If three helium nuclei combine to form a carbon nucleus then the energy released in this reaction is _________ MeV. (Given 1u = 931 MeV/, atomic mass of helium = 4.002603u).

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Solution

Q.
If three helium nuclei combine to form a carbon nucleus then the energy released in this reaction is _________ $\times 10^{- 2}$ MeV. (Given 1u = 931 MeV/ $c^{2}$ , atomic mass of helium = 4.002603u). [2024]

Ans.
(727) Reaction: $3 {}_{2}^{4}H e \to_{6}^{12} C + Q$

Energy released

$Q = [3 M_{He} - M_{C}] c^{2} = [3 \times 4.002603 - 12] [931]$

$Q = 727 \times 10^{- 2} MeV$

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