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NCERT Class 12 Physics Textbook Chapter 13 With Answer PDF Free Download
Chapter 13: Nuclei
13.1 Introduction
In the previous chapter, we have learned that in every atom, the positive charge and mass are densely concentrated at the center of the atom forming its nucleus.
The overall dimensions of a nucleus are much smaller than those of an atom.
Experiments on the scattering of α-particles demonstrated that the radius of a nucleus was smaller than the radius of an atom by a factor of about 104.
This means the volume of a nucleus is about 10–12 times the volume of the atom. In other words, an atom is almost empty.
If an atom is enlarged to the size of a classroom, the nucleus would be the size of a pinhead. Nevertheless, the nucleus contains most (more than 99.9%) of the mass of an atom.
Does the nucleus have a structure, just as the atom does? If so, what are the constituents of the nucleus? How are these held together?
In this chapter, we shall look for answers to such questions.
We shall discuss various properties of nuclei such as their size, mass, and stability, and also associated with nuclear phenomena such as radioactivity, fission, and fusion
Discovery of Neutron
Since the nuclei of deuterium and tritium are isotopes of hydrogen, they must contain only one proton each.
But the masses of the nuclei of hydrogen, deuterium, and tritium are in the ratio of 1:2:3. Therefore, the nuclei of deuterium and tritium must contain, in addition to a proton, some neutral matter.
The amount of neutral matter present in the nuclei of these isotopes, expressed in units of mass of a proton, is approximately equal to one and two, respectively.
This fact indicates that the nuclei of
atoms contain, in addition to protons, neutral matter in multiples of a basic unit.
This hypothesis was verified in 1932 by James Chadwick who observed the emission of neutral radiation when beryllium nuclei were bombarded with alpha particles. (α-particles are helium nuclei, to be discussed in a later section).
It was found that this neutral radiation could knock out protons from light nuclei such as those helium, carbon, and nitrogen.
The only neutral radiation known at that time was photons (electromagnetic radiation).
Application of the principles of conservation of energy and momentum showed that if the neutral radiation consisted of photons, the energy of photons would have to be much higher than is
available from the bombardment of beryllium nuclei with α-particles.
The clue to this puzzle, which Chadwick satisfactorily solved, was to assume that neutral radiation consists of a new type of neutral particles called neutrons.
From the conservation of energy and momentum, he was able to determine the mass of the new particle ‘as very nearly the same as the mass of a proton.
| Author | NCERT |
| Language | English |
| No. of Pages | 29 |
| PDF Size | 735 KB |
| Category | Physics |
| Source/Credits | ncert.nic.in |
NCERT Solutions Class 12 Physics Chapter 13 Nuclei
Q 13.3: Obtain the binding energy in MeV of a nitrogen nucleus _{7}^{14}{N}714N, given m(_{7}^{14}{N}714N)=14.00307 u
Ans:
Atomic mass of nitrogen _{7}^{14}{N}714N, m = 14.00307 u
A nucleus of _{7}^{14}{N}714Nnitrogen contains 7 neutrons and 7 protons.
∆m = 7mH + 7mn − m is the mass defect the nucleus
Where,
Mass of a proton, mH = 1.007825 u
Mass of a neutron, mn = 1.008665 u
∆m = 7 × 1.007825 + 7 × 1.008665 − 14.00307
= 7.054775 + 7.06055 − 14.00307
= 0.11236 u
But 1 u = 931.5 MeV/c2
∆m = 0.11236 × 931.5 MeV/c2
Eb = ∆mc2 is the binding energy of the nucleus
Where, c =Speed of lightE_b = 0.11236 \times 931.5 \left ( \frac{MeV}{c^{2}} \right ) \times c^{2}Eb=0.11236×931.5(c2MeV)×c2
= 104.66334 Mev
Therefore, 104.66334 MeV is the binding energy of the nitrogen nucleus.
Q 13.4: Obtain the binding energy of the nuclei _{26}^{56}{Fe}2656Fe and _{83}^{209}{Bi}83209Bi in units of MeV from the following
data:
m (_{23}^{56}{Fe}2356Fe) = 55.934939 u
m(_{83}^{209}{Bi}83209Bi) = 208.980388 u
Ans :Atomic mass of _{26}^{56}{Fe}2656Fe, m1 = 55.934939 u_{26}^{56}{Fe}2656Fe nucleus has 26 protons and (56 − 26) = 30 neutrons
Hence, the mass defect of the nucleus, ∆m = 26 × mH + 30 × mn − m1
Where,
Mass of a proton, mH = 1.007825 u
Mass of a neutron, mn = 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 MeV/c2
∆m = 0.528461 × 931.5 MeV/c2E_{b_{1}}Eb1 = ∆mc2 is the binding energy of the nucleus.
Where, c = Speed of lightE_{b_{1}} = 0.528461 \times 931.5 \left ( \frac{MeV}{c^{2}} \right ) \times c^{2}Eb1=0.528461×931.5(c2MeV)×c2
= 492.26 MeV
Average binding energy per nucleon = \frac{492.26}{56} = 8.79 MeV56492.26=8.79MeV
Atomic mass of _{83}^{209}{Bi}83209Bi, m2 = 208.980388 u_{83}^{209}{Bi}83209Bi nucleus has 83 protons and (209 − 83) 126 neutrons.
The mass defect of the nucleus is given as:
∆m’ = 83 × mH + 126 × mn − m2
Where,
Mass of a proton, mH = 1.007825 u
Mass of a neutron, mn = 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 MeV/c2
∆m’ = 1.760877 × 931.5 MeV/c2
Eb2 = ∆m’c2 is the binding energy of the nucleus.
= 1.760877 × 931.5 \left ( \frac{MeV}{c^{2}} \right ) \times c^{2}(c2MeV)×c2
= 1640.26 MeV
Average binding energy per nucleon = 1640.26/209 = 7.848 MeV
Q 13.5: A given coin has a mass of 3.0 g. Calculate the nuclear energy that would be required to separate all the neutrons and protons from each other. For simplicity assume that the coin is entirely made of _{29}^{63}{Cu}2963Cu with mass = 62.92960 u.
Ans :
Mass of a copper coin, m’ = 3 g
Atomic mass of _{29}^{63}{Cu}2963Cu atom, m = 62.92960 u
The total number of _{29}^{63}{Cu}2963Cu atoms in the coin, N = \frac{N_A \times m’}{Mass \; number}N=MassnumberNA×m’
Where,
NA = Avogadro’s number = 6.023 × 1023 atoms /g
Mass number = 63 gN = \frac{6.023 \times 10^{23}\times 3}{63}N=636.023×1023×3 = 2.868 x 1022 atoms_{29}^{63}{Cu}2963Cu nucleus has 29 protons and (63 − 29) 34 neutrons
Mass defect of this nucleus, ∆m’ = 29 × mH + 34 × mn − m
Where,
Mass of a proton, mH = 1.007825 u
Mass of a neutron, mn = 1.008665 u
∆m’ = 29 × 1.007825 + 34 × 1.008665 − 62.9296
= 0.591935 u
Mass defect of all the atoms present in the coin, ∆m = 0.591935 × 2.868 × 1022
= 1.69766958 × 1022 u
But 1 u = 931.5 MeV/c2
∆m = 1.69766958 × 1022 × 931.5 MeV/c2
Eb = ∆mc2 is the binding energy of the nuclei of the coin
= 1.69766958 × 1022 × 931.5 MeV/c2 × c2
= 1.581 × 1025 MeV
But 1 MeV = 1.6 × 10−13 J
Eb = 1.581 × 1025 × 1.6 × 10−13
= 2.5296 × 1012 J
Therefore, the energy required to separate all the neutrons and protons from the given coin is 2.5296 × 1012 J
NCERT Class 12 Physics Textbook Chapter 13 With Answer PDF Free Download