Q75 / 90 · JEE Main 2022 (24 Jun Shift 1) · Physics

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:

  • A
    0.8MeV
  • B
    275MeV
  • C
    220MeV
  • Correct answer
    D
    176MeV

Solution

The total binding energy of nucleus A is,

EA=220×5.6=1232 MeV

The total binding energy of B and C combined is,

EB+EC=105×6.4+115×6.4=1408 MeV

Therefore, the energy released per fission is,

Q=EB+EC-EA=1408-1232=176 MeV

Hence, the correct answer is option: 'D'.

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