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Obtain the bindig energy of the nuclie ""_(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) = 208.980388 u |
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Answer» SOLUTION :`BE = [(Nm_p + (A - Z)m_n) - M]c^2` `1u = 931 MEV` `:. Be " of" ""_(26)^(56) Fe = [(26 xx 1.007825 + 30 xx 1.008662) - 55.934939]931 MeV = 492 MeV` `:. BE//"nucleon" = (492)/(56) = 8.79 MeV` Similarly BE/nucleon of `""_(83)^(209)Bi = 7.84 MeV`. |
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