Q143 / 180 · COMEDK 2025 (10 May Shift 1) · Physics
In a nuclear fusion reaction, two nuclei, $A$ and $B$ fuse to produce a nucleus $C$, releasing an amount of energy $\Delta \mathrm{E}$ 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 is true? ( $c$ is the speed of light).
- A$\Delta M_A-\Delta M_B=\Delta M_C+\dfrac{\Delta E}{c^2}$
- Correct answerB$\Delta M_A+\Delta M_B=\Delta M_C-\dfrac{\Delta E}{c^2}$
- C$\Delta M_A+\Delta M_B=\Delta M_C+\dfrac{\Delta E}{c^2}$
- D$\Delta M_A-\Delta M_B=\Delta M_C-\dfrac{\Delta E}{c^2}$
Solution
Energy released in fusion: $\Delta E = BE_C - (BE_A + BE_B)$
Since $BE = \Delta M \cdot c^2$:
$\Delta E = (\Delta M_C)c^2 - [(\Delta M_A)c^2 + (\Delta M_B)c^2]$
Dividing by $c^2$:
$\dfrac{\Delta E}{c^2} = \Delta M_C - \Delta M_A - \Delta M_B$
$\Delta M_A + \Delta M_B = \Delta M_C - \dfrac{\Delta E}{c^2}$
Since $BE = \Delta M \cdot c^2$:
$\Delta E = (\Delta M_C)c^2 - [(\Delta M_A)c^2 + (\Delta M_B)c^2]$
Dividing by $c^2$:
$\dfrac{\Delta E}{c^2} = \Delta M_C - \Delta M_A - \Delta M_B$
$\Delta M_A + \Delta M_B = \Delta M_C - \dfrac{\Delta E}{c^2}$
Hence, the correct answer is option: 'B'.