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QJEEM22Phb4df: JEE Main 2022 Physics - Class 12 JEEM22Phb4df Solution: Nucleus A is having mass number 220 and its binding energy per…

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 :

Multiple Choice Options

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

Step-by-Step Verified Solution

220A → 105B + 115C

⇒ Q = [105 × 6.4 + 115 × 6.4] − [220 × 5.6] MeV

⇒ Q = 176 MeV

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