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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` |
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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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