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Solve · Q38

Q.Calculate the energy in MeV released in the nuclear reaction
174-77-Ir → 170-75-Re + 4-2-He
Atomic masses : Ir = 173.97 u, Re = 169.96 u and He = 4.0026 u
(Ans. 6.89 MeV)

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Step 1. The reaction 174-77-Ir → 170-75-Re + 4-2-He is an alpha decay (mass number drops by 4, atomic number drops by 2, consistent with the alpha-decay rule of section 13.6.1).

Step 2. Mass defect: Δm = (mass of Ir) - (mass of Re + mass of He) = 173.97 - (169.96 + 4.0026) = 173.97 - 173.9626 = 0.0074 u.

Step 3. Energy released: E = Δm x 931.4 = 0.0074 x 931.4 ≈ 6.892 MeV. …

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