What is the mass number of a nucleus with radius \( 5.4 \times 10^{-15} \, \text{m} \)? (Given \( R_0 = 1.2 \times 10^{-
**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. R = R₀ A¹/³ . 5.4 × 10⁻¹⁵ = 1.2 × 10⁻¹⁵ × A¹/³ . A¹/³ = (5.4/1.2) = 4.5 . A = (4.5)³ = 91.125 ≈ 91 . 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 > Nuclear Reactions - Fission, Fusion and Energy Release