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QJEEM19Ph3eae: JEE Main 2019 Physics - Class 12 JEEM19Ph3eae Solution: Consider the nuclear fission Ne 20 →  2He 4 + C 12 Given…

Consider the nuclear fission

Ne20 → 2He4 + C12

Given that the binding energy/ nucleon of Ne20, He4 and C12 are, respectively, 8.03 MeV, 7.07 MeV and 7.86 MeV, identify the correct statement -

Multiple Choice Options

  • A 8.3 MeV energy will be released
  • B energy of 11.9 MeV has to be supplied
  • C energy of 12.4 MeV will be supplied
  • D energy of 3.6 MeV will be released

Step-by-Step Verified Solution

Ne20 →  2He4 + C12

Q – value, EB = (BE)react − (BE)product

= (20 × 8.03) – ((2 × 7.07 × 4) + 7.86 × 12)

= 9.72 MeV

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