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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 speec of light.
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The correct answer is:
$\Delta M_A+\Delta M_B=\Delta M_C-\Delta E / c^2$
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
$A+B \rightarrow C$
Energy released ,
$\Delta E=$ Binding energy of $C-$ (Binding energy of $A+$ Binding energy of $B)$
$\begin{aligned} & =\Delta M_C c^2-\left(\Delta M_A c^2+\Delta M_B c^2\right) \\ & \frac{\Delta E}{c^2}=\Delta M_C-\left(\Delta M_A+\Delta M_B\right) \\ & \Delta M_A+\Delta M_B=\Delta M_C-\frac{\Delta E}{c^2}\end{aligned}$
Binding energy of $B=\Delta M_B c^2$
Binding energy of $C=\Delta M_C c^2$
The nuclear reaction is given by
$A+B \rightarrow C$
Energy released ,
$\Delta E=$ Binding energy of $C-$ (Binding energy of $A+$ Binding energy of $B)$
$\begin{aligned} & =\Delta M_C c^2-\left(\Delta M_A c^2+\Delta M_B c^2\right) \\ & \frac{\Delta E}{c^2}=\Delta M_C-\left(\Delta M_A+\Delta M_B\right) \\ & \Delta M_A+\Delta M_B=\Delta M_C-\frac{\Delta E}{c^2}\end{aligned}$
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