Question

In the reaction ${}_{1}^{2}H$ + ${}_{1}^{3}H$ ®${}_{2}^{4}He$ + ${}_{0}^{1}n$ if the binding energies of ${}_{1}^{2}H$, ${}_{1}^{3}H$ and ${}_{2}^{4}He$ are respectively a, b and c (in MeV), then the energy (in MeV) released in this reaction is-

• A. a + b + c
• B. a + b – c
• C. c – a – b
• D. c + a – b

Answer 1

Answer: (C)

c – a – b

Answer 2

₁H² + ₁H³ ® ₂He⁴ + ₀n¹

B.E. a b c

D E = c – a – b