1.

Calculate binding energy per nucleon of `._83Bi^(209)`. Given that m `(._83Bi^(209))=208.980388am u` `m("neutron") = 1.008665 am u` `m ("proton") = 1.007825 am u`

Answer» In `.83Bi^(209)`, number of proton=83
number of neutrons `=209-83=126`
Mass defect,
`Delta m=83xxm_p+126xxm_n-M(Bi)`
`=83xx1.007825+126xx1.008665-208.980388`
`=83.6494475xx127.091790-208.980388`
`=1.760877 am u`
`B.E. //"nucleon"=(1.760877xx931)/209`
`=7.85MeV//N`


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