COMEDK20269 May 2026Morning ShiftPhysicsNuclear PhysicsActual
A nucleus of uranium-235 absorbs a slow neutron and undergoes nuclear fission according to the reaction: ²³⁵₉₂U + ¹₀n ¹⁴¹₅₆Ba + ⁹²₃₆Kr + 3 ¹₀n + Q If the average energy released per fission is 202 MeV, the energy released when 2.35 g of U²³⁵ undergoes complete fission is approximately; [Given 1eV = 1.6 10⁻¹⁹ J , Avogadro number = 6.02 10²³ ]
Options
- A1.945 10¹² J
- B19.45 10¹² J
- C19.45 10¹⁰ J
- D1.945 10¹⁰ J
Correct answer
C. 19.45 10¹⁰ J
Step-by-step solution
Number of moles of U²³⁵ = 2.35 235 = 0.01 mol Number of atoms of U²³⁵ = 0.01 6.02 10²³ = 6.02 10²¹ Energy released per fission = 202 MeV = 202 10^6 1.6 10⁻¹⁹ J = 3.232 10⁻¹¹ J Total energy released = 6.02 10²¹ 3.232 10⁻¹¹ J Total energy released = 19.456 10¹⁰ J Answer: 19.45 10¹⁰ J