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 $\Delta M_{C}$, and respectively, then which of the following relations holds? Here $c$ 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 nuclei $A=\Delta M_{A} C^{2}$
Binding energy of nuclei $B=\Delta M_{B} C^{2}$
Binding energy of nuclei $C=\Delta M_{C} C^{2}$
According to question $A+B \rightarrow C$
So, released energy $\Delta E=B E$ of $C$
$(B E$ of $A$ and $B )$
$=\Delta M_{C} C^{2}-\left(\Delta M_{A} C^{2}+\Delta M_{B} C^{2}\right)$
or $\frac{\Delta E}{c^{2}}=\Delta M_{C}-\left(\Delta M_{A}+\Delta M_{B}\right)$
$\Delta M_{A}+\Delta M_{B}=\Delta M_{C}-\frac{\Delta E}{c^{2}}$