COMEDK · Physics · 29. Nuclear 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 \(\Delta M_A-\Delta M_B=\Delta M_C+\dfrac{\Delta E}{c^2}\)
- B \(\Delta M_A+\Delta M_B=\Delta M_C-\dfrac{\Delta E}{c^2}\)
- C \(\Delta M_A+\Delta M_B=\Delta M_C+\dfrac{\Delta E}{c^2}\)
- D \(\Delta M_A-\Delta M_B=\Delta M_C-\dfrac{\Delta E}{c^2}\)
Answer & Solution
Correct Answer
(B) \(\Delta M_A+\Delta M_B=\Delta M_C-\dfrac{\Delta E}{c^2}\)
Step-by-step Solution
Detailed explanation
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 \(c^2\): \(\dfrac{\Delta E}{c^2} = \Delta M_C - \Delta M_A - \Delta M_B\)…
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