Assume each uranium-235 fission releases 200 MeV. Take 1 eV = 1.6×10⁻¹⁹ J. If 6.25×10¹⁹ fissions occur, the total energy released is N×10⁹ J. Find N.

Question

G1 Physics · Accepted Answer

Energy per fission: 200 MeV = 200×10⁶ eV. Using 1 eV = 1.6×10⁻¹⁹ J, energy per fission = 200×10⁶×1.6×10⁻¹⁹ = 3.2×10⁻¹¹ J. For 6.25×10¹⁹ fissions: E = 6.25×10¹⁹×3.2×10⁻¹¹ = 20×10⁸ J = 2.0×10⁹ J. Therefore N = 2.