclear binding energy systematics.

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

Why is nuclear fusion more likely to release energy when light nuclei are involved?

**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. In light nuclei (A < 30), the binding energy per nucleon is lower. When they fuse into a heavier nucleus, the binding energy per nucleon increases, releasing energy as the final system is more tightly bound. 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,

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