Q. The binding energy of two nuclei $P^{n}\ and\ Q^{2n}$ are $x$ joule and $y$ joule respectively. If $2\ x > y$, then the energy released in the reaction $P^{n} + P^{n} = Q^{2n}$ will be
A) $2\ x + y$
B) $2\ x - y$
C) $xy$
D) $x + y$
Explanation
(b) Energy released
$=$initial BE$-$final BE $= 2x - y$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.