If three helium nuclei combine to form a carbon nucleus then the energy released in this reaction is ________ × 10⁻² MeV. (Given 1u = 931 MeV/c², atomic mass of helium = 4.002603u) Numerical answer type. Work it out, then open the answer.

Answer: 727. 3 ⁴He → ¹²C, with mass of ¹²C = 12 u exactly. Δ m = 3×4.002603 - 12 = 0.007809 u. E = 0.007809×931 ≈ 7.27 MeV = 727×10⁻² MeV.

If three helium nuclei combine to form a carbon nucleus then the… | JEE Main 2024 (5 Apr, Shift 1)

Q12JEE Main 2024 · 5 APR · SHIFT 1Medium

Nuclear Fission & Fusion

If three helium nuclei combine to form a carbon nucleus then the energy released in this reaction is ________ $\times 10^{-2}\text{ MeV}$ . (Given $1\text{u} = 931\text{ MeV/c}^{2}$ , atomic mass of helium $= 4.002603\text{u}$ )

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

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Answer: 727

$3\,{}^4\text{He} \to {}^{12}\text{C}$ , with mass of ${}^{12}\text{C} = 12\text{ u}$ exactly.
$\Delta m = 3\times4.002603 - 12 = 0.007809\text{ u}$ .
$E = 0.007809\times931 \approx 7.27\text{ MeV} = 727\times10^{-2}\text{ MeV}$ .

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