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Nucleus A is having mass number 220 and its binding energy per nucleon is 5 . 6 MeV . It splits in two fragments B and C of mass numbers 105 and 115 . The binding energy of nucleons in B and C is 6 . 4 MeV per nucleon. The energy Q released per fission will be:

Options

  1. A0 . 8 MeV
  2. B275 MeV
  3. C220 MeV
  4. D176 MeV

Correct answer

D. 176 MeV

Step-by-step solution

The total binding energy of nucleus A is, E A = 220 × 5 . 6 = 1232   MeV The total binding energy of B and C combined is, E B + E C = 105 × 6 . 4 + 115 × 6 . 4 = 1408   MeV Therefore, the energy released per fission is, Q = E B + E C - E A = 1408 - 1232 = 176   MeV

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