The binding energy of two nuclei\[{{P}^{n}}\] and \[{{Q}^{2n}}\]are \[x-\]joule and \[y-\]joule, respectively. Then the energy released in the reaction \[{{P}^{n}}+{{P}^{n}}={{Q}^{2n}}\]will be
A)\[2x+y\]
B)\[y-2x\]
C) \[xy\]
D) \[x+y\]
Correct Answer:
B
Solution :
Energy released =final B.E.-initial B.E \[=y-2x\] Hence, the correction option is (b).