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13.1 INTRODUCTION
In the previous chapter, we have learnt that in every atom, the positive
charge and mass are densely concentrated at the centre of the atom
forming its nucleus. The overall dimensions of a nucleus are much smaller
than those of an atom. Experiments on scattering of
α
-particles
demonstrated that the radius of a nucleus was smaller than the radius
of an atom by a factor of about 10
4
. 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 of the size of 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 nuclear phenomena such as radioactivity, fission and fusion.
13.2 ATOMIC MASSES AND COMPOSITION OF NUCLEUS
The mass of an atom is very small, compared to a kilogram; for example,
the mass of a carbon atom,
12
C, is 1.992647 × 10
–26
kg. Kilogram is not
a very convenient unit to measure such small quantities. Therefore, a
Chapter Thirteen
NUCLEI
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different mass unit is used for expressing atomic masses. This unit is the
atomic mass unit (u), defined as 1/12
th
of the mass of the carbon (
12
C)
atom. According to this definition
12
mass of one C atom
1u =
12
26
1.992647 10 kg
12
−
×
=
27
1.660539 10 kg
−
= ×
(13.1)
The atomic masses of various elements expressed in atomic mass
unit (u) are close to being integral multiples of the mass of a hydrogen
atom. There are, however, many striking exceptions to this rule. For
example, the atomic mass of chlorine atom is 35.46 u.
Accurate measurement of atomic masses is carried out with a mass
spectrometer, The measurement of atomic masses reveals the existence
of different types of atoms of the same element, which exhibit the same
chemical properties, but differ in mass. Such atomic species of the same
element differing in mass are called isotopes. (In Greek, isotope means
the same place, i.e. they occur in the same place in the periodic table of
elements.) It was found that practically every element consists of a mixture
of several isotopes. The relative abundance of different isotopes differs
from element to element. Chlorine, for example, has two isotopes having
masses 34.98 u and 36.98 u, which are nearly integral multiples of the
mass of a hydrogen atom. The relative abundances of these isotopes are
75.4 and 24.6 per cent, respectively. Thus, the average mass of a chlorine
atom is obtained by the weighted average of the masses of the two
isotopes, which works out to be
=
75.4 34.98 24.6 36.98
100
× + ×
= 35.47 u
which agrees with the atomic mass of chlorine.
Even the lightest element, hydrogen has three isotopes having masses
1.0078 u, 2.0141 u, and 3.0160 u. The nucleus of the lightest atom of
hydrogen, which has a relative abundance of 99.985%, is called the
proton. The mass of a proton is
27
1.00727 u 1.67262 10 kg
p
m
−
= = ×
(13.2)
This is equal to the mass of the hydrogen atom (= 1.00783u), minus
the mass of a single electron (m
e
= 0.00055 u). The other two isotopes of
hydrogen are called deuterium and tritium. Tritium nuclei, being
unstable, do not occur naturally and are produced artificially in
laboratories.
The positive charge in the nucleus is that of the protons. A proton
carries one unit of fundamental charge and is stable. It was earlier thought
that the nucleus may contain electrons, but this was ruled out later using
arguments based on quantum theory. All the electrons of an atom are
outside the nucleus. We know that the number of these electrons outside
the nucleus of the atom is Z, the atomic number. The total charge of the
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atomic electrons is thus (–Ze), and since the atom is neutral, the charge
of the nucleus is (+Ze
). The number of protons in the nucleus of the atom
is, therefore, exactly Z, the atomic number.
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 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 of 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 the neutral radiation consists of a new type of neutral
particles called neutrons. From conservation of energy and momentum,
he was able to determine the mass of new particle ‘as very nearly the
same as mass of proton’.
The mass of a neutron is now known to a high degree of accuracy. It is
m
n
= 1.00866 u = 1.6749×10
–27
kg (13.3)
Chadwick was awarded the 1935 Nobel Prize in Physics for his
discovery of the neutron.
A free neutron, unlike a free proton, is unstable. It decays into a
proton, an electron and a antineutrino (another elementary particle), and
has a mean life of about 1000s. It is, however, stable inside the nucleus.
The composition of a nucleus can now be described using the following
terms and symbols:
Z - atomic number = number of protons [13.4(a)]
N - neutron number = number of neutrons [13.4(b)]
A - mass number = Z + N
= total number of protons and neutrons [13.4(c)]
One also uses the term nucleon for a proton or a neutron. Thus the
number of nucleons in an atom is its mass number A.
Nuclear species or nuclides are shown by the notation
X
A
Z
where X is
the chemical symbol of the species. For example, the nucleus of gold is
denoted by
197
79
Au
. It contains 197 nucleons, of which 79 are protons
and the rest118 are neutrons.
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The composition of isotopes of an element can now be readily
explained. The nuclei of isotopes of a given element contain the same
number of protons, but differ from each other in their number of neutrons.
Deuterium,
2
1
H
, which is an isotope of hydrogen, contains one proton
and one neutron. Its other isotope tritium,
3
1
H
, contains one proton and
two neutrons. The element gold has 32 isotopes, ranging from A =173 to
A = 204. We have already mentioned that chemical properties of elements
depend on their electronic structure. As the atoms of isotopes have
identical electronic structure they have identical chemical behaviour and
are placed in the same location in the periodic table.
All nuclides with same mass number A are called isobars. For
example, the nuclides
3
1
H
and
3
2
He
are isobars. Nuclides with same
neutron number N but different atomic number Z, for example
198
80
Hg
and
197
79
Au
, are called isotones.
13.3 SIZE OF THE NUCLEUS
As we have seen in Chapter 12, Rutherford was the pioneer who
postulated and established the existence of the atomic nucleus. At
Rutherford’s suggestion, Geiger and Marsden performed their classic
experiment: on the scattering of
α
-particles from thin gold foils. Their
experiments revealed that the distance of closest approach to a gold
nucleus of an
α
-particle of kinetic energy 5.5 MeV is about 4.0 × 10
–14
m.
The scattering of
α
-particle by the gold sheet could be understood by
Rutherford by assuming that the coulomb repulsive force was solely
responsible for scattering. Since the positive charge is confined to the
nucleus, the actual size of the nucleus has to be less than 4.0 × 10
–14
m.
If we use
α
-particles of higher energies than 5.5 MeV, the distance of
closest approach to the gold nucleus will be smaller and at some point
the scattering will begin to be affected by the short range nuclear forces,
and differ from Rutherford’s calculations. Rutherford’s calculations are
based on pure coulomb repulsion between the positive charges of the
α
-
particle and the gold nucleus. From the distance at which deviations set
in, nuclear sizes can be inferred.
By performing scattering experiments in which fast electrons, instead
of α-particles, are projectiles that bombard targets made up of various
elements, the sizes of nuclei of various elements have been accurately
measured.
It has been found that a nucleus of mass number A has a radius
R = R
0
A
1/3
(13.5)
where R
0
= 1.2 × 10
–15
m (=1.2 fm; 1 fm = 10
–15
m). This means the volume
of the nucleus, which is proportional to R
3
is proportional to A. Thus the
density of nucleus is a constant, independent of A, for all nuclei. Different
nuclei are like a drop of liquid of constant density. The density of nuclear
matter is approximately 2.3 × 10
17
kg m
–3
. This density is very large
compared to ordinary matter, say water, which is 10
3
kg m
–3
. This is
understandable, as we have already seen that most of the atom is empty.
Ordinary matter consisting of atoms has a large amount of empty space.
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EXAMPLE 13.2
Example 13.1 Given the mass of iron nucleus as 55.85u and A=56,
find the nuclear density?
Solution
m
Fe
= 55.85, u = 9.27 × 10
–26
kg
Nuclear density =
mass
volume
=
26
15 3
9.27 10
1
56
(4 /3)(1.2 10 )
−
−
×
×
π ×
= 2.29 × 10
17
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
13.4.1 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.
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
3×10
8
m s
–1
.
Example 13.2 Calculate the energy equivalent of 1 g of substance.
Solution
Energy, E = 10
–3
× ( 3 × 10
8
)
2
J
E = 10
–3
× 9 × 10
16
= 9 × 10
13
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. They form the subject
matter of the next few sections.
13.4.2 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. However,
EXAMPLE 13.1
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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 × 1.00866 u
Mass of 8 protons = 8 × 1.00727 u
Mass of 8 electrons = 8 × 0.00055 u
Therefore the expected mass of
16
8
O
nucleus
= 8 × 2.01593 u = 16.12744 u.
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 15.99053 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, ∆M, 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. If one wants to break the oxygen nucleus into 8 protons
and 8 neutrons, this extra energy ∆M c
2
, has to supplied. This energy
required E
b
is related to the mass defect by
E
b
= ∆ M c
2
(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/c
2
.
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 × 10
8
)
2
kg m
2
/s
2
= 1.4924 × 10
–10
J
=
10
19
1.4924 10
eV
1.602 10
−
−
×
×
= 0.9315 × 10
9
eV
= 931.5 MeV
or, 1u = 931.5 MeV/c
2
For
16
8
O
, ∆M = 0.13691 u = 0.13691×931.5 MeV/c
2
= 127.5 MeV/c
2
The energy needed to separate
16
8
O
into its constituents is thus
127.5 MeV/c
2
.
If a certain number of neutrons and protons are brought together to
form a nucleus of a certain charge and mass, an energy E
b
will be released
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in the process. The energy E
b
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 E
b
, 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, E
bn
, which is the ratio of the binding energy E
b
of a
nucleus to the number of the nucleons, A, in that nucleus:
E
bn
= E
b
/ 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 E
bn
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, E
bn
, 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 8.75 MeV for A = 56
and has a value of 7.6 MeV
for A = 238.
(ii) E
bn
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. 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. 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 13.1 The binding energy per nucleon
as a function of mass number.
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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.
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. The binding energy per nucleon of the fused heavier nuclei
is more than the binding energy per nucleon of the lighter nuclei.
This means that the final system is more tightly bound than the initial
system. Again energy would be released in such a process of
fusion. This is the energy source of sun, to be discussed later in
Section 13.7.3.
13.5 NUCLEAR FORCE
The force that determines the motion of atomic electrons is the familiar
Coulomb force. In Section 13.4, we have seen that for average mass
nuclei the binding energy per nucleon is approximately 8 MeV, which is
much larger than the binding energy in atoms. Therefore, to bind a
nucleus together there must be a strong attractive force of a totally
different kind. It must be strong enough to overcome the repulsion
between the (positively charged) protons and to bind both protons and
neutrons into the tiny nuclear volume. We have already seen
that the constancy of binding energy per nucleon can be
understood in terms of its short-range. Many features of the
nuclear binding force are summarised below. These are
obtained from a variety of experiments carried out during 1930
to 1950.
(i) The nuclear force is much stronger than the Coulomb force
acting between charges or the gravitational forces between
masses. The nuclear binding force has to dominate over
the Coulomb repulsive force between protons inside the
nucleus. This happens only because the nuclear force is
much stronger than the coulomb force. The gravitational
force is much weaker than even Coulomb force.
(ii) The nuclear force between two nucleons falls rapidly to
zero as their distance is more than a few femtometres. This
leads to saturation of forces in a medium or a large-sized
nucleus, which is the reason for the constancy of the
binding energy per nucleon.
A rough plot of the potential energy between two nucleons
as a function of distance is shown in the Fig. 13.2. The
potential energy is a minimum at a distance r
0
of about
0.8 fm. This means that the force is attractive for distances larger
than 0.8 fm and repulsive if they are separated by distances less
than 0.8 fm.
FIGURE 13.2 Potential energy
of a pair of nucleons as a
function of their separation.
For a separation greater
than r
0
, the force is attractive
and for separations less
than r
0
, the force is
strongly repulsive.
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(iii) The nuclear force between neutron-neutron, proton-neutron and
proton-proton is approximately the same. The nuclear force does not
depend on the electric charge.
Unlike Coulomb’s law or the Newton’s law of gravitation there is no
simple mathematical form of the nuclear force.
13.6 RADIOACTIVITY
A. H. Becquerel discovered radioactivity in 1896 purely by accident. While
studying the fluorescence and phosphorescence of compounds irradiated
with visible light, Becquerel observed an interesting phenomenon. After
illuminating some pieces of uranium-potassium sulphate with visible
light, he wrapped them in black paper and separated the package from a
photographic plate by a piece of silver. When, after several hours of
exposure, the photographic plate was developed, it showed blackening
due to something that must have been emitted by the compound and
was able to penetrate both black paper and the silver.
Experiments performed subsequently showed that radioactivity was
a nuclear phenomenon in which an unstable nucleus undergoes a decay.
This is referred to as radioactive decay. Three types of radioactive decay
occur in nature :
(i) α-decay in which a helium nucleus
4
2
He
is emitted;
(ii) β-decay in which electrons or positrons (particles with the same mass
as electrons, but with a charge exactly opposite to that of electron)
are emitted;
(iii) γ-decay in which high energy (hundreds of keV or more) photons are
emitted.
Each of these decay will be considered in subsequent sub-sections.
13.6.1 Law of radioactive decay
In any radioactive sample, which undergoes α, β or γ-decay, it is found
that the number of nuclei undergoing the decay per unit time is
proportional to the total number of nuclei in the sample. If N is the
number of nuclei in the sample and ∆N undergo decay in time ∆t then
N
N
t
∆
∝
∆
or, ∆N/∆t = λN, (13.10)
where λ is called the radioactive decay constant or disintegration constant.
The change in the number of nuclei in the sample* is dN = – ∆N in
time ∆t. Thus the rate of change of N is (in the limit ∆t → 0)
d
–
d
N
N
t
λ
=
* ∆N is the number of nuclei that decay, and hence is always positive. dN is the
change in N, which may have either sign. Here it is negative, because out of
original N nuclei, ∆N have decayed, leaving (N–∆N) nuclei.
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or,
d
N
N
= –λdt
Now, integrating both sides of the above equation,we get,
(13.11)
or, ln N − ln N
0
= −
λ
(t – t
0
) (13.12)
Here N
0
is the number of radioactive nuclei in the sample at some
arbitrary time t
0
and N is the number of radioactive nuclei at any
subsequent time t. Setting t
0
= 0 and rearranging Eq. (13.12) gives us
ln
0
N
t
N
= −
λ
(13.13)
which gives
N(t) = N
0
e
−
λ
t
(13.14)
Note, for example, the light bulbs follow no such exponential decay law.
If we test 1000 bulbs for their life (time span before they burn out or
fuse), we expect that they will ‘decay’ (that is, burn out) at more or less
the same time. The decay of radionuclides follows quite a different law,
the law of radioactive decay represented by Eq. (13.14).
The total decay rate R of a sample is the number of nuclei
disintegrating per unit time. Suppose in a time interval dt, the decay
count measured is ∆N. Then dN = – ∆N.
The positive quantity R is then defined as
R = –
d
d
N
t
Differentiating Eq. (13.14), we get
R = λN
0
e
−
λ
t
or, R = R
0
e
–
λ
t
(13.15)
This is equivalant to the law of radioactivity decay,
since you can integrate Eq. (13.15) to get back Eq.
(13.14). Clearly, R
0
= λN
0
is the decay rate at t = 0. The
decay rate R at a certain time t and the number of
undecayed nuclei N at the same time are related by
R = λN (13.16)
The decay rate of a sample, rather than the number of radioactive
nuclei, is a more direct experimentally measurable quantity and is given
a specific name: activity. The SI unit for activity is becquerel, named
after the discoverer of radioactivity, Henry Becquerel.
FIGURE 13.3 Exponential decay of a
radioactive species. After a lapse of
T
1/2
, population of the given species
drops by a factor of 2.
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EXAMPLE 13.4
1 becquerel is simply equal to 1 disintegration or decay per second.
There is also another unit named “curie” that is widely used and is related
to the SI unit as:
1 curie = 1 Ci = 3.7 × 10
10
decays per second
= 3.7 × 10
10
Bq
Different radionuclides differ greatly in their rate of
decay. A common way to characterize this feature is
through the notion of half-life. Half-life of a radionuclide
(denoted by T
1/2
) is the time it takes for a sample that has
initially, say N
0
radionuclei to reduce to N
0
/2. Putting
N = N
0
/2 and t = T
1/2
in Eq. (13.14), we get
T
1/2
=
ln 2
λ
=
0.693
λ
(13.17)
Clearly if N
0
reduces to half its value in time T
1/2
, R
0
will also reduce to half its value in the same time according
to Eq. (13.16).
Another related measure is the average or mean life
τ
. This again can be obtained from Eq. (13.14). The
number of nuclei which decay in the time interval t to t +
∆t is R(t)∆t (= λN
0
e
–λt
∆t). Each of them has lived for time
t. Thus the total life of all these nuclei would be t λN
0
e
–λt
∆t. It is clear that some nuclei may live for a short time
while others may live longer. Therefore to obtain the mean
life, we have to sum (or integrate) this expression over all
times from 0 to
∞
, and divide by the total number N
0
of
nuclei at t = 0. Thus,
–
0
–
0
0
0
d
d
t
t
N te t
te t
N
λ
λ
λ
τ λ
∞
∞
= =
∫
∫
One can show by performing this integral that
τ
= 1/
λ
We summarise these results with the following:
T
1/2
=
ln 2
λ
=
τ
ln 2 (13.18)
Radioactive elements (e.g., tritium, plutonium) which are short-lived
i.e., have half-lives much less than the age of the universe (
∼
15 billion
years) have obviously decayed long ago and are not found in nature.
They can, however, be produced artificially in nuclear reactions.
Example 13.4 The half-life of
238
92
U
undergoing
α
-decay is 4.5 × 10
9
years. What is the activity of 1g sample of
238
92
U
?
Solution
T
1/2
= 4.5 × 10
9
y
= 4.5 × 10
9
y x 3.16 x 10
7
s/y
= 1.42 × 10
17
s
MARIE SKLODOWSKA CURIE (1867-1934)
Marie Sklodowska Curie
(1867-1934) Born in
Poland. She is recognised
both as a physicist and as
a chemist. The discovery of
radioactivity by Henri
Becquerel in 1896 inspired
Marie and her husband
Pierre Curie in their
researches and analyses
which led to the isolation of
radium and polonium
elements. She was the first
person to be awarded two
Nobel Prizes- for Physics in
1903 and for Chemistry
in 1911.
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EXAMPLE 13.5 EXAMPLE 13.4
One kmol of any isotope contains Avogadro’s number of atoms, and
so 1g of
238
92
U
contains
–3
10
kmol
238
× 6.025 × 10
26
atoms/kmol
= 25.3 × 10
20
atoms.
The decay rate R is
R =
λ
N
=
1/ 2
0.693
N
T
=
20
1
17
0.693 25.3 10
1.42 10
s
−
× ×
×
= 1.23 × 10
4
s
–1
= 1.23 × 10
4
Bq
Example 13.5 Tritium has a half-life of 12.5 y undergoing beta decay.
What fraction of a sample of pure tritium will remain undecayed
after 25 y.
Solution
By definition of half-life, half of the initial sample will remain
undecayed after 12.5 y. In the next 12.5 y, one-half of these nuclei
would have decayed. Hence, one fourth of the sample of the initial
pure tritium will remain undecayed.
13.6.2 Alpha decay
A well-known example of alpha decay is the decay of uranium
238
92
U
to
thorium
234
90
Th
with the emission of a helium nucleus
4
2
He
238
92
U
→
234
90
Th
+
4
2
He
(α-decay) (13.19)
In α-decay, the mass number of the product nucleus (daughter
nucleus) is four less than that of the decaying nucleus (parent nucleus),
while the atomic number decreases by two. In general, α-decay of a parent
nucleus
A
Z
X
results in a daughter nucleus
A 4
Z 2
Y
−
−
A
Z
X
→
A 4
Z 2
Y
−
−
+
4
2
He
(13.20)
From Einstein’s mass-energy equivalance relation [Eq. (13.6)] and
energy conservation, it is clear that this spontaneous decay is possible
only when the total mass of the decay products is less than the mass of
the initial nucleus. This difference in mass appears as kinetic energy of
the products. By referring to a table of nuclear masses, one can check
that the total mass of
234
90
Th
and
4
2
He
is indeed less than that of
238
92
U
.
The disintegration energy or the Q-value of a nuclear reaction is the
difference between the initial mass energy and the total mass energy of
the decay products. For α-decay
Q = (m
X
– m
Y
– m
He
) c
2
(13.21)
Q is also the net kinetic energy gained in the process or, if the initial
nucleus X is at rest, the kinetic energy of the products. Clearly, Q> 0 for
exothermic processes such as α-decay.
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EXAMPLE
13.6
Example 13.6 We are given the following atomic masses:
238
92
U
= 238.05079 u
4
2
He
= 4.00260 u
234
90
Th
= 234.04363 u
1
1
H
= 1.00783 u
237
91
Pa
= 237.05121 u
Here the symbol Pa is for the element protactinium (Z = 91).
(a) Calculate the energy released during the alpha decay of
238
92
U
.
(b) Show that
238
92
U
can not spontaneously emit a proton.
Solution
(a) The alpha decay of
238
92
U
is given by Eq. (13.20). The energy released
in this process is given by
Q = (m
U
– m
Th
– m
He
) c
2
Substituting the atomic masses as given in the data, we find
Q = (238.05079 – 234.04363 – 4.00260)u × c
2
= (0.00456 u) c
2
= (0.00456 u) (931.5 MeV/u)
= 4.25 MeV.
(b) If
238
92
U
spontaneously emits a proton, the decay process would be
238
92
U
→
237
91
Pa
+
1
1
H
The Q for this process to happen is
= (m
U
– m
Pa
– m
H
) c
2
= (238.05079 – 237.05121 – 1.00783) u × c
2
= (– 0.00825 u) c
2
= – (0.00825 u)(931.5 MeV/u)
= – 7.68 MeV
Thus, the Q of the process is negative and therefore it cannot proceed
spontaneously. We will have to supply an energy of 7.68 MeV to a
238
92
U
nucleus to make it emit a proton.
13.6.3 Beta decay
In beta decay, a nucleus spontaneously emits an electron (
β
−−
−−
−
decay) or a
positron (
β
+
decay). A common example of
β
−−
−−
−
decay is
32 32
15 16
P S e
ν
−
→ + +
(13.22)
and that of
β
+
decay is
22 22
11 10
Na Ne e
ν
+
→ + +
(13.23)
The decays are governed by the Eqs. (13.14) and (13.15), so that one
can never predict which nucleus will undergo decay, but one can
characterize the decay by a half-life T
1/2
. For example, T
1/2
for the decays
above is respectively 14.3 d and 2.6y. The emission of electron in
β
−−
−−
−
decay
is accompanied by the emission of an antineutrino (
ν
); in
β
+
decay, instead,
a neutrino (ν) is generated. Neutrinos are neutral particles with very small
(possiblly, even zero) mass compared to electrons. They have only weak
interaction with other particles. They are, therefore, very difficult to detect,
since they can penetrate large quantity of matter (even earth) without any
interaction.
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In both
β
−−
−−
−
and
β
+
decay, the mass number A remains unchanged. In
β
−−
−−
−
decay, the atomic number Z of the nucleus goes up by 1, while in
β
+
decay Z goes down by 1. The basic nuclear process underlying
β
−−
−−
−
decay
is the conversion of neutron to proton
n → p + e
–
+
ν
(13.24)
while for
β
+
decay, it is the conversion of proton into neutron
p → n + e
+
+
ν
(13.25)
Note that while a free neutron decays to proton, the decay of proton to
neutron [Eq. (13.25)] is possible only inside the nucleus, since proton
has smaller mass than neutron.
13.6.4 Gamma decay
Like an atom, a nucleus also has discrete energy levels - the ground
state and excited states. The scale of energy is, however, very different.
Atomic energy level spacings are of the order of eV, while the difference in
nuclear energy levels is of the order of MeV. When a
nucleus in an excited state spontaneously decays
to its ground state (or to a lower energy state), a
photon is emitted with energy equal to the difference
in the two energy levels of the nucleus. This is the
so-called gamma decay. The energy (MeV)
corresponds to radiation of extremely short
wavelength, shorter than the hard X-ray region.
Typically, a gamma ray is emitted when a α or β
decay results in a daughter nucleus in an excited
state. This then returns to the ground state by a
single photon transition or successive transitions
involving more than one photon. A familiar example
is the successive emmission of gamma rays of
energies 1.17 MeV and 1.33 MeV from the
deexcitation of
60
28
Ni
nuclei formed from
β
−−
−−
−
decay
of
60
27
Co
.
13.7 NUCLEAR ENERGY
The curve of binding energy per nucleon E
bn
, given in Fig. 13.1, has
a long flat middle region between A = 30 and A = 170. In this region
the binding energy per nucleon is nearly constant (8.0 MeV). For
the lighter nuclei region, A < 30, and for the heavier nuclei region,
A > 170, the binding energy per nucleon is less than 8.0 MeV, as we
have noted earlier. Now, the greater the binding energy, the less is the
total mass of a bound system, such as a nucleus. Consequently, if nuclei
with less total binding energy transform to nuclei with greater binding
energy, there will be a net energy release. This is what happens when a
heavy nucleus decays into two or more intermediate mass fragments
(fission) or when light nuclei fuse into a havier nucleus (fusion.)
Exothermic chemical reactions underlie conventional energy sources
such as coal or petroleum. Here the energies involved are in the range of
FIGURE 13.4 β-decay of
28
60
Ni
nucleus
followed by emission of two
γ
rays
from deexcitation of the daughter
nucleus
28
60
Ni
.
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electron volts. On the other hand, in a nuclear reaction, the energy release
is of the order of MeV. Thus for the same quantity of matter, nuclear
sources produce a million times more energy than a chemical source.
Fission of 1 kg of uranium, for example, generates 10
14
J of energy;
compare it with burning of 1 kg of coal that gives 10
7
J.
13.7.1 Fission
New possibilities emerge when we go beyond natural radioactive decays
and study nuclear reactions by bombarding nuclei with other nuclear
particles such as proton, neutron, α-particle, etc.
A most important neutron-induced nuclear reaction is fission. An
example of fission is when a uranium isotope
235
92
U
bombarded with a
neutron breaks into two intermediate mass nuclear fragments
1 235 236 144 89 1
0 92 92 56 36 0
n U U Ba Kr 3 n
+ → → + +
(13.26)
The same reaction can produce other pairs of intermediate mass
fragments
1 235 236 133 99 1
0 92 92 51 41 0
n U U Sb Nb 4 n
+ → → + +
(13.27)
Or, as another example,
1 235 140 94 1
0 92 54 38 0
n U Xe Sr 2 n
+ → + +
(13.28)
The fragment products are radioactive nuclei; they emit
β
particles in
succession to achieve stable end products.
The energy released (the Q value ) in the fission reaction of nuclei like
uranium is of the order of 200 MeV per fissioning nucleus. This is
estimated as follows:
Let us take a nucleus with A = 240 breaking into two fragments each
of A = 120. Then
E
bn
for A = 240 nucleus is about 7.6 MeV,
E
bn
for the two A = 120 fragment nuclei is about 8.5 MeV.
∴ Gain in binding energy for nucleon is about 0.9 MeV.
Hence the total gain in binding energy is 240×0.9 or 216 MeV.
The disintegration energy in fission events first appears as the kinetic
energy of the fragments and neutrons. Eventually it is transferred to the
surrounding matter appearing as heat. The source of energy in nuclear
reactors, which produce electricity, is nuclear fission. The enormous
energy released in an atom bomb comes from uncontrolled nuclear fission.
We discuss some details in the next section how a nuclear reactor
functions.
13.7.2 Nuclear reactor
Notice one fact of great importance in the fission reactions given in Eqs.
(13.26) to (13.28). There is a release of extra neutron (s) in the fission
process. Averagely, 2½ neutrons are released per fission of uranium
nucleus. It is a fraction since in some fission events 2 neutrons are
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produced, in some 3, etc. The extra neutrons in turn can initiate fission
processes, producing still more neutrons, and so on. This leads to the
possibility of a chain reaction, as was first suggested by Enrico Fermi. If
the chain reaction is controlled suitably, we can get a steady energy
INDIA’S ATOMIC ENERGY PROGRAMME
The atomic energy programme in India was launched around the time of independence
under the leadership of Homi J. Bhabha (1909-1966). An early historic achievement
was the design and construction of the first nuclear reactor in India (named Apsara)
which went critical on August 4, 1956. It used enriched uranium as fuel and water as
moderator. Following this was another notable landmark: the construction of CIRUS
(Canada India Research U.S.) reactor in 1960. This 40 MW reactor used natural uranium
as fuel and heavy water as moderator. Apsara and CIRUS spurred research in a wide
range of areas of basic and applied nuclear science. An important milestone in the first
two decades of the programme was the indigenous design and construction of the
plutonium plant at Trombay, which ushered in the technology of fuel reprocessing
(separating useful fissile and fertile nuclear materials from the spent fuel of a reactor) in
India. Research reactors that have been subsequently commissioned include ZERLINA,
PURNIMA (I, II and III), DHRUVA and KAMINI. KAMINI is the country’s first large research
reactor that uses U-233 as fuel. As the name suggests, the primary objective of a research
reactor is not generation of power but to provide a facility for research on different aspects
of nuclear science and technology. Research reactors are also an excellent source for
production of a variety of radioactive isotopes that find application in diverse fields:
industry, medicine and agriculture.
The main objectives of the Indian Atomic Energy programme are to provide safe and
reliable electric power for the country’s social and economic progress and to be self-
reliant in all aspects of nuclear technology. Exploration of atomic minerals in India
undertaken since the early fifties has indicated that India has limited reserves of uranium,
but fairly abundant reserves of thorium. Accordingly, our country has adopted a three-
stage strategy of nuclear power generation. The first stage involves the use of natural
uranium as a fuel, with heavy water as moderator. The Plutonium-239 obtained from
reprocessing of the discharged fuel from the reactors then serves as a fuel for the second
stage — the fast breeder reactors. They are so called because they use fast neutrons for
sustaining the chain reaction (hence no moderator is needed) and, besides generating
power, also breed more fissile species (plutonium) than they consume. The third stage,
most significant in the long term, involves using fast breeder reactors to produce fissile
Uranium-233 from Thorium-232 and to build power reactors based on them.
India is currently well into the second stage of the programme and considerable
work has also been done on the third — the thorium utilisation — stage. The country
has mastered the complex technologies of mineral exploration and mining, fuel
fabrication, heavy water production, reactor design, construction and operation, fuel
reprocessing, etc. Pressurised Heavy Water Reactors (PHWRs) built at different sites in
the country mark the accomplishment of the first stage of the programme. India is now
more than self-sufficient in heavy water production. Elaborate safety measures both in
the design and operation of reactors, as also adhering to stringent standards of
radiological protection are the hallmark of the Indian Atomic Energy Programme.
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output. This is what happens in a nuclear reactor. If the chain reaction is
uncontrolled, it leads to explosive energy output, as in a nuclear bomb.
There is, however, a hurdle in sustaining a chain reaction, as described
here. It is known experimentally that slow neutrons (thermal neutrons)
are much more likely to cause fission in
235
92
U
than fast neutrons. Also
fast neutrons liberated in fission would escape instead of causing another
fission reaction.
The average energy of a neutron produced in fission of
235
92
U
is 2 MeV.
These neutrons unless slowed down will escape from the reactor without
interacting with the uranium nuclei, unless a very large amount of
fissionable material is used for sustaining the chain reaction. What one
needs to do is to slow down the fast neutrons by elastic scattering with
light nuclei. In fact, Chadwick’s experiments showed that in an elastic
collision with hydrogen the neutron almost comes to rest and proton
carries away the energy. This is the same situation as when a marble hits
head-on an identical marble at rest. Therefore, in reactors, light nuclei
called moderators are provided along with the fissionable nuclei for slowing
down fast neutrons. The moderators commonly used are water, heavy
water (D
2
O) and graphite. The Apsara reactor at the Bhabha Atomic
Research Centre (BARC), Mumbai, uses water as moderator. The other
Indian reactors, which are used for power production, use heavy water
as moderator.
Because of the use of moderator, it is possible that the ratio, K, of
number of fission produced by a given generation of neutrons to the
number of fission of the preceeding generation may be greater than one.
This ratio is called the multiplication factor; it is the measure of the growth
rate of the neutrons in the reactor. For K = 1, the operation of the reactor
is said to be critical, which is what we wish it to be for steady power
operation. If K becomes greater than one, the reaction rate and the reactor
power increases exponentially. Unless the factor K is brought down very
close to unity, the reactor will become supercritical and can even explode.
The explosion of the Chernobyl reactor in Ukraine in 1986 is a sad
reminder that accidents in a nuclear reactor can be catastrophic.
The reaction rate is controlled through control-rods made out of
neutron-absorbing material such as cadmium. In addition to control rods,
reactors are provided with safety rods which, when required, can be
inserted into the reactor and K can be reduced rapidly to less than unity.
The more abundant isotope
238
92
U
in naturally occurring uranium is
non-fissionable. When it captures a neutron, it produces the highly
radioactive plutonium through these reactions
238 239 239 –
92 92 93
U + n U Np + e
ν
→ → +
239 239 –
93 94
Np Pu+e
ν
→ +
(13.29)
Plutonium undergoes fission with slow neutrons.
Figure 13.5 shows the schematic diagram of a nuclear reactor based
on thermal neutron fission. The core of the reactor is the site of nuclear
Nuclear power plants in India
http://www.npcil.nic.in/main/AllProjectOperationDisplay.aspx
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Nuclei
fission. It contains the fuel elements in suitably fabricated form. The fuel
may be say enriched uranium (i.e., one that has greater abundance of
235
92
U
than naturally occurring uranium). The core contains a moderator
to slow down the neutrons. The core is surrounded by a reflector to reduce
leakage. The energy (heat) released in fission is continuously removed by
a suitable coolant. A containment vessel prevents the escape of radioactive
fission products. The whole assembly is shielded to check harmful
radiation from coming out. The reactor can be shut down by means of
rods (made of, for example, cadmium) that have high absorption of
neutrons. The coolant transfers heat to a working fluid which in turn
may produce stream. The steam drives turbines and generates electricity.
Like any power reactor, nuclear reactors generate considerable waste
products. But nuclear wastes need special care for treatment since they
are radioactive and hazardous. Elaborate safety measures, both for reactor
operation as well as handling and reprocessing the spent fuel, are
required. These safety measures are a distinguishing feature of the Indian
Atomic Energy programme. An appropriate plan is being evolved to study
the possibility of converting radioactive waste into less active and short-
lived material.
13.7.3 Nuclear fusion – energy generation in stars
When two light nuclei fuse to form a larger nucleus, energy is released,
since the larger nucleus is more tightly bound, as seen from the binding
energy curve in Fig.13.1. Some examples of such energy liberating nuclear
fusion reactions are :
1 1 2
1 1 1
H H H
+ →
+ e
+
+
ν
+ 0.42 MeV [13.29(a)]
2 2 3
1 1 2
H H He
+ →
+ n + 3.27 MeV [13.29(b)]
2 2 3 1
1 1 1 1
H H H H
+ → +
+ 4.03 MeV [13.29(c)]
A simplified online simulation of a nuclear reactor
http://esa21.kennesaw.edu/activities/nukeenergy/nuke.htm
FIGURE 13.5 Schematic diagram of a nuclear reactor based on
thermal neutron fission.
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In the first reaction, two protons combine to form a deuteron and
a positron with a release of 0.42 MeV energy. In reaction [13.29(b)], two
deuterons combine to form the light isotope of helium. In reaction
(13.29c), two deuterons combine to form a triton and a proton. For
fusion to take place, the two nuclei must come close enough so that
attractive short-range nuclear force is able to affect them. However,
since they are both positively charged particles, they experience coulomb
repulsion. They, therefore, must have enough energy to overcome this
coulomb barrier. The height of the barrier depends on the charges and
radii of the two interacting nuclei. It can be shown, for example, that
the barrier height for two protons is ~ 400 keV, and is higher for nuclei
with higher charges. We can estimate the temperature at which two
protons in a proton gas would (averagely) have enough energy to
overcome the coulomb barrier:
(3/2)k T = K
400 keV, which gives T ~ 3 × 10
9
K.
When fusion is achieved by raising the temperature of the system so
that particles have enough kinetic energy to overcome the coulomb
repulsive behaviour, it is called thermonuclear fusion.
Thermonuclear fusion is the source of energy output in the interior
of stars. The interior of the sun has a temperature of 1.5×10
7
K, which
is considerably less than the estimated temperature required for fusion
of particles of average energy. Clearly, fusion in the sun involves protons
whose energies are much above the average energy.
The fusion reaction in the sun is a multi-step process in which the
hydrogen is burned into helium. Thus, the fuel in the sun is the hydrogen
in its core. The proton-proton (p, p) cycle by which this occurs is
represented by the following sets of reactions:
1 1 2
1 1 1
H H H
+ →
+ e
+
+
ν
+ 0.42 MeV (i)
e
+
+ e
–
→
γ
+
γ
+ 1.02 MeV (ii)
2 1 3
1 1 2
H H He
+ →
+
γ
+ 5.49 MeV (iii)
+ → + +
3 3 4 1 1
2 2 2 1 1
He He He H H
+ 12.86 MeV (iv) (13.30)
For the fourth reaction to occur, the first three reactions must occur
twice, in which case two light helium nuclei unite to form ordinary helium
nucleus. If we consider the combination 2(i) + 2(ii) + 2(iii) +(iv), the net
effect is
1 4
1 2
4 H 2 He 2 6 26.7 MeV
e
ν γ
−
+ → + + +
or
1 4
1 2
(4 H 4 ) ( He 2 ) 2 6 26.7 MeV
e e
ν γ
− −
+ → + + + +
(13.31)
Thus, four hydrogen atoms combine to form an
4
2
He
atom with a
release of 26.7 MeV of energy.
Helium is not the only element that can be synthesized in the interior
of a star. As the hydrogen in the core gets depleted and becomes helium,
the core starts to cool. The star begins to collapse under its own gravity
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which increases the temperature of the core. If this temperature increases
to about 10
8
K, fusion takes place again, this time of helium nuclei into
carbon. This kind of process can generate through fusion higher and
higher mass number elements. But elements more massive than those
near the peak of the binding energy curve in Fig. 13.1 cannot be so
produced.
The age of the sun is about 5×10
9
y and it is estimated that there is
enough hydrogen in the sun to keep it going for another 5 billion years.
After that, the hydrogen burning will stop and the sun will begin to cool
and will start to collapse under gravity, which will raise the core
temperature. The outer envelope of the sun will expand, turning it into
the so called red giant.
NUCLEAR HOLOCAUST
In a single uranium fission about 0.9×235 MeV (≈200 MeV) of energy is liberated. If
each nucleus of about 50 kg of
235
U undergoes fission the amount of energy involved is
about 4 × 10
15
J. This energy is equivalent to about 20,000 tons of TNT, enough for a
superexplosion. Uncontrolled release of large nuclear energy is called an atomic explosion.
On August 6, 1945 an atomic device was used in warfare for the first time. The US
dropped an atom bomb on Hiroshima, Japan. The explosion was equivalent to 20,000
tons of TNT. Instantly the radioactive products devastated 10 sq km of the city which
had 3,43,000 inhabitants. Of this number 66,000 were killed and 69,000 were injured;
more than 67% of the city’s structures were destroyed.
High temperature conditions for fusion reactions can be created by exploding a fission
bomb. Super-explosions equivalent to 10 megatons of explosive power of TNT were tested
in 1954. Such bombs which involve fusion of isotopes of hydrogen, deuterium and tritium
are called hydrogen bombs. It is estimated that a nuclear arsenal sufficient to destroy
every form of life on this planet several times over is in position to be triggered by the
press of a button. Such a nuclear holocaust will not only destroy the life that exists now
but its radioactive fallout will make this planet unfit for life for all times. Scenarios based
on theoretical calculations predict a long nuclear winter, as the radioactive waste will
hang like a cloud in the earth’s atmosphere and will absorb the sun’s radiation.
13.7.4 Controlled thermonuclear fusion
The natural thermonuclear fusion process in a star is replicated in a
thermonuclear fusion device. In controlled fusion reactors, the aim is to
generate steady power by heating the nuclear fuel to a temperature in the
range of 10
8
K. At these temperatures, the fuel is a mixture of positive
ions and electrons (plasma). The challenge is to confine this plasma, since
no container can stand such a high temperature. Several countries
around the world including India are developing techniques in this
connection. If successful, fusion reactors will hopefully supply almost
unlimited power to humanity.
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EXAMPLE 13.7
Example 13.7 Answer the following questions:
(a) Are the equations of nuclear reactions (such as those given in
Section 13.7) ‘balanced’ in the sense a chemical equation (e.g.,
2H
2
+ O
2
→ 2 H
2
O) is? If not, in what sense are they balanced on
both sides?
(b) If both the number of protons and the number of neutrons are
conserved in each nuclear reaction, in what way is mass converted
into energy (or vice-versa) in a nuclear reaction?
(c) A general impression exists that mass-energy interconversion
takes place only in nuclear reaction and never in chemical
reaction. This is strictly speaking, incorrect. Explain.
Solution
(a) A chemical equation is balanced in the sense that the number of
atoms of each element is the same on both sides of the equation.
A chemical reaction merely alters the original combinations of
atoms. In a nuclear reaction, elements may be transmuted. Thus,
the number of atoms of each element is not necessarily conserved
in a nuclear reaction. However, the number of protons and the
number of neutrons are both separately conserved in a nuclear
reaction. [Actually, even this is not strictly true in the realm of
very high energies – what is strictly conserved is the total charge
and total ‘baryon number’. We need not pursue this matter here.]
In nuclear reactions (e.g., Eq. 13.26), the number of protons and
the number of neutrons are the same on the two sides of the equation.
(b) We know that the binding energy of a nucleus gives a negative
contribution to the mass of the nucleus (mass defect). Now, since
proton number and neutron number are conserved in a nuclear
reaction, the total rest mass of neutrons and protons is the same
on either side of a reaction. But the total binding energy of nuclei
on the left side need not be the same as that on the right hand
side. The difference in these binding energies appears as energy
released or absorbed in a nuclear reaction. Since binding energy
contributes to mass, we say that the difference in the total mass
of nuclei on the two sides get converted into energy or vice-versa.
It is in these sense that a nuclear reaction is an example of mass-
energy interconversion.
(c) From the point of view of mass-energy interconversion, a chemical
reaction is similar to a nuclear reaction in principle. The energy
released or absorbed in a chemical reaction can be traced to the
difference in chemical (not nuclear) binding energies of atoms and
molecules on the two sides of a reaction. Since, strictly speaking,
chemical binding energy also gives a negative contribution (mass
defect) to the total mass of an atom or molecule, we can equally
well say that the difference in the total mass of atoms or molecules,
on the two sides of the chemical reaction gets converted into energy
or vice-versa. However, the mass defects involved in a chemical
reaction are almost a million times smaller than those in a nuclear
reaction.This is the reason for the general impression, (which is
incorrect) that mass-energy interconversion does not take place
in a chemical reaction.
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SUMMARY
1. An atom has a nucleus. The nucleus is positively charged. The radius
of the nucleus is smaller than the radius of an atom by a factor of
10
4
. More than 99.9% mass of the atom is concentrated in the nucleus.
2. On the atomic scale, mass is measured in atomic mass units (u). By
definition, 1 atomic mass unit (1u) is 1/12
th
mass of one atom of
12
C;
1u = 1.660563 × 10
–27
kg.
3. A nucleus contains a neutral particle called neutron. Its mass is almost
the same as that of proton
4. The atomic number Z is the number of protons in the atomic nucleus
of an element. The mass number A is the total number of protons and
neutrons in the atomic nucleus; A = Z+N; Here N denotes the number
of neutrons in the nucleus.
A nuclear species or a nuclide is represented as
X
A
Z
, where X is the
chemical symbol of the species.
Nuclides with the same atomic number Z, but different neutron number
N are called isotopes. Nuclides with the same A are isobars and those
with the same N are isotones.
Most elements are mixtures of two or more isotopes. The atomic mass
of an element is a weighted average of the masses of its isotopes and
calculated in accordance to the relative abundances of the isotopes.
5. A nucleus can be considered to be spherical in shape and assigned a
radius. Electron scattering experiments allow determination of the
nuclear radius; it is found that radii of nuclei fit the formula
R = R
0
A
1/3
,
where R
0
= a constant = 1.2 fm. This implies that the nuclear density
is independent of A. It is of the order of 10
17
kg/m
3
.
6. Neutrons and protons are bound in a nucleus by the short-range strong
nuclear force. The nuclear force does not distinguish between neutron
and proton.
7. The nuclear mass M is always less than the total mass, Σm, of its
constituents. The difference in mass of a nucleus and its constituents
is called the mass defect,
∆M = (Z m
p
+ (A – Z)m
n
) – M
Using Einstein’s mass energy relation, we express this mass difference
in terms of energy as
∆E
b
= ∆M c
2
The energy ∆E
b
represents the binding energy of the nucleus. In the
mass number range A = 30 to 170, the binding energy per nucleon is
nearly constant, about 8 MeV/nucleon.
8. Energies associated with nuclear processes are about a million times
larger than chemical process.
9. The Q-value of a nuclear process is
Q = final kinetic energy – initial kinetic energy.
Due to conservation of mass-energy, this is also,
Q = (sum of initial masses – sum of final masses)c
2
10. Radioactivity is the phenomenon in which nuclei of a given species
transform by giving out α or
β
or
γ
rays;
α
-rays are helium nuclei;
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β
-rays are electrons.
γ
-rays are electromagnetic radiation of wavelengths
shorter than X-rays;
11. Law of radioactive decay : N (t) = N(0) e
–
λ
t
where
λ
is the decay constant or disintegration constant.
The half-life T
1/2
of a radionuclide is the time in which N has been
reduced to one-half of its initial value. The mean life
τ
is the time at
which N has been reduced to e
–1
of its initial value
1/2
ln 2
ln 2
T
τ
λ
= =
12. Energy is released when less tightly bound nuclei are transmuted into
more tightly bound nuclei. In fission, a heavy nucleus like
235
92
U
breaks
into two smaller fragments, e.g.,
235 1 133 99 1
92 0 51 41 0
U+ n Sb Nb + 4 n
→ +
13. The fact that more neutrons are produced in fission than are consumed
gives the possibility of a chain reaction with each neutron that is
produced triggering another fission. The chain reaction is uncontrolled
and rapid in a nuclear bomb explosion. It is controlled and steady in
a nuclear reactor. In a reactor, the value of the neutron multiplication
factor k is maintained at 1.
14. In fusion, lighter nuclei combine to form a larger nucleus. Fusion of
hydrogen nuclei into helium nuclei is the source of energy of all stars
including our sun.
Physical Quantity Symbol Dimensions Units Remarks
Atomic mass unit [M] u Unit of mass for
expressing atomic or
nuclear masses. One
atomic mass unit equals
1/12
th
of the mass of
12
C
atom.
Disintegration or
λ
[T
–1
] s
–1
decay constant
Half-life T
1/2
[T] s Time taken for the decay
of one-half of the initial
number of nuclei present
in a radioactive sample.
Mean life τ [T] s Time at which number of
nuclei has been reduced to
e
–1
of its initial value
Activity of a radio- R [T
–1
] Bq Measure of the activity
active sample of a radioactive source.
POINTS TO PONDER
1. The density of nuclear matter is independent of the size of the nucleus.
The mass density of the atom does not follow this rule.
2. The radius of a nucleus determined by electron scattering is found to
be slightly different from that determined by alpha-particle scattering.
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This is because electron scattering senses the charge distribution of
the nucleus, whereas alpha and similar particles sense the nuclear
matter.
3. After Einstein showed the equivalence of mass and energy, E = mc
2
,
we cannot any longer speak of separate laws of conservation of mass
and conservation of energy, but we have to speak of a unified law of
conservation of mass and energy. The most convincing evidence that
this principle operates in nature comes from nuclear physics. It is
central to our understanding of nuclear energy and harnessing it as a
source of power. Using the principle, Q of a nuclear process (decay or
reaction) can be expressed also in terms of initial and final masses.
4. The nature of the binding energy (per nucleon) curve shows that
exothermic nuclear reactions are possible, when two light nuclei fuse
or when a heavy nucleus undergoes fission into nuclei with intermediate
mass.
5. For fusion, the light nuclei must have sufficient initial energy to
overcome the coulomb potential barrier. That is why fusion requires
very high temperatures.
6. Although the binding energy (per nucleon) curve is smooth and slowly
varying, it shows peaks at nuclides like
4
He,
16
O etc. This is considered
as evidence of atom-like shell structure in nuclei.
7. Electr
ons and positron ar
e a particle-antiparticle pair. They are
identical in mass; their charges are equal in magnitude and opposite.
(It is found that when an electron and a positron come together, they
annihilate each other giving energy in the form of gamma-ray photons.)
8. In â
-
-decay (electron emission), the particle emitted along with electron
is anti-neutrino (
ν
). On the other hand, the particle emitted in
β
+
-
decay (positron emission) is neutrino (ν). Neutrino and anti-neutrino
are a particle-antiparticle pair. There are anti particles associated
with every particle. What should be antiproton which is the anti
particle of the proton?
9. A free neutron is unstable (
–
n p e
ν
→ + +
). But a similar free proton
decay is not possible, since a proton is (slightly) lighter than a neutron.
10. Gamma emission usually follows alpha or beta emission. A nucleus
in an excited (higher) state goes to a lower state by emitting a gamma
photon. A nucleus may be left in an excited state after alpha or beta
emission. Successive emission of gamma rays from the same nucleus
(as in case of
60
Ni, Fig. 13.4) is a clear proof that nuclei also have
discrete energy levels as do the atoms.
11. Radioactivity is an indication of the instability of nuclei. Stability
requires the ratio of neutron to proton to be around 1:1 for light
nuclei. This ratio increases to about 3:2 for heavy nuclei. (More
neutrons are required to overcome the effect of repulsion among the
protons.) Nuclei which are away from the stability ratio, i.e., nuclei
which have an excess of neutrons or protons are unstable. In fact,
only about 10% of knon isotopes (of all elements), are stable. Others
have been either artificially produced in the laboratory by bombarding
α
, p, d, n or other particles on targets of stable nuclear species or
identified in astronomical observations of matter in the universe.
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EXERCISES
You may find the following data useful in solving the exercises:
e = 1.6×10
–19
C N = 6.023×10
23
per mole
1/(4πε
0
) = 9 × 10
9
N m
2
/C
2
k = 1.381×10
–23
J K
–1
1 MeV = 1.6×10
–13
J 1 u = 931.5 MeV/c
2
1 year = 3.154×10
7
s
m
H
= 1.007825 u m
n
= 1.008665 u
m(
4
2
He
) = 4.002603 u m
e
= 0.000548 u
13.1 (a) Two stable isotopes of lithium
6
3
Li
and
7
3
Li
have respective
abundances of 7.5% and 92.5%. These isotopes have masses
6.01512 u and 7.01600 u, respectively. Find the atomic mass
of lithium.
(b) Boron has two stable isotopes,
10
5
B
and
11
5
B
. Their respective
masses are 10.01294 u and 11.00931 u, and the atomic mass of
boron is 10.811 u. Find the abundances of
10
5
B
and
11
5
B
.
13.2 The three stable isotopes of neon:
20 21 22
10 10 10
Ne, Ne and Ne
have
respective abundances of 90.51%, 0.27% and 9.22%. The atomic
masses of the three isotopes are 19.99 u, 20.99 u and 21.99 u,
respectively. Obtain the average atomic mass of neon.
13.3 Obtain the binding energy (in MeV) of a nitrogen nucleus
(
)
14
7
N
,
given m
(
)
14
7
N
=14.00307 u
13.4 Obtain the binding energy of the nuclei
56
26
Fe
and
209
83
Bi
in units of
MeV from the following data:
m (
56
26
Fe
) = 55.934939 u m (
209
83
Bi
) = 208.980388 u
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
63
29
Cu
atoms (of mass 62.92960 u).
13.6 Write nuclear reaction equations for
(i)
α
-decay of
226
88
Ra
(ii)
α
-decay of
242
94
Pu
(iii)
β
–
-decay of
32
15
P
(iv)
β
–
-decay of
210
83
Bi
(v)
β
+
-decay of
11
6
C
(vi)
β
+
-decay of
97
43
Tc
(vii) Electron capture of
120
54
Xe
13.7 A radioactive isotope has a half-life of T years. How long will it take
the activity to reduce to a) 3.125%, b) 1% of its original value?
13.8 The normal activity of living carbon-containing matter is found to
be about 15 decays per minute for every gram of carbon. This activity
arises from the small proportion of radioactive
14
6
C
present with the
stable carbon isotope
12
6
C
. When the organism is dead, its interaction
with the atmosphere (which maintains the above equilibrium activity)
ceases and its activity begins to drop. From the known half-life (5730
years) of
14
6
C
, and the measured activity, the age of the specimen
can be approximately estimated. This is the principle of
14
6
C
dating
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used in archaeology. Suppose a specimen from Mohenjodaro gives
an activity of 9 decays per minute per gram of carbon. Estimate the
approximate age of the Indus-Valley civilisation.
13.9 Obtain the amount of
60
27
Co
necessary to provide a radioactive source
of 8.0 mCi strength. The half-life of
60
27
Co
is 5.3 years.
13.10 The half-life of
90
38
Sr
is 28 years. What is the disintegration rate of
15 mg of this isotope?
13.11 Obtain approximately the ratio of the nuclear radii of the gold isotope
197
79
Au
and the silver isotope
107
47
Ag
.
13.12 Find the Q-value and the kinetic energy of the emitted
α
-particle in
the
α
-decay of (a)
226
88
Ra
and (b)
220
86
Rn
.
Given m (
226
88
Ra
) = 226.02540 u, m (
222
86
Rn
) = 222.01750 u,
m (
220
86
Rn
) = 220.01137 u, m (
216
84
Po
) = 216.00189 u.
13.13 The radionuclide
11
C decays according to
11 11 +
6 5 1/2
C B + + : =20.3 min
e T
ν
→
The maximum energy of the emitted positron is 0.960 MeV.
Given the mass values:
m (
11
6
C
) = 11.011434 u and m (
11
6
B
) = 11.009305 u,
calculate Q and compare it with the maximum energy of the positron
emitted.
13.14 The nucleus
23
10
Ne
decays by
β
–
emission. Write down the
β
-decay
equation and determine the maximum kinetic energy of the
electrons emitted. Given that:
m (
23
10
Ne
) = 22.994466 u
m (
23
11
Na
) = 22.989770 u.
13.15 The Q value of a nuclear reaction A + b → C + d is defined by
Q = [ m
A
+ m
b
– m
C
– m
d
]c
2
where the masses refer to the respective nuclei. Determine from the
given data the Q-value of the following reactions and state whether
the reactions are exothermic or endothermic.
(i)
1 3 2 2
1 1 1 1
H+ H H+ H
→
(ii)
12 12 20 4
6 6 10 2
C+ C Ne+ He
→
Atomic masses are given to be
m (
2
1
H
) = 2.014102 u
m (
3
1
H
) = 3.016049 u
m (
12
6
C
) = 12.000000 u
m (
20
10
Ne
) = 19.992439 u
13.16 Suppose, we think of fission of a
56
26
Fe
nucleus into two equal
fragments,
28
13
Al
. Is the fission energetically possible? Argue by
working out Q of the process. Given m (
56
26
Fe
) = 55.93494 u and
m (
28
13
Al
) = 27.98191 u.
13.17 The fission properties of
239
94
Pu
are very similar to those of
235
92
U
. The
average energy released per fission is 180 MeV. How much energy,
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in MeV, is released if all the atoms in 1 kg of pure
239
94
Pu
undergo
fission?
13.18 A 1000 MW fission reactor consumes half of its fuel in 5.00 y. How
much
235
92
U
did it contain initially? Assume that the reactor operates
80% of the time, that all the energy generated arises from the fission
of
235
92
U
and that this nuclide is consumed only by the fission process.
13.19 How long can an electric lamp of 100W be kept glowing by fusion of
2.0 kg of deuterium? Take the fusion reaction as
2 2 3
1 1 2
H+ H He+n +3.27 MeV
→
13.20 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.)
13.21 From the relation R = R
0
A
1/3
, where R
0
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).
13.22 For the
β
+
(positron) emission from a nucleus, there is another
competing process known as electron capture (electron from an inner
orbit, say, the K–shell, is captured by the nucleus and a neutrino is
emitted).
1
A A
Z Z
e X Y
ν
+
−
+ → +
Show that if
β
+
emission is energetically allowed, electron capture
is necessarily allowed but not vice–versa.
ADDITIONAL EXERCISES
13.23 In a periodic table the average atomic mass of magnesium is given
as 24.312 u. The average value is based on their relative natural
abundance on earth. The three isotopes and their masses are
24
12
Mg
(23.98504u),
25
12
Mg
(24.98584u) and
26
12
Mg
(25.98259u). The natural
abundance of
24
12
Mg
is 78.99% by mass. Calculate the abundances
of other two isotopes.
13.24 The neutron separation energy is defined as the energy required to
remove a neutron from the nucleus. Obtain the neutron separation
energies of the nuclei
41
20
Ca
and
27
13
Al
from the following data:
m(
40
20
Ca
) = 39.962591 u
m(
41
20
Ca
) = 40.962278 u
m(
26
13
Al
) = 25.986895 u
m(
27
13
Al
) = 26.981541 u
13.25 A source contains two phosphorous radio nuclides
32
15
P
(T
1/2
= 14.3d)
and
33
15
P
(T
1/2
= 25.3d). Initially, 10% of the decays come from
33
15
P
.
How long one must wait until 90% do so?
13.26 Under certain circumstances, a nucleus can decay by emitting a
particle more massive than an
α
-particle. Consider the following
decay processes:
223 209 14
88 82 6
Ra Pb C
→ +
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223 219 4
88 86 2
Ra Rn He
→ +
Calculate the Q-values for these decays and determine that both
are energetically allowed.
13.27 Consider the fission of
238
92
U
by fast neutrons. In one fission event,
no neutrons are emitted and the final end products, after the beta
decay of the primary fragments, are
140
58
Ce
and
99
44
Ru
. Calculate Q
for this fission process. The relevant atomic and particle masses
are
m(
238
92
U
) =238.05079 u
m(
140
58
Ce
) =139.90543 u
m(
99
44
Ru
) = 98.90594 u
13.28 Consider the D–T reaction (deuterium–tritium fusion)
2 3 4
1 1 2
H H He n
+ → +
(a) Calculate the energy released in MeV in this reaction from the
data:
m(
2
1
H
)=2.014102 u
m(
3
1
H
) =3.016049 u
(b) Consider the radius of both deuterium and tritium to be
approximately 2.0 fm. What is the kinetic energy needed to
overcome the coulomb repulsion between the two nuclei? To what
temperature must the gas be heated to initiate the reaction?
(Hint: Kinetic energy required for one fusion event =average
thermal kinetic energy available with the interacting particles
= 2(3kT/2); k = Boltzman’s constant, T = absolute temperature.)
13.29 Obtain the maximum kinetic energy of
β
-particles, and the radiation
frequencies of
γ
decays in the decay scheme shown in Fig. 13.6. You
are given that
m(
198
Au) = 197.968233 u
m(
198
Hg) =197.966760 u
FIGURE13.6
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13.30 Calculate and compare the energy released by a) fusion of 1.0 kg of
hydrogen deep within Sun and b) the fission of 1.0 kg of
235
U in a
fission reactor.
13.31 Suppose India had a target of producing by 2020 AD, 200,000 MW
of electric power, ten percent of which was to be obtained from nuclear
power plants. Suppose we are given that, on an average, the efficiency
of utilization (i.e. conversion to electric energy) of thermal energy
produced in a reactor was 25%. How much amount of fissionable
uranium would our country need per year by 2020? Take the heat
energy per fission of
235
U to be about 200MeV.
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