Question

In a nuclear fusion reaction, two nuclei, $A$ and $B$ fuse to produce a nucleus $C$, releasing an amount of energy $\Delta$E in the process. If the mass defects of the three nuclei are $\Delta M_A,\Delta M_B$ and $\Delta M_C$ 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

Answer: (A)

Binding energy of A =$\Delta M_{A^{C^2}}$
Binding energy of B =$\Delta M_{B^{C^2}}$
Binding energy of C =$\Delta M_{C^{C^2}}$
The nuclear reaction is given by
A + B $\to$ C
Energy released,
$\Delta$E = Binding energy of C - (Binding energy of A + Binding energy of B)
$=\Delta M_{C^{C^2}}-\left(\Delta M_{A^{C^2}}+\Delta M_{B^{C^2}}\right)$
$\frac{\Delta E}{C^2}=\Delta M_C-\left(\Delta M_A+\Delta M_B\right)$
$\Delta M_A+\Delta M_B=\Delta M_C-\frac{\Delta E}{C^2}$