Practice question
Question
What is the energy equivalent of \( 0.005 \, \text{kg} \) of matter in Joules? (Given \( c = 3 \times
10^8 \, \text{m/s} \))
Explanation
**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. E = m c² . m = 0.005 kg , c² = 9 × 10¹⁶ m²/s² . E = 0.005 × 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
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