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
$\quad$$\quad$A + B $\to$ C
Energy released,
$\Delta$E = Binding energy of C - (Binding energy of A + Binding energy of B)
$=\Delta M_{C^{C^2}}-\left(\Delta M_{A^{C^2}}+\Delta M_{B^{C^2}}\right)$
$\frac{\Delta E}{C^{2}}=\Delta M_{C}-\left(\Delta M_{A}+\Delta M_{B}\right)$
$\Delta M_{A}+\Delta M_{B}=\Delta M_{C}-\frac{\Delta E}{C^{2}}$