Question

A) An atom of iron-56, which consists of 30 neutrons, 26 protons, and 26 electrons, has a mass of 55.9349375 u. Calculate the mass defect for iron-56.

B) Convert the mass defect to energy.

C) Determine the average binding energy per nucleon for iron 56 (in MeV/nucleon).

Answer 1

Part A.

Mass defect is given by:

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

Given that

Atomic mass of iron = 55.9349375 u

Mass of 1 proton = 1.007276467 u

Mass of 1 electron = 9.11*10^-31 kg = 0.000549 amu

Mass of 1 neutron = 1.008664916 u

Number of protons in iron = 26

Number of neutrons in iron = 56 - 26 = 30

Number of electrons in iron = 26

So,

dm = 26*mp + 30*mn + 26*me - M_Fe

dm = 26*1.007276467 + 30*1.008664916 +26*0.000549 - 55.9349375

(1 amu = 1.6605*10^-27 kg)

dm = mass defect = 0.528472122 amu = 8.775*10^-28 kg

Part B.

Binding energy is given by:

E = dm*c^2

c^2 = 931.5 Mev/u

So,

E = (0.528472122 u)*(931.5 MeV/u) = 0.528472122*931.5

E = Binding energy = 492.271782 MeV = 492.3 MeV

Part C.

Binding energy per nucleon will be:

BE per nucleon = BE/total number of nucleons

Total number of nucleons = 56

BE per nucleon = BE/total number of nucleons = 492.271782/56

Binding energy per nucleon = 8.79 MeV/nucleon

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