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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

  1. A22.8 10^5
  2. B22.8 10^7
  3. C11.4 10^5
  4. 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

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