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Question
Calculate the energy in MeV released in the nuclear reaction \[\ce{^174_77Ir -> ^170_75Re + ^4_2He}\]
Atomic mass: Ir = 173.97 u, Re = 169.96 u and He 4.0026 u
Solution
Given: mIr = 173.97 u, mRe = 169.96 u, mHe = 4.0026 u
To find: Energy released
Formulae:
- ∆m = (mass of 174Ir) - (mass of 170Re + mass of 4He)
- E = ∆m × 931.4 MeV
Calculation:
- ∆m = (mass of 174Ir) - (mass of 170Re + mass of 4He)
= 173.97 - (169.96 + 4.0026)
= 7.4 × 10-3 u - E = ∆m × 931.4
= 7.4 × 10-3 × 931.4
= 6.89236 MeV ≈ 6.892 MeV
The energy released in given nuclear reaction is 6.892 MeV.
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