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9 | 2739-2742 | Solution
mFe = 55 85, u = 9 27 × 10–26 kg
Nuclear density =
mass
volume =
26
15 3
9 27 10
1
56
(4 /3)(1 |
9 | 2740-2743 | 85, u = 9 27 × 10–26 kg
Nuclear density =
mass
volume =
26
15 3
9 27 10
1
56
(4 /3)(1 2
10
)
−
−
×
×
π
×
= 2 |
9 | 2741-2744 | 27 × 10–26 kg
Nuclear density =
mass
volume =
26
15 3
9 27 10
1
56
(4 /3)(1 2
10
)
−
−
×
×
π
×
= 2 29 × 1017 kg m–3
The density of matter in neutron stars (an astrophysical object) is
comparable to this density |
9 | 2742-2745 | 27 10
1
56
(4 /3)(1 2
10
)
−
−
×
×
π
×
= 2 29 × 1017 kg m–3
The density of matter in neutron stars (an astrophysical object) is
comparable to this density This shows that matter in these objects
has been compressed to such an extent that they resemble a big nucleus |
9 | 2743-2746 | 2
10
)
−
−
×
×
π
×
= 2 29 × 1017 kg m–3
The density of matter in neutron stars (an astrophysical object) is
comparable to this density This shows that matter in these objects
has been compressed to such an extent that they resemble a big nucleus 13 |
9 | 2744-2747 | 29 × 1017 kg m–3
The density of matter in neutron stars (an astrophysical object) is
comparable to this density This shows that matter in these objects
has been compressed to such an extent that they resemble a big nucleus 13 4 MASS-ENERGY AND NUCLEAR BINDING ENERGY
13 |
9 | 2745-2748 | This shows that matter in these objects
has been compressed to such an extent that they resemble a big nucleus 13 4 MASS-ENERGY AND NUCLEAR BINDING ENERGY
13 4 |
9 | 2746-2749 | 13 4 MASS-ENERGY AND NUCLEAR BINDING ENERGY
13 4 1 Mass – Energy
Einstein showed from his theory of special relativity that it is necessary
to treat mass as another form of energy |
9 | 2747-2750 | 4 MASS-ENERGY AND NUCLEAR BINDING ENERGY
13 4 1 Mass – Energy
Einstein showed from his theory of special relativity that it is necessary
to treat mass as another form of energy Before the advent of this theory
of special relativity it was presumed that mass and energy were conserved
separately in a reaction |
9 | 2748-2751 | 4 1 Mass – Energy
Einstein showed from his theory of special relativity that it is necessary
to treat mass as another form of energy Before the advent of this theory
of special relativity it was presumed that mass and energy were conserved
separately in a reaction However, Einstein showed that mass is another
form of energy and one can convert mass-energy into other forms of
energy, say kinetic energy and vice-versa |
9 | 2749-2752 | 1 Mass – Energy
Einstein showed from his theory of special relativity that it is necessary
to treat mass as another form of energy Before the advent of this theory
of special relativity it was presumed that mass and energy were conserved
separately in a reaction However, Einstein showed that mass is another
form of energy and one can convert mass-energy into other forms of
energy, say kinetic energy and vice-versa Einstein gave the famous mass-energy equivalence relation
E = mc 2
(13 |
9 | 2750-2753 | Before the advent of this theory
of special relativity it was presumed that mass and energy were conserved
separately in a reaction However, Einstein showed that mass is another
form of energy and one can convert mass-energy into other forms of
energy, say kinetic energy and vice-versa Einstein gave the famous mass-energy equivalence relation
E = mc 2
(13 6)
Here the energy equivalent of mass m is related by the above equation
and c is the velocity of light in vacuum and is approximately equal to
3×108 m s–1 |
9 | 2751-2754 | However, Einstein showed that mass is another
form of energy and one can convert mass-energy into other forms of
energy, say kinetic energy and vice-versa Einstein gave the famous mass-energy equivalence relation
E = mc 2
(13 6)
Here the energy equivalent of mass m is related by the above equation
and c is the velocity of light in vacuum and is approximately equal to
3×108 m s–1 Example 13 |
9 | 2752-2755 | Einstein gave the famous mass-energy equivalence relation
E = mc 2
(13 6)
Here the energy equivalent of mass m is related by the above equation
and c is the velocity of light in vacuum and is approximately equal to
3×108 m s–1 Example 13 2 Calculate the energy equivalent of 1 g of substance |
9 | 2753-2756 | 6)
Here the energy equivalent of mass m is related by the above equation
and c is the velocity of light in vacuum and is approximately equal to
3×108 m s–1 Example 13 2 Calculate the energy equivalent of 1 g of substance Solution
Energy, E = 10–3 × ( 3 × 108)2 J
E = 10–3 × 9 × 1016 = 9 × 1013 J
Thus, if one gram of matter is converted to energy, there is a release
of enormous amount of energy |
9 | 2754-2757 | Example 13 2 Calculate the energy equivalent of 1 g of substance Solution
Energy, E = 10–3 × ( 3 × 108)2 J
E = 10–3 × 9 × 1016 = 9 × 1013 J
Thus, if one gram of matter is converted to energy, there is a release
of enormous amount of energy Experimental verification of the Einstein’s mass-energy relation has
been achieved in the study of nuclear reactions amongst nucleons, nuclei,
electrons and other more recently discovered particles |
9 | 2755-2758 | 2 Calculate the energy equivalent of 1 g of substance Solution
Energy, E = 10–3 × ( 3 × 108)2 J
E = 10–3 × 9 × 1016 = 9 × 1013 J
Thus, if one gram of matter is converted to energy, there is a release
of enormous amount of energy Experimental verification of the Einstein’s mass-energy relation has
been achieved in the study of nuclear reactions amongst nucleons, nuclei,
electrons and other more recently discovered particles In a reaction the
conservation law of energy states that the initial energy and the final
energy are equal provided the energy associated with mass is also
included |
9 | 2756-2759 | Solution
Energy, E = 10–3 × ( 3 × 108)2 J
E = 10–3 × 9 × 1016 = 9 × 1013 J
Thus, if one gram of matter is converted to energy, there is a release
of enormous amount of energy Experimental verification of the Einstein’s mass-energy relation has
been achieved in the study of nuclear reactions amongst nucleons, nuclei,
electrons and other more recently discovered particles In a reaction the
conservation law of energy states that the initial energy and the final
energy are equal provided the energy associated with mass is also
included This concept is important in understanding nuclear masses
and the interaction of nuclei with one another |
9 | 2757-2760 | Experimental verification of the Einstein’s mass-energy relation has
been achieved in the study of nuclear reactions amongst nucleons, nuclei,
electrons and other more recently discovered particles In a reaction the
conservation law of energy states that the initial energy and the final
energy are equal provided the energy associated with mass is also
included This concept is important in understanding nuclear masses
and the interaction of nuclei with one another They form the subject
matter of the next few sections |
9 | 2758-2761 | In a reaction the
conservation law of energy states that the initial energy and the final
energy are equal provided the energy associated with mass is also
included This concept is important in understanding nuclear masses
and the interaction of nuclei with one another They form the subject
matter of the next few sections 13 |
9 | 2759-2762 | This concept is important in understanding nuclear masses
and the interaction of nuclei with one another They form the subject
matter of the next few sections 13 4 |
9 | 2760-2763 | They form the subject
matter of the next few sections 13 4 2 Nuclear binding energy
In Section 13 |
9 | 2761-2764 | 13 4 2 Nuclear binding energy
In Section 13 2 we have seen that the nucleus is made up of neutrons
and protons |
9 | 2762-2765 | 4 2 Nuclear binding energy
In Section 13 2 we have seen that the nucleus is made up of neutrons
and protons Therefore it may be expected that the mass of the nucleus
is equal to the total mass of its individual protons and neutrons |
9 | 2763-2766 | 2 Nuclear binding energy
In Section 13 2 we have seen that the nucleus is made up of neutrons
and protons Therefore it may be expected that the mass of the nucleus
is equal to the total mass of its individual protons and neutrons However,
EXAMPLE 13 |
9 | 2764-2767 | 2 we have seen that the nucleus is made up of neutrons
and protons Therefore it may be expected that the mass of the nucleus
is equal to the total mass of its individual protons and neutrons However,
EXAMPLE 13 1
Rationalised 2023-24
311
Nuclei
EXAMPLE 13 |
9 | 2765-2768 | Therefore it may be expected that the mass of the nucleus
is equal to the total mass of its individual protons and neutrons However,
EXAMPLE 13 1
Rationalised 2023-24
311
Nuclei
EXAMPLE 13 3
the nuclear mass M is found to be always less than this |
9 | 2766-2769 | However,
EXAMPLE 13 1
Rationalised 2023-24
311
Nuclei
EXAMPLE 13 3
the nuclear mass M is found to be always less than this For example, let
us consider 16
8 O ; a nucleus which has 8 neutrons and 8 protons |
9 | 2767-2770 | 1
Rationalised 2023-24
311
Nuclei
EXAMPLE 13 3
the nuclear mass M is found to be always less than this For example, let
us consider 16
8 O ; a nucleus which has 8 neutrons and 8 protons We
have
Mass of 8 neutrons = 8 × 1 |
9 | 2768-2771 | 3
the nuclear mass M is found to be always less than this For example, let
us consider 16
8 O ; a nucleus which has 8 neutrons and 8 protons We
have
Mass of 8 neutrons = 8 × 1 00866 u
Mass of 8 protons = 8 × 1 |
9 | 2769-2772 | For example, let
us consider 16
8 O ; a nucleus which has 8 neutrons and 8 protons We
have
Mass of 8 neutrons = 8 × 1 00866 u
Mass of 8 protons = 8 × 1 00727 u
Mass of 8 electrons = 8 × 0 |
9 | 2770-2773 | We
have
Mass of 8 neutrons = 8 × 1 00866 u
Mass of 8 protons = 8 × 1 00727 u
Mass of 8 electrons = 8 × 0 00055 u
Therefore the expected mass of 16
= 8 × 2 |
9 | 2771-2774 | 00866 u
Mass of 8 protons = 8 × 1 00727 u
Mass of 8 electrons = 8 × 0 00055 u
Therefore the expected mass of 16
= 8 × 2 01593 u = 16 |
9 | 2772-2775 | 00727 u
Mass of 8 electrons = 8 × 0 00055 u
Therefore the expected mass of 16
= 8 × 2 01593 u = 16 12744 u |
9 | 2773-2776 | 00055 u
Therefore the expected mass of 16
= 8 × 2 01593 u = 16 12744 u 8 O nucleus
The atomic mass of 16
8 O found from mass spectroscopy experiments
is seen to be 15 |
9 | 2774-2777 | 01593 u = 16 12744 u 8 O nucleus
The atomic mass of 16
8 O found from mass spectroscopy experiments
is seen to be 15 99493 u |
9 | 2775-2778 | 12744 u 8 O nucleus
The atomic mass of 16
8 O found from mass spectroscopy experiments
is seen to be 15 99493 u Substracting the mass of 8 electrons (8 × 0 |
9 | 2776-2779 | 8 O nucleus
The atomic mass of 16
8 O found from mass spectroscopy experiments
is seen to be 15 99493 u Substracting the mass of 8 electrons (8 × 0 00055 u)
from this, we get the experimental mass of 16
8 O nucleus to be 15 |
9 | 2777-2780 | 99493 u Substracting the mass of 8 electrons (8 × 0 00055 u)
from this, we get the experimental mass of 16
8 O nucleus to be 15 99053 u |
9 | 2778-2781 | Substracting the mass of 8 electrons (8 × 0 00055 u)
from this, we get the experimental mass of 16
8 O nucleus to be 15 99053 u Thus, we find that the mass of the 16
8 O nucleus is less than the total
mass of its constituents by 0 |
9 | 2779-2782 | 00055 u)
from this, we get the experimental mass of 16
8 O nucleus to be 15 99053 u Thus, we find that the mass of the 16
8 O nucleus is less than the total
mass of its constituents by 0 13691u |
9 | 2780-2783 | 99053 u Thus, we find that the mass of the 16
8 O nucleus is less than the total
mass of its constituents by 0 13691u The difference in mass of a nucleus
and its constituents, DM, is called the mass defect, and is given by
[
(
)
]
p
n
M
Zm
A
Z m
M
∆
=
+
−
−
(13 |
9 | 2781-2784 | Thus, we find that the mass of the 16
8 O nucleus is less than the total
mass of its constituents by 0 13691u The difference in mass of a nucleus
and its constituents, DM, is called the mass defect, and is given by
[
(
)
]
p
n
M
Zm
A
Z m
M
∆
=
+
−
−
(13 7)
What is the meaning of the mass defect |
9 | 2782-2785 | 13691u The difference in mass of a nucleus
and its constituents, DM, is called the mass defect, and is given by
[
(
)
]
p
n
M
Zm
A
Z m
M
∆
=
+
−
−
(13 7)
What is the meaning of the mass defect It is here that Einstein’s
equivalence of mass and energy plays a role |
9 | 2783-2786 | The difference in mass of a nucleus
and its constituents, DM, is called the mass defect, and is given by
[
(
)
]
p
n
M
Zm
A
Z m
M
∆
=
+
−
−
(13 7)
What is the meaning of the mass defect It is here that Einstein’s
equivalence of mass and energy plays a role Since the mass of the oxygen
nucleus is less that the sum of the masses of its constituents (8 protons
and 8 neutrons, in the unbound state), the equivalent energy of the oxygen
nucleus is less than that of the sum of the equivalent energies of its
constituents |
9 | 2784-2787 | 7)
What is the meaning of the mass defect It is here that Einstein’s
equivalence of mass and energy plays a role Since the mass of the oxygen
nucleus is less that the sum of the masses of its constituents (8 protons
and 8 neutrons, in the unbound state), the equivalent energy of the oxygen
nucleus is less than that of the sum of the equivalent energies of its
constituents If one wants to break the oxygen nucleus into 8 protons
and 8 neutrons, this extra energy DM c2, has to supplied |
9 | 2785-2788 | It is here that Einstein’s
equivalence of mass and energy plays a role Since the mass of the oxygen
nucleus is less that the sum of the masses of its constituents (8 protons
and 8 neutrons, in the unbound state), the equivalent energy of the oxygen
nucleus is less than that of the sum of the equivalent energies of its
constituents If one wants to break the oxygen nucleus into 8 protons
and 8 neutrons, this extra energy DM c2, has to supplied This energy
required Eb is related to the mass defect by
Eb = D M c2
(13 |
9 | 2786-2789 | Since the mass of the oxygen
nucleus is less that the sum of the masses of its constituents (8 protons
and 8 neutrons, in the unbound state), the equivalent energy of the oxygen
nucleus is less than that of the sum of the equivalent energies of its
constituents If one wants to break the oxygen nucleus into 8 protons
and 8 neutrons, this extra energy DM c2, has to supplied This energy
required Eb is related to the mass defect by
Eb = D M c2
(13 8)
Example 13 |
9 | 2787-2790 | If one wants to break the oxygen nucleus into 8 protons
and 8 neutrons, this extra energy DM c2, has to supplied This energy
required Eb is related to the mass defect by
Eb = D M c2
(13 8)
Example 13 3 Find the energy equivalent of one atomic mass unit,
first in Joules and then in MeV |
9 | 2788-2791 | This energy
required Eb is related to the mass defect by
Eb = D M c2
(13 8)
Example 13 3 Find the energy equivalent of one atomic mass unit,
first in Joules and then in MeV Using this, express the mass defect
of 16
8 O in MeV/c2 |
9 | 2789-2792 | 8)
Example 13 3 Find the energy equivalent of one atomic mass unit,
first in Joules and then in MeV Using this, express the mass defect
of 16
8 O in MeV/c2 Solution
1u = 1 |
9 | 2790-2793 | 3 Find the energy equivalent of one atomic mass unit,
first in Joules and then in MeV Using this, express the mass defect
of 16
8 O in MeV/c2 Solution
1u = 1 6605 × 10–27 kg
To convert it into energy units, we multiply it by c 2 and find that
energy equivalent = 1 |
9 | 2791-2794 | Using this, express the mass defect
of 16
8 O in MeV/c2 Solution
1u = 1 6605 × 10–27 kg
To convert it into energy units, we multiply it by c 2 and find that
energy equivalent = 1 6605 × 10–27 × (2 |
9 | 2792-2795 | Solution
1u = 1 6605 × 10–27 kg
To convert it into energy units, we multiply it by c 2 and find that
energy equivalent = 1 6605 × 10–27 × (2 9979 × 108)2 kg m2/s2
= 1 |
9 | 2793-2796 | 6605 × 10–27 kg
To convert it into energy units, we multiply it by c 2 and find that
energy equivalent = 1 6605 × 10–27 × (2 9979 × 108)2 kg m2/s2
= 1 4924 × 10–10 J
=
10
1 |
9 | 2794-2797 | 6605 × 10–27 × (2 9979 × 108)2 kg m2/s2
= 1 4924 × 10–10 J
=
10
1 4924 1019
eV
1 |
9 | 2795-2798 | 9979 × 108)2 kg m2/s2
= 1 4924 × 10–10 J
=
10
1 4924 1019
eV
1 602 10
−
−
×
×
= 0 |
9 | 2796-2799 | 4924 × 10–10 J
=
10
1 4924 1019
eV
1 602 10
−
−
×
×
= 0 9315 × 109 eV
= 931 |
9 | 2797-2800 | 4924 1019
eV
1 602 10
−
−
×
×
= 0 9315 × 109 eV
= 931 5 MeV
or, 1u = 931 |
9 | 2798-2801 | 602 10
−
−
×
×
= 0 9315 × 109 eV
= 931 5 MeV
or, 1u = 931 5 MeV/c2
For 16
8 O , DM = 0 |
9 | 2799-2802 | 9315 × 109 eV
= 931 5 MeV
or, 1u = 931 5 MeV/c2
For 16
8 O , DM = 0 13691 u = 0 |
9 | 2800-2803 | 5 MeV
or, 1u = 931 5 MeV/c2
For 16
8 O , DM = 0 13691 u = 0 13691×931 |
9 | 2801-2804 | 5 MeV/c2
For 16
8 O , DM = 0 13691 u = 0 13691×931 5 MeV/c2
= 127 |
9 | 2802-2805 | 13691 u = 0 13691×931 5 MeV/c2
= 127 5 MeV/c 2
The energy needed to separate 16
8 O into its constituents is thus
127 |
9 | 2803-2806 | 13691×931 5 MeV/c2
= 127 5 MeV/c 2
The energy needed to separate 16
8 O into its constituents is thus
127 5 MeV/c2 |
9 | 2804-2807 | 5 MeV/c2
= 127 5 MeV/c 2
The energy needed to separate 16
8 O into its constituents is thus
127 5 MeV/c2 If a certain number of neutrons and protons are brought together to
form a nucleus of a certain charge and mass, an energy Eb will be released
Rationalised 2023-24
Physics
312
in the process |
9 | 2805-2808 | 5 MeV/c 2
The energy needed to separate 16
8 O into its constituents is thus
127 5 MeV/c2 If a certain number of neutrons and protons are brought together to
form a nucleus of a certain charge and mass, an energy Eb will be released
Rationalised 2023-24
Physics
312
in the process The energy Eb is called the binding energy of the nucleus |
9 | 2806-2809 | 5 MeV/c2 If a certain number of neutrons and protons are brought together to
form a nucleus of a certain charge and mass, an energy Eb will be released
Rationalised 2023-24
Physics
312
in the process The energy Eb is called the binding energy of the nucleus If we separate a nucleus into its nucleons, we would have to supply a
total energy equal to Eb, to those particles |
9 | 2807-2810 | If a certain number of neutrons and protons are brought together to
form a nucleus of a certain charge and mass, an energy Eb will be released
Rationalised 2023-24
Physics
312
in the process The energy Eb is called the binding energy of the nucleus If we separate a nucleus into its nucleons, we would have to supply a
total energy equal to Eb, to those particles Although we cannot tear
apart a nucleus in this way, the nuclear binding energy is still a convenient
measure of how well a nucleus is held together |
9 | 2808-2811 | The energy Eb is called the binding energy of the nucleus If we separate a nucleus into its nucleons, we would have to supply a
total energy equal to Eb, to those particles Although we cannot tear
apart a nucleus in this way, the nuclear binding energy is still a convenient
measure of how well a nucleus is held together A more useful measure
of the binding between the constituents of the nucleus is the binding
energy per nucleon, Ebn, which is the ratio of the binding energy Eb of a
nucleus to the number of the nucleons, A, in that nucleus:
Ebn = Eb / A
(13 |
9 | 2809-2812 | If we separate a nucleus into its nucleons, we would have to supply a
total energy equal to Eb, to those particles Although we cannot tear
apart a nucleus in this way, the nuclear binding energy is still a convenient
measure of how well a nucleus is held together A more useful measure
of the binding between the constituents of the nucleus is the binding
energy per nucleon, Ebn, which is the ratio of the binding energy Eb of a
nucleus to the number of the nucleons, A, in that nucleus:
Ebn = Eb / A
(13 9)
We can think of binding energy per nucleon as the average energy
per nucleon needed to separate a nucleus into its individual nucleons |
9 | 2810-2813 | Although we cannot tear
apart a nucleus in this way, the nuclear binding energy is still a convenient
measure of how well a nucleus is held together A more useful measure
of the binding between the constituents of the nucleus is the binding
energy per nucleon, Ebn, which is the ratio of the binding energy Eb of a
nucleus to the number of the nucleons, A, in that nucleus:
Ebn = Eb / A
(13 9)
We can think of binding energy per nucleon as the average energy
per nucleon needed to separate a nucleus into its individual nucleons Figure 13 |
9 | 2811-2814 | A more useful measure
of the binding between the constituents of the nucleus is the binding
energy per nucleon, Ebn, which is the ratio of the binding energy Eb of a
nucleus to the number of the nucleons, A, in that nucleus:
Ebn = Eb / A
(13 9)
We can think of binding energy per nucleon as the average energy
per nucleon needed to separate a nucleus into its individual nucleons Figure 13 1 is a plot of the
binding energy per nucleon Ebn
versus the mass number A for a
large number of nuclei |
9 | 2812-2815 | 9)
We can think of binding energy per nucleon as the average energy
per nucleon needed to separate a nucleus into its individual nucleons Figure 13 1 is a plot of the
binding energy per nucleon Ebn
versus the mass number A for a
large number of nuclei We notice
the following main features of
the plot:
(i)
the binding energy per
nucleon, Ebn, is practically
constant, i |
9 | 2813-2816 | Figure 13 1 is a plot of the
binding energy per nucleon Ebn
versus the mass number A for a
large number of nuclei We notice
the following main features of
the plot:
(i)
the binding energy per
nucleon, Ebn, is practically
constant, i e |
9 | 2814-2817 | 1 is a plot of the
binding energy per nucleon Ebn
versus the mass number A for a
large number of nuclei We notice
the following main features of
the plot:
(i)
the binding energy per
nucleon, Ebn, is practically
constant, i e practically
independent of the atomic
number for nuclei of middle
mass number ( 30 < A < 170) |
9 | 2815-2818 | We notice
the following main features of
the plot:
(i)
the binding energy per
nucleon, Ebn, is practically
constant, i e practically
independent of the atomic
number for nuclei of middle
mass number ( 30 < A < 170) The curve has a maximum of
about 8 |
9 | 2816-2819 | e practically
independent of the atomic
number for nuclei of middle
mass number ( 30 < A < 170) The curve has a maximum of
about 8 75 MeV for A = 56
and has a value of 7 |
9 | 2817-2820 | practically
independent of the atomic
number for nuclei of middle
mass number ( 30 < A < 170) The curve has a maximum of
about 8 75 MeV for A = 56
and has a value of 7 6 MeV
for A = 238 |
9 | 2818-2821 | The curve has a maximum of
about 8 75 MeV for A = 56
and has a value of 7 6 MeV
for A = 238 (ii) Ebn is lower for both light
nuclei (A<30) and heavy
nuclei (A>170) |
9 | 2819-2822 | 75 MeV for A = 56
and has a value of 7 6 MeV
for A = 238 (ii) Ebn is lower for both light
nuclei (A<30) and heavy
nuclei (A>170) We can draw some conclusions from these two observations:
(i)
The force is attractive and sufficiently strong to produce a binding
energy of a few MeV per nucleon |
9 | 2820-2823 | 6 MeV
for A = 238 (ii) Ebn is lower for both light
nuclei (A<30) and heavy
nuclei (A>170) We can draw some conclusions from these two observations:
(i)
The force is attractive and sufficiently strong to produce a binding
energy of a few MeV per nucleon (ii) The constancy of the binding energy in the range 30 < A < 170 is a
consequence of the fact that the nuclear force is short-ranged |
9 | 2821-2824 | (ii) Ebn is lower for both light
nuclei (A<30) and heavy
nuclei (A>170) We can draw some conclusions from these two observations:
(i)
The force is attractive and sufficiently strong to produce a binding
energy of a few MeV per nucleon (ii) The constancy of the binding energy in the range 30 < A < 170 is a
consequence of the fact that the nuclear force is short-ranged Consider
a particular nucleon inside a sufficiently large nucleus |
9 | 2822-2825 | We can draw some conclusions from these two observations:
(i)
The force is attractive and sufficiently strong to produce a binding
energy of a few MeV per nucleon (ii) The constancy of the binding energy in the range 30 < A < 170 is a
consequence of the fact that the nuclear force is short-ranged Consider
a particular nucleon inside a sufficiently large nucleus It will be under
the influence of only some of its neighbours, which come within the
range of the nuclear force |
9 | 2823-2826 | (ii) The constancy of the binding energy in the range 30 < A < 170 is a
consequence of the fact that the nuclear force is short-ranged Consider
a particular nucleon inside a sufficiently large nucleus It will be under
the influence of only some of its neighbours, which come within the
range of the nuclear force If any other nucleon is at a distance more
than the range of the nuclear force from the particular nucleon it will
have no influence on the binding energy of the nucleon under
consideration |
9 | 2824-2827 | Consider
a particular nucleon inside a sufficiently large nucleus It will be under
the influence of only some of its neighbours, which come within the
range of the nuclear force If any other nucleon is at a distance more
than the range of the nuclear force from the particular nucleon it will
have no influence on the binding energy of the nucleon under
consideration If a nucleon can have a maximum of p neighbours
within the range of nuclear force, its binding energy would be
proportional to p |
9 | 2825-2828 | It will be under
the influence of only some of its neighbours, which come within the
range of the nuclear force If any other nucleon is at a distance more
than the range of the nuclear force from the particular nucleon it will
have no influence on the binding energy of the nucleon under
consideration If a nucleon can have a maximum of p neighbours
within the range of nuclear force, its binding energy would be
proportional to p Let the binding energy of the nucleus be pk, where
k is a constant having the dimensions of energy |
9 | 2826-2829 | If any other nucleon is at a distance more
than the range of the nuclear force from the particular nucleon it will
have no influence on the binding energy of the nucleon under
consideration If a nucleon can have a maximum of p neighbours
within the range of nuclear force, its binding energy would be
proportional to p Let the binding energy of the nucleus be pk, where
k is a constant having the dimensions of energy If we increase A by
adding nucleons they will not change the binding energy of a nucleon
inside |
9 | 2827-2830 | If a nucleon can have a maximum of p neighbours
within the range of nuclear force, its binding energy would be
proportional to p Let the binding energy of the nucleus be pk, where
k is a constant having the dimensions of energy If we increase A by
adding nucleons they will not change the binding energy of a nucleon
inside Since most of the nucleons in a large nucleus reside inside it
and not on the surface, the change in binding energy per nucleon
would be small |
9 | 2828-2831 | Let the binding energy of the nucleus be pk, where
k is a constant having the dimensions of energy If we increase A by
adding nucleons they will not change the binding energy of a nucleon
inside Since most of the nucleons in a large nucleus reside inside it
and not on the surface, the change in binding energy per nucleon
would be small The binding energy per nucleon is a constant and is
approximately equal to pk |
9 | 2829-2832 | If we increase A by
adding nucleons they will not change the binding energy of a nucleon
inside Since most of the nucleons in a large nucleus reside inside it
and not on the surface, the change in binding energy per nucleon
would be small The binding energy per nucleon is a constant and is
approximately equal to pk The property that a given nucleon
FIGURE 13 |
9 | 2830-2833 | Since most of the nucleons in a large nucleus reside inside it
and not on the surface, the change in binding energy per nucleon
would be small The binding energy per nucleon is a constant and is
approximately equal to pk The property that a given nucleon
FIGURE 13 1 The binding energy per nucleon
as a function of mass number |
9 | 2831-2834 | The binding energy per nucleon is a constant and is
approximately equal to pk The property that a given nucleon
FIGURE 13 1 The binding energy per nucleon
as a function of mass number Rationalised 2023-24
313
Nuclei
influences only nucleons close to it is also referred to as saturation
property of the nuclear force |
9 | 2832-2835 | The property that a given nucleon
FIGURE 13 1 The binding energy per nucleon
as a function of mass number Rationalised 2023-24
313
Nuclei
influences only nucleons close to it is also referred to as saturation
property of the nuclear force (iii) A very heavy nucleus, say A = 240, has lower binding energy per
nucleon compared to that of a nucleus with A = 120 |
9 | 2833-2836 | 1 The binding energy per nucleon
as a function of mass number Rationalised 2023-24
313
Nuclei
influences only nucleons close to it is also referred to as saturation
property of the nuclear force (iii) A very heavy nucleus, say A = 240, has lower binding energy per
nucleon compared to that of a nucleus with A = 120 Thus if a
nucleus A = 240 breaks into two A = 120 nuclei, nucleons get more
tightly bound |
9 | 2834-2837 | Rationalised 2023-24
313
Nuclei
influences only nucleons close to it is also referred to as saturation
property of the nuclear force (iii) A very heavy nucleus, say A = 240, has lower binding energy per
nucleon compared to that of a nucleus with A = 120 Thus if a
nucleus A = 240 breaks into two A = 120 nuclei, nucleons get more
tightly bound This implies energy would be released in the process |
9 | 2835-2838 | (iii) A very heavy nucleus, say A = 240, has lower binding energy per
nucleon compared to that of a nucleus with A = 120 Thus if a
nucleus A = 240 breaks into two A = 120 nuclei, nucleons get more
tightly bound This implies energy would be released in the process It has very important implications for energy production through
fission, to be discussed later in Section 13 |
9 | 2836-2839 | Thus if a
nucleus A = 240 breaks into two A = 120 nuclei, nucleons get more
tightly bound This implies energy would be released in the process It has very important implications for energy production through
fission, to be discussed later in Section 13 7 |
9 | 2837-2840 | This implies energy would be released in the process It has very important implications for energy production through
fission, to be discussed later in Section 13 7 1 |
9 | 2838-2841 | It has very important implications for energy production through
fission, to be discussed later in Section 13 7 1 (iv) Consider two very light nuclei (A ≤ 10) joining to form a heavier
nucleus |
Subsets and Splits