Q. In the reaction $\ _{1}^{2}H + \ _{1}^{3}H \rightarrow \ _{2}^{4}He + \ _{0}^{1}n$ 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) $c + a - b$
B) $c - a - b$
C) $a + b + c$
D) $a + b - c$
Explanation

(b) During fusion binding energy of daughter nucleus is always greater than the total energy of the parent nuclei so energy released $= c - (a + b) = c - a - b$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.

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