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NEET 1995 (AIPMT) · Nuclear Fission & Fusion

Nuclear Fission & Fusion question from NEET 1995 (AIPMT)

Q42NEET 1995 · AIPMTEasy

Nuclear Fission & Fusion

The binding energies per nucleon for a deuteron and an α\alpha-particle are x1x_{1} and x2x_{2} respectively. The energy QQ released in the reaction

2H1+2H1→4He2+Q,{}^{2}\text{H}_{1} + {}^{2}\text{H}_{1} \rightarrow {}^{4}\text{He}_{2} + Q,

is

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Answer: (b)

  1. Total BE of each deuteron =2x1= 2x_1, so two deuterons have 4x14x_1; total BE of 4He=4x2{}^4\text{He} = 4x_2.
  2. Q=4x2−4x1=4(x2−x1)Q = 4x_2 - 4x_1 = 4(x_2 - x_1).
  3. So the answer is (b) 4(x2−x1)4(x_2 - x_1).
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