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 $A+B\to C$
So, released energy $\Delta E=BE$ of $C$
$(BE$ of $A$ and $B)$
$=\Delta M_C{C}^2-\left(\Delta M_A{C}^2+\Delta M_B{C}^2\right)$
or $\frac{\Delta E}{c^2}=\Delta M_C-\left(\Delta M_A+\Delta M_B\right)$
$\Delta M_A+\Delta M_B=\Delta M_C-\frac{\Delta E}{c^2}$