Question

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

`"m"(""_83^209"Bi")`= 208.980388 u

Answer 1

Atomic mass of `""_56^26"Fe"`, m₁ = 55.934939 u

`""_56^26"Fe"` nucleus has 26 protons and (56 − 26) = 30 neutrons

Hence, the mass defect of the nucleus, Δm = 26 × m_H + 30 × m_n − m₁

Where,

Mass of a proton, m_H = 1.007825 u

Mass of a neutron, m_n = 1.008665 u

∴ Δm = 26 × 1.007825 + 30 × 1.008665 − 55.934939

= 26.20345 + 30.25995 − 55.934939

= 0.528461 u

But 1 u = 931.5 MeV/c²

∴ Δm = 0.528461 × 931.5 MeV/c²

The binding energy of this nucleus is given as:

E_b1 = Δmc²

Where,

c = Speed of light

∴ E_b1 = 0.528461 × 931.5 `(("MeV")/"c"^2) xx "c"^2`

= 492.26 MeV

Average binding energy per nucleon = `492.26/56` = 8.79 MeV

Atomic mass of `""_83^209"Bi"`, m₂ = 208.980388 u

`""_83^209"Bi"` nucleus has 83 protons and (209 − 83) 126 neutrons.

Hence, the mass defect of this nucleus is given as:

Δm' = 83 × m_H + 126 × m_n − m₂

Where,

Mass of a proton, m_H = 1.007825 u

Mass of a neutron, m_n = 1.008665 u

∴ Δm' = 83 × 1.007825 + 126 × 1.008665 − 208.980388

= 83.649475 + 127.091790 − 208.980388

= 1.760877 u

But 1 u = 931.5 MeV/c²

∴ Δm' = 1.760877 × 931.5 MeV/c²

Hence, the binding energy of this nucleus is given as:

E_b2 = Δm'c²

= 1.760877 × 931.5 `(("MeV")/"c"^2) xx "c"^2`

= 1640.26 MeV

Average binding energy per nucleon = `1640.26/209` = 7.848 MeV