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COMEDK202510 May 2025Morning ShiftPhysicsNuclear PhysicsActual

In a nuclear fusion reaction, two nuclei, A and B fuse to produce a nucleus C , releasing an amount of energy E in the process. If the mass defects of the three nuclei are M_A, M_B and M_C respectively, then which of the following relations is true? ( c is the speed of light).

Options

  1. AM_A- M_B= M_C+ E c^2
  2. BM_A+ M_B= M_C- E c^2
  3. CM_A+ M_B= M_C+ E c^2
  4. DM_A- M_B= M_C- E c^2

Correct answer

B. M_A+ M_B= M_C- E c^2

Step-by-step solution

Energy released in fusion: E = BE_C - (BE_A + BE_B) Since BE = M c^2 : E = ( M_C)c^2 - [( M_A)c^2 + ( M_B)c^2] Dividing by c^2 : E c^2 = M_C - M_A - M_B M_A + M_B = M_C - E c^2

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