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Obtain the Binding Energy of the Nuclei 2656fe5626fe and 20983bi83209biin Units of Mev from the Following Data: - Physics

Obtain the binding energy of the nuclei `""_56^26Fe` and `""_83^209Bi`in units of MeV from the following data:

`m(""_26^56Fe)` = 55.934939 u `m(""_83^209Bi)`= 208.980388 u

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Solution

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

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

Hence, the mass defect of the nucleus, Δ= 26 × mH + 30 × mn − m1

Where,

Mass of a proton, mH = 1.007825 u

Mass of a neutron, mn = 1.008665 u

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

= 26.20345 + 30.25995 − 55.934939

= 0.528461 u

But 1 u = 931.5 MeV/c2

∴Δ= 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 `""_83^209Bi`, 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 bindingenergy per nucleon = `1640.26/209 = 7.848 MeV`

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APPEARS IN

NCERT Class 12 Physics Textbook
Chapter 13 Nuclei
Q 4 | Page 462
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