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
- AM_A+ M_B= M_C- E / c^2
- BM_A+ M_B= M_C+ E / c^2
- CM_A- M_B= M_C- E / c^2
- 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