NCERT Class 12 Physics Chapter 13 Nuclei Solutions, NCERT Solutions For Class 12 Physics to each chapter is provided in the list so that you can easily browse throughout different chapters NCERT Class 12 Physics Chapter 13 Nuclei Question Answer and select needs one.
NCERT Class 12 Physics Chapter 13 Nuclei
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Nuclei
Chapter: 13
| Part – II |
EXERCISE
1.
Ans: Given:
Z = 7
N = 7
mp = 1.00728 u
mn = 1.00867 u
M = 14.00307 u
c = 3 × 108 m/s
Total mass of nucleons:
Mnucleons = 7 × mp + 7 × mn
= 7 × 1.007825 u + 7 × 1.008665 u
= 7.054775 u + 7.060655 u
= 14.11542 u
The mass defect Δm:
= 14.11543 u – 14.00307 u
= 0.11236 u
Convert the mass defect to energy:
Binding energy (Eb) is:
Eb = Δm × 931.5 MeV
= 0.11236 u × 931.5 MeV / u
= 104.7 MeV.
2.
Ans:
3.
Ans: Given:
Calculate the number of copper atoms in the coin:
u = 1.66054 × 10-27 kg
Mass of the coin to kg
= 3.0g = 3.0 × 10-3 kg
Single copper atom in kg:
62.92960 u = 62.92960 × 1.66054 × 10-27 kg
= 1.04442 × 10-25 kg
N = (3.0 × 10-3) / 1.04442 × 10-25
= 2.873 × 1022 atoms
Total binding energy per atom:
Etotal = 63 nucleons × 8.7 MeV
= 548.1 MeV
Total energy required:
Ecoin = 548.1 × 2.873 × 1022 × 1.60218 × 10-13 J
= 2.52 × 1013 J
Convert into MeV:
= (2.52 × 1013 J) × (1.60218 × 10-13 J /MeV)
= 1.573 × 1026 MeV.
4.
Ans: Given:
Nuclear radius of gold / Nuclear radius of silver
= (197) ⅓ / (107) ⅓
= (1.841)⅓
= 1.23.
5.
Ans: Reaction (i):
Total mass of reactants:
mreactants = 1.007825 u + 3.016049 u = 4.023874 u
mproducts = 2 (2.014102) = 4.28204 u
Mass difference:
Δm = mreactants – mproducts = 4.023874 – 4.28204
= – 0.00433 u
Q = (- 0.00433 u × 931.5 MeV/u)
= – 4.03 MeV.
Reaction (ii):
mreactants = 2 × 12.000000 = 24.000000 u
mproducts = 19.992439 u + 4.002603 u = 23.995042 u
Mass difference:
Δm = mreactants – mproducts = 24.000000 – 23.995042
= 0.004958 u
Q = 0.004958 u × 931.5 MeV/u
= 4.62 MeV.
6.
Ans: Given:
= 55.93494 u
= 27.98191 u
Q = (55.93494 u – 2 × 27.98191 u ) × 931.5 MeV/u
= – 29.90 MeV.
7.
Ans: Given:
Molar mass of plutonium (pu – 239): 239 g/mol
Avogardro’s number: 6.022 × 1023 atoms/gol
Moles in 1 kg of (pu – 239)
1000 g / 239 g/mol = 4.184 moles
Number of atoms in 1 kg of pu – 239:
4.184 × 6.022 × 1023
= 2.52 × 1024
Total energy:
2.52 × 1024 × 180 MeV/atom
= 4.54 × 1026 MeV.
8.
Ans: Given:
Number of deuterium atoms in 2 kg approximately 2.014 u
Mass units (u) = 1.66054 × 10-27 kg
= 2.014 u × 1.66054 × 10-27 kg/u = 3.345
= Deuterium atoms in 2 kg is:
N = 2kg / 3.345 × 10-27 kg
= 5.98 × 1026 atoms
Number of fusion reactions:
= 5.98 × 1026 / 2
= 2.99 × 1026 reactions
Total energy released:
Energy per reaction: 3.2 MeV × 1.60218 × 10-13 J/MeV
= 5.127 × 10-13
Total energy = 2.99 × 1026 × 5.127 × 10-13 J
= 1.533 × 1014 J
Time = Total energy power
= 1.533 × 1014 J / 100 J/s
= 1.533 × 1012 Seconds
Convert time to years
Time in years = 1.533 × 1012 / (60 × 60 × 24 × 365.25)
= 48,600 years.
9. Calculate the height of the potential barrier for a head on collision of two deuterons. (Hint: The height of the potential barrier is given by the Coulomb repulsion between the two deuterons when they just touch each other. Assume that they can be taken as hard spheres of radius 2.0 fm.)
Ans: Given:
e = 1.6 × 10-19 C
R = 2.0 fm = 2.0 × 10-15 m
U = K + K
= 9 × 109 × (1.6 × 10-19)2 / (4 × 2 × 10-15) J
= 9 × 109 × (1.6 × 10-19)2 / (8 × 10-15 × 1.6 × 10-16) KeV
= 180 KeV.
10. From the relation R = R0 A1/3, where R0 is a constant and A is the mass number of a nucleus, show that the nuclear matter density is nearly constant (i.e. independent of A).
Ans: Given the relation: R = R0 A1/3
V ∝ R3
Substituting R from the given relation:
V ∝ R0 A(1/3)3
V ∝ R03 A
M ∝ A
The density (ρ) of the nucleus is given by:
ρ = M/V
ρ ∝ A / (R03 A)
ρ ∝ 1 / R03
Therefore, the nuclear matter density is nearly constant, independent of the mass number A.
