²³⁸₉₂A → ²³⁴₉₀B + ⁴₂D + Q In the given nuclear reaction, the approximate amount of energy released will be: [Given, mass of ²³⁸₉₂A = 238.05079 × 931.5 MeV/c², mass of ²³⁴₉₀B = 234.04363 × 931.5 MeV/c², mass of ⁴₂D = 4.00260 × 931.5 MeV/c²] (a) 2.12 MeV (b) 4.25 MeV (c) 3.82 MeV (d) 5.9 MeV

Answer: (b). Δ m = 238.05079 - (234.04363 + 4.00260) = 0.00456 u Q = 0.00456 × 931.5 ≈ 4.25 MeV So the answer is (b) 4.25 MeV.

²³⁸₉₂A → ²³⁴₉₀B + ⁴₂D + Q In the given nuclear reaction, the… | JEE Main 2023 (13 Apr, Shift 1)

Q8JEE Main 2023 · 13 APR · SHIFT 1Medium

Nuclear Reactions & Q-value

${}^{238}_{92}A \rightarrow {}^{234}_{90}B + {}^{4}_{2}D + Q$

In the given nuclear reaction, the approximate amount of energy released will be: [Given, mass of ${}^{238}_{92}\text{A} = 238.05079 \times 931.5\text{ MeV/c}^{2},$

mass of ${}^{234}_{90}B = 234.04363 \times 931.5\text{ MeV/c}^{2},$

mass of ${}^{4}_{2}D = 4.00260 \times 931.5\text{ MeV/c}^{2}]$

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Answer: (b)

$\Delta m = 238.05079 - (234.04363 + 4.00260) = 0.00456\ \text{u}$
$Q = 0.00456 \times 931.5 \approx 4.25\ \text{MeV}$
So the answer is (b) $4.25\ \text{MeV}$ .

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Nuclear Reactions & Q-value question from JEE Main 2023 (13 Apr, Shift 1)

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