Question

Find the total binding energy for Manganese, 55 Mn (atomic mass = 54.938047 u).
MeV

Answer 1

Mass defect is given by:

dm = Mass of Protons in Mn + mass of neutrons in Mn - atomic Mass of Mn

Given that

Atomic mass of Mn = 54.938047 u

Mass of 1 proton = 1.007276 u

Mass of 1 neutron = 1.008664 u

Number of protons in Pb = 25

Number of neutrons in Pb = 55 - 25 = 30

So,

dm = 25*mp + 30*mn - M_Mn

dm = 25*1.007276 + 30*1.008664 - 54.938047

dm = mass defect = 0.503773 amu

Binding energy is given by:

E = dm*c^2

c^2 = 931.5 Mev/u

So,

E = (0.503773 u)*(931.5 MeV/u) = 0.503773*931.5

E = Binding energy = 469.26 MeV

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