Question

In a nuclear fusion reaction, two nuclei, $A$ & $B$, fuse to produce a nucleus $C$, releasing an atom of energy $\Delta E$ in the process. If the mass defects on the three nuclie are $\Delta M_A,\Delta M_B,\Delta M_C$ respectively, then which of the following relations holds? Here $c$ is the speed of light:-

• A. $\Delta M_A+\Delta M_B=\Delta M_C-\Delta E/c^2$
• B. $\Delta M_A+\Delta M_B=\Delta M_C+\Delta E/c^2$
• C. $\Delta M_A-\Delta M_B=\Delta M_C-\Delta E/c^2$
• D. $\Delta M_A-\Delta M_B=\Delta M_C+\Delta E/c^2$

Answer 1

The correct option is A $\Delta M_A+\Delta M_B=\Delta M_C-\Delta E/c^2$

$\begin{array}{c}\text{Mass of photon is Difference in}\\ \text{Mass defects of products and Reactants}\\ \therefore \left[\Delta M_C\right]-[\Delta M_A+\Delta M_B]=\frac{\Delta E^D}{C^2}\\ △M_A+\Delta M_B=\Delta M_C-\frac{\Delta E}{C^2}\end{array}$