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JEE Main 2023 (12 Apr, Shift 1) · Nuclear Reactions & Q-value

Nuclear Reactions & Q-value question from JEE Main 2023 (12 Apr, Shift 1)

Q9JEE Main 2023 · 12 APR · SHIFT 1Medium

Nuclear Reactions & Q-value

A common example of alpha decay is 92238U⟶90234Th+2He4+Q{}^{238}_{92}\text{U} \longrightarrow {}^{234}_{90}\text{Th} + {}_{2}\text{He}^{4} + \text{Q}

Given:

92238U=238.05060 u{}^{238}_{92}\text{U} = 238.05060\text{ u},

90234Th=234.04360 u{}^{234}_{90}\text{Th} = 234.04360\text{ u},

24He=4.00260 u{}^{4}_{2}\text{He} = 4.00260\text{ u} and

1u=931.5MeVc21\text{u} = 931.5\frac{\text{MeV}}{c^{2}}

The energy released (Q)(Q) during the alpha decay of 92238U{}^{238}_{92}\text{U} is __________ MeV

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 4

  1. Δm=238.05060−(234.04360+4.00260)=0.00440 u\Delta m = 238.05060 - (234.04360 + 4.00260) = 0.00440\ \text{u}
  2. Q=0.00440×931.5=4.10 MeVQ = 0.00440 \times 931.5 = 4.10\ \text{MeV}
  3. So Q≈4 MeVQ \approx 4\ \text{MeV}.
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