Q. In the reaction ²₁H + ³₁H → ⁴₂He + ¹₀n, if the binding energies of ²₁H, ³₁H and ⁴₂He are respectively a, b and c (in MeV), then the energy (in MeV) released in this reaction is :
A) c + a − b
B) c − a − b
C) a + b + c
D) a + b − c
Explanation
[Solution from source] Energy released = total binding energy of products − total binding energy of reactants = c − (a + b). The free neutron has zero binding energy. a + b − c has the sign reversed. Answer (2).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.