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92 238 A → 90 234 B + D 2 4 + Q In the given nuclear reaction, the approximate amount of energy released will be: [Given, mass of 92 238 A = 238 . 05079 × 931 . 5 MeV c - 2 , mass of B 90 234 = 234 . 04363 × 931 . 5 MeV c - 2 , mass of D 2 4 = 4 . 00260 × 931 . 5 Me V c - 2 ]

Options

  1. A3 . 82 MeV
  2. B5 . 9 MeV
  3. C2 . 12 MeV
  4. D4 . 25 MeV

Correct answer

D. 4 . 25 MeV

Step-by-step solution

The value of Q is Q = ( m A - m B - m D ) × 931 . 5   MeV ⇒ Q = ( 238 . 05079 - 234 . 04363 - 4 . 00260 ) × 931 . 5   MeV ⇒ Q = 4 . 25   MeV

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