Nuclei Easy Questions and Answers | Page - 9

Nuclei

A nucleus _ZX^A has mass represented by M(A, Z). If M_p and M_n denote the mass of proton and neutron respectively and B.E. the binding energy in MeV, then

1. B.E. = [ZM_p + (A – Z)M_p – M(A, Z)]c²
1. B.E. = [ZM_p + ZMn – M(A, Z)]c²
1. B.E. = M(A, Z) – ZM_p – (A – Z)M_n
1. B.E. = [M(A, Z) – ZM_p – (A – Z)M_n]c²

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The difference in mass of a nucleus and its constituents, ΔM, is called the mass defect and is given by
ΔM = [ZM_p + (A - Z)M_n] - M
and binding energy = ΔMc²
[{ZM_p + (A - Z)M_n} - M]c²

Correct Option: A

The difference in mass of a nucleus and its constituents, ΔM, is called the mass defect and is given by
ΔM = [ZM_p + (A - Z)M_n] - M
and binding energy = ΔMc²
[{ZM_p + (A - Z)M_n} - M]c²

If M (A; Z), M_p and M_n denote the masses of the nucleus _ZX^A proton and neutron respectively in units of u ( 1u = 931.5 MeV/c²) and BE represents its bonding energy in MeV, then

1. M (A, Z) = ZM_p + (A – Z) M_n –BE/c²
1. M (A, Z) = ZM_p + ( A–Z) M_n + BE
1. M (A, Z) = ZM_p + (A – Z) M_n – BE
1. M (A, Z) = ZM_p + (A – Z)M_n + BE/c²

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Mass defect = ZM_p + (A –Z)M_n–M(A,Z)
or, B.E
c²

= ZM_p + (A–Z) M_n–M(A,Z)
∴ M (A, Z) = ZM_p + (A–Z)M_n – B.E
c²

Correct Option: A

Mass defect = ZM_p + (A –Z)M_n–M(A,Z)
or, B.E
c²

= ZM_p + (A–Z) M_n–M(A,Z)
∴ M (A, Z) = ZM_p + (A–Z)M_n – B.E
c²

The binding energy per nucleon in deuterium and helium nuclei are 1.1 MeV and 7.0 MeV, respectively. When two deuterium nuclei fuse to form a helium nucleus the energy released in the fusion is :

1. 30.2 MeV
1. 23.6 MeV
1. 2.2 MeV
1. 28.0 MeV

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Binding energy of two ₁H² nuclei
= 2(1.1 × 2) = 4.4 MeV
Binding energy of one ₂He⁴ nucelus
= 4 × 7.0 =28 MeV
∴ Energy released
= 28 – 4.4 = 23.6 MeV

Correct Option: B

Binding energy of two ₁H² nuclei
= 2(1.1 × 2) = 4.4 MeV
Binding energy of one ₂He⁴ nucelus
= 4 × 7.0 =28 MeV
∴ Energy released
= 28 – 4.4 = 23.6 MeV

The mass of a ₃Li⁷ nucleus is 0.042 u less than the sum of the masses of all its nucleons. The binding energy per nucleon of ₃Li⁷ nucleus is nearly

1. 46 MeV
1. 5.6 MeV
1. 3.9 MeV
1. 23 MeV

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B.E. = 0.042 × 931 ≃ 42 MeV
Number of nucleons in
7 Li is 7.
3

∴ B.E./ nucleon = 42/7 = 6 MeV ≃ 5.6 MeV

Correct Option: B

B.E. = 0.042 × 931 ≃ 42 MeV
Number of nucleons in
7 Li is 7.
3

∴ B.E./ nucleon = 42/7 = 6 MeV ≃ 5.6 MeV

Fusion reaction takes place at high temperature because

1. nuclei break up at high temperature
1. atoms get ionised at high temperature
1. kinetic energy is high enough to overcome the coulomb repulsion between nuclei
1. molecules break up at high temperature

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When the coulomb repulsion between the nuclei is overcome then nuclear fusion reaction takes place. This is possible when temperature is too high.

Correct Option: C

When the coulomb repulsion between the nuclei is overcome then nuclear fusion reaction takes place. This is possible when temperature is too high.