# Obtain the binding energy of the nuclei Fe2656Fe and Bi83209Bi in units of MeV from the following data: mFem(2656Fe) = 55.934939 u mBim(83209Bi)= 208.980388 u - Physics

Numerical

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

#### Solution

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 × mH + 30 × mn − m1

Where,

Mass of a proton, mH = 1.007825 u

Mass of a neutron, mn = 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:

Eb1 = Δmc2

Where,

c = Speed of light

∴ Eb1 = 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 × mH + 126 × mn − m2

Where,

Mass of a proton, mH = 1.007825 u

Mass of a neutron, mn = 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:

Eb2 = 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

Concept: Mass-energy and Nuclear Binding Energy - Nuclear Binding Energy
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#### APPEARS IN

NCERT Physics Part 1 and 2 Class 12
Chapter 13 Nuclei
Exercise | Q 13.4 | Page 462
NCERT Class 12 Physics Textbook
Chapter 13 Nuclei
Exercise | Q 4 | Page 462

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