NCERT solution Class 12 nuclei

NCERT Solutions

Class 12 Physics

Nuclei

NCERT Questions

Q.1
Q.2
Q.3
Q.4
Q.5
Q.6
Q.7
Q.8
Q.9
Q.10
Q.11
Q.12
Q.13
Q.14
Q.15
Q.16
Q.17
Q.18
Q.19
Q.20
Q.21
Q.22

Q.4

Obtain the binding energy of the nuclei ⁵⁶₂₆Fe and ²⁰⁹₈₃ Bi in units of MeV from the following data: m (⁵⁶₂₆Fe) = 55.934939 u m (²⁰⁹₈₃Bi) = 208.980388 u

Given:

Atomic mass of (⁵⁶₂₆Fe), m_Fe = 55.934939 u

(⁵⁶₂₆Fe) nucleus has 26 protons and (56 − 26) = 30 neutrons

Hence, the mass defect of the nucleus, ∆m = ((26 × m_H) + (30 × m_n) – m_Fe)

where,

Mass of a proton, m_H = 1.007825 u

Mass of a neutron, m_n = 1.008665 u

∆m = ((26 × 1.007825 + 30 × 1.008665) − 55.934939)

= (26.20345 + 30.25995 − 55.934939)

= 0.528461 u

But 1 u = 931.5

Therefore, ∆m = (0.528461 × 931.5)

The binding energy of this nucleus is given as:

E_b1 = ∆mc²

Where, c =Speed of light

E_b1 = (0.528461 × 931.5 c²)

= 492.26 MeV

Average binding energy per nucleon= (492.26/56) =8.76 MeV

Atomic mass of (²⁰⁹₈₃Bi), m₂ = 208.980388 u

(²⁰⁹₈₃Bi) nucleus has 83 protons and (209 − 83) = 126 neutrons.

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

∆m' = (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

∆m' = ((83 × 1.007825) + (126 × 1.008665) – 208.980388)

= 83.649475 + 127.091790 − 208.980388

= 1.760877 u

But 1 u = 931.5

Therefore, ∆m' = (1.760877 × 931.5) x c²

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

E_b2 = ∆m'c²

= (1.760877 × 931.5 x c²)

= 1640.26 MeV

Average binding energy per nucleon = =7.848MeV

Nuclei Class 12 Physics NCERT Chapter 13 NCERT Solutions

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