Skip to content

#mass-energy conversion

4 public questions tagged with this topic.

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

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

What is the energy released when \( 2 \, \text{g} \) of matter is completely converted into energy? (Given \( c = 3 \tim

**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. E = m c² . m = 2 × 10⁻³ kg , c² = 9 × 10¹⁶ m²/s² . E = 2 × 10⁻³ × 9 × 10¹⁶ = 1.8 × 10¹⁴ J . Using E_n = -13.6/n² eV, r_n = n² a₀, L =

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

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

What is the energy equivalent of \( 1 \, \text{kg} \) 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 = 1 kg , c² = 9 × 10¹⁶ m²/s² . E = 1 × 9 × 10¹⁶ = 9 × 10¹⁶ 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 × 10¹⁶ J, consistent with Bohr

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