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NEET2018PhysicsNuclear 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 holds? Here, c is the speec 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

A. M_A+ M_B= M_C- E / c^2

Step-by-step solution

Binding energy of A= M_A c^2 Binding energy of B= M_B c^2 Binding energy of C= M_C c^2 The nuclear reaction is given by A+B C Energy released , E= Binding energy of C- (Binding energy of A+ Binding energy of B) aligned & = M_C c^2- ( M_A c^2+ M_B c^2 ) & E c^2 = M_C- ( M_A+ M_B ) & M_A+ M_B= M_C- E c^2 aligned

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