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One Line Answer
Binding energy per nucleon for helium nucleus (2 He) is 7.0 MeV Find value of mass defect for helium nucleus
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Solution
BE for `""_2^4"He"0` = 7.0 Me V per nucleon
If Δm is the mass defect, then total BE = Δm x 931.5 MeV
∴ BE per nucleon = `(Δm xx 931.5)/(4)` MeV
`(Δm xx 931.5)/(4)` = 7.0 MeV
Δm = `( 4 xx 7.0 )/931.5` u
Δm = 0.030059 u
Concept: Mass-energy and Nuclear Binding Energy - Nuclear Binding Energy
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