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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 and m(" "_(83)^(209)Bi) = 208.980388 u. |
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Answer» Solution :BINDING energy of the nucleus `" "_(26)^(56)Fe`: `(E_(B))_(Fe) = [26 m_(H) + (56 - 26) m_(n) - m_(Fe)] xx 931.5 MeV = [26 xx 1.007825 + 30 xx 1.008665 - 55.934939] xx 931.5 MeV = 492.3 MeV` and binding energy per nucleon of `" "_(26)^(56)Fe`: `(E_(bn))_(Fe) = 492.3/56=8.79` MeV /nucleon Similarly binding energy of the nucleus of `" "_(83)^(209)Bi` `(E_(b))_(Bi) = [83 m_(H) + (209 -83)m_(n) - m_(Bi) ] xx 931.5 MeV= [83 xx 1.007825 + 126 xx 1.008665 - 208.980388] xx 931.5 MeV = 1640.3 MeV` and binding energy per nucleon of `" "_(83)^(209)Bi, (E_(bn))_(Bi) = 1640.3/209 =7.84` MeV /nucleon. |
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