Q82 / 90 · JEE Main 2019 (10 Jan Shift 2) · Physics

Consider the nuclear fission, Ne20→2He4+C12. Given that the binding energy/nucleon of Ne20, He4 and C12 are 8.03 MeV, 7.86 MeV, respectively. Identify the correct statement:

  • A
    Energy of 12.4 MeV will be supplied.
  • Correct answer
    B
    Energy of 9.72 MeV has to be supplied.
  • C
    Energy of 3.6 MeV will be released.
  • D
    8.3 MeV energy will be released.

Solution

Q=Ef-Ei
=12×7.86+4×7.07+4×7.07-(20×8.03)
Q=-9.72 MeV

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

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