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Obtain the binding energy of the nuclei ...

Obtain the binding energy of the nuclei `._26^56Fe` and `._83^209Bi` in units of MeV from the following data .
`m_H`=1.007825 u, `m_n`=1.008665 u
`m(._26^56Fe) =55.934939 u " " m(._83^209Bi)` = 208.980388 u
Which nucleus has greater binding energy per nucleon ?

Text Solution

Verified by Experts

For `._26^56Fe`, mass defect ,
`Deltam=Zm_H + (A-Z) m_n-m(._26^56Fe)`
`Deltam`=26 x 1.007825 + (56-26) x 1.008665 - 55.934939
=0.528461 u
`therefore` Binding energy `=Deltamxx931.2` MeV = 492.2 MeV
`therefore ` Binding energy per nucleon `=492.2/56`=8.79 MeV
For `._83^209Bi`, mass defect,
`Deltam`=83 x 1.007858 + (209-83) x 1.008665-208.980388
=1.760877 u
`therefore` Binding energy = 1.760877 x 931.5 MeV = 1640. 25 MeV
`therefore` Binding energy per nucleon `=1640.26/209`=7.85 MeV
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