(a) In a nuclear reaction `""_(2)^(3)He +""_(2)^(3)He to ""_(2)^(4)He+""_(1)^(1)H +""_(1)^(1)H +12.86` MeV, though the number of nucleons is conserved on both sides of the reaction, yet the energy is released. How? Explain.
(b) Draw a plot of potential energy between a pair of nucleons as a function of their separation. Mark the regions where potential energy is (i) positive and (ii) negative.

Explanation

(a) In a nuclear reaction, the sum of the masses of the target nuclues ` ""_(2)^(3)He` and the bombarding particle ` ""_(2)^(3)He` may be greater or less the sum of the masses of the product nucleus ` ""_(2)^(4)He` and the ` ""_(1)^(1)He`. So from the law of conservation of mass energy some energy some energy (12.86 MeV) is envolved in nuclear reaction . This energy is called Q-value of the nuclear reaction . The binding energy of the nucleus on the left side is not equal to the right side. The difference in the binding energies on two sides appears as energy released or absorbed in the nuclear reaction.
(b) The potential energy is minimum at `r_(0)`. For distances larger than `r_(0)` the negative potential energy goes on decreasing and for the distances less than `r_(0)` the negative potential energy decreases to zero and then becomes positive and increases abruptly. Thus, A to B is the positive potential energy region and B to C is the negative potential energy region.