AP EAMCET202421 May 2024Morning ShiftPhysicsNuclear PhysicsActual
If the energy released per fission of a ₉₂²³⁵ U nucleus is 200 Me V . the energy released in the fission of 0.1 kg of ₉₂²³⁵ U in kilowatt - hour is.
Options
- A22.8 10^5
- B22.8 10^7
- C11.4 10^5
- D850 10¹⁰
Correct answer
A. 22.8 10^5
Step-by-step solution
Energy released per fission per atom, E =200 MeV Number of atoms in 0.1 kg of ₉₂²³⁵ U is N = ( m ~m ) N _ A = 0.1 6.023 10²³ 235 10⁻³ =25.63 10²² Energy released per fission is aligned & E/fission = NE & = 25.63 10²² 200 10^6 1.6 10⁻¹⁹ 3.6 10^6 & =22.8 10^5 kwh aligned