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Ordinary iron has a high binding energy per nucleon. a) Calculate this binding energy per nucleon (Me V/nucleon) of 56Fe. b) How much energy is needed to take away the last proton of 56Fe. This is the so-called separation energy. From tables: M(56 Fe) = 55.934939,mn = 1.008665 u, M (1 H) = 1.007825 u, M (55 Mn) = 54.938046 u Will need step-by-step instructions and answer to this question, many thanks. Will be rewarded asap.

Explanation / Answer

a)

56Fe has 26 protons and 30 neutrons.

mass of 30 neutrons = 30x(1.008665 u) = 30.25995 u

mass of 26 protons = 26x(1.007825) = 26.20345 u

Mass defect = [Mn + Mp]-M(26Fe) = (30.25995 + 26.20345) - 55.934939 = 0.528461 u

binding energy = m.c2

1 u = 931.5 MeV

hence, binding energy = 0.528461 x 931.5MeV = 492.26 MeV

b)separation energy for proton = [M(55Mn) + mp] - [M(56Fe)] = 0.010932 u

                                           = (0.010932 u) x 931.5 MeV = 10.1832 MeV

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