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Obtain the binding energy of the nuclei ...

Obtain the binding energy of the nuclei `_(26)^(56)Fe" and "_(83)^(209)Bi` in units of MeV from the following data:
`m(""_(26)^(56)Fe)=55.934939" "u" "m(""_(83)^(209)Bi)=208.9803388" "u`

Text Solution

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Mass defect in `._26Fe^56`= 30 x 1.008665 + 26 x 1.007825 - 55.934939
=0.528461 amu
`therefore` Total binding energy =0.528461 x 931.5 MeV
=492.26 MeV
`therefore` B.E. per nucleon=`492.26/56`=8.760 MeV
Mass defect in `._83Bi^209` nucleus = 83 x 1.007825 + 126 x 1.008665 - 208.980388
=1.760872
Total B.E.=1.760872 x 931.5
=1640.26 MeV
`therefore` B.E. per nucleon =`1640.26/209`=7.848 MeV
`._26Fe^56` has greater B.E. per nucleon than `._83Bi^209`.
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