Hello everyone and welcome back to this MOOCs
course titled fundamentals of nuclear
power generation. Now, today we are in to
the second lecture of module 1, where we are
discussing about the very fundamentals of
this course. We are just preparing the
backdrop to the course through some kind of
informal discussions.
I hope you enjoyed the previous lecture, where
we had just to general talk about the
introduction to the topic of nuclear power.
We discussed about the global and national
energy scenarios through some projections
and stats and thereby started tried to establish
the importance of nuclear power in context
of the present day energy demand and also
considering the depleting reserve of fossil
fuels and then, we also discussed a bit about
the historical development of a nuclear power.
Today we shall be trying to enter a bit to
the technical details and there by complete
this
particular module, so that we can move forward
to more inverted chapters. Just a brief
summary of whatever we did in the previous
lecture.
We had a discussion on the global and national
energy scenarios and we briefly
discussed about the historical development
by mentioning the works by some of the
eminent scientists and then we discussed the
Rutherford-Bohr model of atoms, which is
quite similar to the solar system having a
heavy nucleus at the center and tiny electrons
moving around that in the orbits. Now the,
well there are possibility of having several
kinds of sub atomic particles at the nucleus,
but for the purpose of this course we are
considering only two kinds of particles in
the nucleus, one is the proton which is
positively charged and other is neutron which
is having more or less the similar mass of
proton, but it is electrically neutral.
The electrons are having the same amount of
charge as the proton, but it is of opposite
sense and its mass is extremely small compared
to that mass of proton and neutron. And
we also discussed the very important concept
of isotopes. So, from that, we know that it
is possible to have for any natural element
to have different kinds of nucleus, where
the
atomic number remains the same; that is the
total number of protons inside the nucleus
is
the same, but their mass number varies because
they may be having different number of
neutrons and as the atomic number that is
the number of protons reside the chemical
properties, so different isotopes of the same
element, more or less have the same kind of
chemical nature , but the nuclear properties
depends strongly on the number of neutrons
and the basically on the mass number.
So, different isotopes of the same element
may have different kind of reactions or
response to nuclear reactions. Like, for example,
for Uranium which is probably the
element that you are going to discuss the
most in this course has 3 natural elements
or 3
natural isotopes I should say. Out of them
Uranium 235 give is the one, which is the
most common nuclear fuel that is used worldwide.
It is strongly reactive to the nuclear
reactions. So, it has strong nuclear behavior,
but the two other isotopes U-234 and U-
238, hardly has any kind of a nuclear related
behavior, we shall discuss in more detail
on
that.
Now, the question with which we have to start
today discussion is how such heavy
amount of mass can stay in a small volume
inside the nucleus. From our previous
discussion we know that, inside an atom the
most of the space is actually void because
when the radius of an atom can be of the order
of 10 to the power minus 9 to minus 11
meter that of a nucleus is typically of the
order of 10 to the power minus 16 meter and
that entire amount of void space is occupied
only by tiny electrons.
So, we can think about an atom, where almost
all the mass are basically lumped into the
nucleus in an extremely small volume, but
from electrostatics we know that opposite
charges attract each other, whereas like charges
repel each other and amount of or
magnitude of such attraction or repulsion
force can be given by the Coulombs law of
electrostatic attraction or electrostatic
repulsion. Just this one, where q1 and q2
are the
charges of the two particles that we are talking
about and r refers to the distance between
these two particles and k is the coulombs
constant. When you are talking about the
repulsive force between two protons q1, q2
both of them are equal and the magnitude is
1.602 into 10 to the power of minus 19 coulomb,
as I had mentioned in the previous
lecture.
Now, when you talk about any nucleus, particularly
those of heavier kind, there are
several protons present inside the nucleus
and all of them are facing this repulsive
force
because of the presence of other neighboring
protons. Neutrons are electrically neutral
and therefore, they have electrostatic attraction
towards the proton. So, it can be
understood that, neutrons stay in contact
with the protons.
Electrons are oppositely charged to protons,
but they are orbiting around the nucleus and
therefore corresponding centrifugal force
repulse their attraction towards the nucleus
and
that allows a electrons to stay in the orbits
and not jump back to the nucleus, but it is
very difficult to understand how protons can
stay there. Just think about Uranium,
Uranium is the heaviest naturally occurring
nucleus and it has 92 protons inside its
nucleus.
Now, therefore, each of those protons are
facing strong repulsion from all the
neighboring 91 protons and that can have huge
magnitude, particularly when you
consider this r that we are talking about
here, is of the order of 10 to the power minus
16
meters. So, the magnitude of the corresponding
repulsive force can be huge. That means,
there has to be some kind of attractive force,
which keeps the nucleus together, keeps all
those nucleons together inside very, very
small volume.
Then, let us check what can be the possible
forces that can be present inside the nucleus.
Gravity we can definitely be there, whenever
you are talking about mass, gravitational
attraction is there. Now, to calculate the
magnitude of gravity we need to take the help
from Newtons laws of gravitation, which is
given by this particular form. Here m one
and m two are the mass of the two particles
that you are talking about and r is the same
the distance between this two particles.
Now, when we put this particular relation
for protons and also compare that with the
amount of repulsive force calculated using
the Coulombs relation, then it can be seen
that
while for any given value of r, while the
magnitude of the repulsive force is of the
order
of 10 to the power of minus 29, the magnitude
of the corresponding gravitation attraction
can be more than 10 to the power 30 times
lower. So, gravity definitely is negligible,
or
the presence of gravitational attraction can
safely be neglected, whenever we are talking
about the nucleus and associated calculations.
Electrostatic repulsion, we are talking about
that one only. Then, there has to have some
kind of nuclear force, which opposes this
electrostatic repulsion and keeps the nucleons
together. That is expected to be a short-range
force, while gravitational and electrostatic
forces are long range that means, they act
over longer distance, this expected nuclear
force has to be a short range one. So, that
its influence is limited, only inside the
nucleus
and it must be strong enough to oppose the
electrostatic repulsion and that is where
the
concept of binding energy comes into the picture.
Just to repeat what I have mentioned, in a
very small volume of the nucleus we have
several strongly charged or positively charged
protons clubbed together and therefore,
they are expecting strong repulsive force.
Still despite such kind of opposing effect,
the existence of the nucleus hints towards
the
presence of a short-range force which is able
to overcome this electrostatic repulsion and
keep the nucleons together. This particular
force is often called the binding force and
the
energy associated with this is called the
binding energy. I repeat, binding force is
a force
which keeps the nucleons together by opposing
the repulsive nature of the proton to
proton interaction and the energy associated
with this binding force is called the binding
energy.
Alternatively, it can be also be viewed as
the amount of energy required to break a
nucleus into it is constituents. Like, when
we have a nucleus and we want to break it,
we
have supply at least the binding energy so
that it is able to break the bonds between
the
nucleons and break apart into the nucleons
or separate sub atomic particles. So, the
concept of binding energy is probably the
most important one to understand the origin
of
the nuclear energy.
Now, the nature of the binding energy need
to be discussed from the point of view of
physics, there are several theories existing
in nuclear physics, some of them talk about
the exchange of quarks between protons and
neutrons, some of them talk about the role
of fermions and bosons, but here in this course
as you are looking more from engineering
point of view, we are not going to enter any
kind of details about the nature of binding
energy, rather we are just going to take it
as the amount of energy associated with the
formation of nucleus or the energy that is
essential to give the nucleons together in
a very
small volume.
Now, by providing binding energy or higher
amount of energy, if we are able break a
nucleus into it is constituents, the most
interesting thing to observe is that the combined
mass of all these components actually exceeds
the mass of the original nucleus itself and
that difference is called mass defect. It
is really fascinating to think about that,
the
nucleus which are forming a nucleus their
combined mass is greater than the nucleus
itself. This mass defect is actually the source
of the nuclear energy.
Let us consider an example, let us see we
have an atom which is having an atomic
number of Z and mass number of A. Therefore,
atomic number, represent the number of
protons, so this particular atom consists
of Z number of protons and also Z number of
electrons to keep it electrically neutral
and also it is having A minus Z number of
neutrons, let us say capital N represent the
number of neutrons.
So, mass defect can be represented as the
difference in mass of the constituents, which
is
this particular one and the mass of the original
nucleus. Here, the mass of the electron,
that means, this mass is generally, extremely
negligible compared to the mass of proton
and neutron. If you refer to the value that
I presented in the previous lecture, the mass
of
electron is at least 10 to the power four
times lower than that of protons and neutrons
and
therefore, while calculating this mass defect
and associated quantities, the mass of
electron is quite often neglected.
Accordingly, by putting the mass of proton
and neutron, we can get this particular
mathematical relation for mass defect, in
terms of amu's, where we have delta m equal
to
1.007825, which is the mass of proton into
the number of protons plus 1.008665 which
is
a mass of neutrons into the number of neutrons
minus the original mass of the nucleus
itself, which can generally be obtained from
mass spectrography or some associated
techniques.
This particular energy associated with such
mass defect is called the binding energy,
which I mentioned earlier. You can think about
just going back to the previous slides,
whenever we are coupling some of the sub atomic
particles that is some of the nucleons
together to form a nucleus, then it seems
that some of the amount of mass is lost because
the mass of the new fully formed nucleus is
less than the combined mass of it is nucleons
and if we can convert this missing amount
of mass to equivalent amount of energy, then
we can think that energy as some kind of glue
which is binding the nucleons together and
that is what we refer as the binding energy.
Following the Einstein's relation, this binding
energy hence can be represented as at the
product of the mass defect and c square where
c is the velocity of light in vacuum and it
is as per present standard is magnitude is
2.9979 into 10 to be power 8 meter per second.
So, delta m is the mass defect and once we
multiply that with c square, then we can get
the equivalent amount of energy. Therefore,
binding energy if we use delta m equal to
1
amu and put the numbers in the earlier relation,
then we get the binding energy
corresponding to a mass defect of 1 amu is
equal to 931 MeV or we can safely say that
1
amu of mass or 1 amu of mass defect is equivalent
to 931 MeV amount of energy, that
means, whenever a few nucleons have combined
together to form a nucleus, if the
difference in the mass between the constituents
and the final product is equal to 1 amu,
then that particular nuclear reaction will
be associated with an energy releases of 931
MeV.
Let us take couple of examples; first we have
the example of n-helium isotope, the
common helium isotope 2 He 4. So, from the
symbols we can say that it is having an
atomic number of 2 and mass number of 4. That
means, this nucleus is having 2 number
of protons and 2 number of neutrons in it
is nucleus. Now, the mass of this one is
4.00277 amu, which is a measured value and
then from here or from this particular
value, if we compare this with the mass of
2 protons and 2 neutrons kept separately,
then
that will be associated with the mass defect
of 0.03021 amu and a binding energy of
28.1255 MeV.
That means, if we want to break apart 1 helium
isotope; a common helium isotope, then
we have to supply at least 28.1255 MeV of
energy and then only we shall be able to
separate out these 4 nucleons. Let us compare
or study the situation of another typical
nuclear reaction; we do not need to understand,
how this reaction is happening. Just see,
what we have here, we have the common hydrogen
isotope 1 H 1. You know, that the
common hydrogen nucleus consists of only a
single proton and therefore, its mass
number and atomic number they are same.
So, this common hydrogen nucleus is absorbing
1 neutron leading to the formation of 1
deuterium. This deuterium is having 1 proton
and now 1 neutron, therefore, its atomic
number remains one, but mass number is now
2. If we put the measured mass of
hydrogen and deuterium here and also we put
the information about the mass of neutron,
then this particular nuclear reaction you
associated with the mass defect of 0.00239
amu
and a binding energy of amount 2.2251 MeV.
So, if we want to break this deuterium, we
need to supply this 2.2251 MeV amount of
energy. Of course, this number is smaller
compare to the example of helium above, but
still we need to supply this amount energy.
This way, we can calculate the binding
energy associated with the formation of any
particular nucleus, just by measuring the
mass of that particular nucleus and using
the information about the concerned number
of
protons and neutrons.
As the number of nucleus keeps on increasing
inside a nucleus, the binding energy for
that nucleus also keeps on growing, mass defect
corresponding to the formation of that
particular nucleus keeps on increasing and
therefore, it becomes very difficult to
compare lighter and heavier isotopes. Like
the binding energy associated with helium
is
written here on this slide. We cannot compare
that with the binding energy associated
with Uranium 235 because, while helium is
having only 4 nucleons in it is nucleus,
Uranium is having to 235 nucleons and therefore,
that binding energy of that Uranium
235 has to be much higher, compare to this
helium, but another measure is found to be
extremely important and extremely useful in
this context, that is binding energy per
nucleon.
We can calculate binding energy per nucleon
by dividing the binding magnitude with the
number of nucleons presents in that particular
nucleus. Like in case of helium, we are
dividing this 28.1255 by 4 and so we are getting
a binding energy per nucleon as 7.0314
MeV, whereas in case of deuterium, we are
dividing this magnitude by 2 because there
are only 2 nucleons and therefore, we are
getting the value as 1.1126. I am having a
mistake on this slide, please ignore that
or correct that, the symbol for MeV should
have
V in capital. So, correct symbol is M small
e V, this is the correct one, this 2 are not
correct, V refers to the volt and that has
to be in capital.
But this binding energy per nucleon is extremely
important from the stability of nucleus
point of view; it has been observed that higher
the value of binding energy per nucleon
more stable is the nucleus. That means, to
break apart even a single nucleon from that
nucleus, we have to supply even higher amount
of energy and this way the binding
energy per nucleon value has been measured
by scientist for different kinds of nucleus
and we get this particular graph which is
very common, you will get this one in any
text
books, more or less the similar version of
this one and also you can get this one over
internet, where we have the mass number on
the horizontal axis and the binding energy
per nucleon on a vertical axis.
Now, if we start with the lightest possible
element that is hydrogen having a mass
number of 1. Can you guess, what is the binding
energy per nucleon for a common
hydrogen atom or hydrogen nucleus? How many
nucleons are there? For a hydrogen
there is only a single nucleon, that is an
hydrogen nucleus consist of only a single
proton.
So, as there is no other proton, it does not
face a kind of repulsive force and therefore,
it
does need any kind of binding energy. Therefore,
binding energy per nucleon for
common hydrogen is equal to 0, but this value
of binding energy per nucleon keeps on
increasing as we increase the number of nucleons
or the mass number increases and this
keeps on growing and reaches a maxima around
this zone of mass number of 56, 58, 60
and then it keeps on coming down steadily.
So, the largest value of binding energy per
nucleon corresponds to a mass number of
around 60. It has been found that among the
naturally occurring isotopes, Nickel 62 is
the one, which is having the highest value
of binding energy per nucleon 8.7945 MeV,
2
isotopes of Irons are also not far behind
Iron 58 and Iron 56, their values are also
very,
very close. So, this, particular group of
isotopes, which generally corresponds to iron,
nickel, cobalt etcetera, which are having
mass number around 60, are characterized by
the highest possible binding energy per nucleon
and therefore, they are the most stable
nucleus, that we can have form nuclei reaction
point of view.
The nucleus which are having mass number far
away from this particular zone, they
generally have binding energy per nucleon
value much lower than this highest value of
approximately 8.8 and hence they are more
unstable. The far we move away from this
particular zone, we get more unstable nucleus
like, when we move towards this particular
direction, mass number increases and binding
energy per nucleon keeps on coming down
and hence the elements are more nuclearly
responsive, particular the elements in these
zone Uranium 235 and as the neighboring nucleus
they are strongly radioactive because
their binding energy per nucleon is significantly
lower compared to this.
The same situation is also prevailing for
lighter elements, like the deuterium, tritium
helium 3 etcetera. They also having very,
very low binding energy and therefore or I
should say binding energy per nucleon and
therefore, they are also strongly radioactive.
Generally, any isotope wants to be at a state
which gives us the largest amount of
stability and maximum amount of stability
from nuclear reaction point of view and
therefore, it would like to have its binding
energy nucleon value as high as possible,
that
means, any nucleus whatever may be its mass
number, once to undergo some kind of
nuclear reaction, which allows it to move
towards this particular band of a binding
energy per nucleon, which is around and mass
number of 60.
Now, a nucleus which is heavier that is something
which is having a mass much higher
than 60, the only way it can move towards
the direction is by getting broken into 2
species, that means, 1 heavier nucleus being
broken into 2 smaller or lighter nucleus.
So,
that each of the newly produced nucleus can
have a mass number close to 60. On the
contrary, the nucleus which are lighter that
is something in this particular zone, the
only
way they can approach that largest stability
zone is by combing with another lighter
element. So, that 2 light elements combining
to a heavier element and the heavier
element is having a mass number, close to
that zone of higher stability.
Accordingly, we get 2 principle kinds of nuclear
reaction, one is the fission. Fission
corresponds to the breakup of one heavy nucleus
into 2 lighter or smaller counterpart and
therefore, elements which are having mass
number higher than 60, generally quite higher
than 60 well, beyond that value. They undergo
this fission kind of reaction, whereas the
elements which are having mass number much
lower than 60, they go through the fusion
reaction. Fusion refers to 2 lighter nucleus
combining together to form a heavier nucleus.
Therefore deuterium, tritium, helium 3 etcetera
this lighter nucleus are most likely to
undergo fusion kind of reaction and the entire
science of nuclear power generation is
based upon controlling this fission and fusion
reaction, while fission is the option that
is
exercised in any commercial nuclear power
plant. Fusion probably is a technology which
is more prospective because fusion is proposed
to generate even higher amount of energy
or much larger power density compared to fission,
but fusion is an technology which
requires much larger amount of control compare
to fission and present day scientist are
yet to have complete control and this particular
technology.
Fission however, has gone through several
phases or experiments and it is possible to
control the fission reaction, so that we can
harness the positive effect of nuclei energy
through this fission reaction.
These are the 2 examples that I just refer
to one is fission, where we are having in
this
particular example we having U 235, which
is getting broken into 2 lighter components,
Krypton having a mass number of 92 and Barium
having a the mass number of 141.
So, this is the fusion reaction, where a heavy
nucleus is getting broken in to 2 smaller
lighter nuclei. The other side, that is right
hand side of your screen we are having the
fusion reaction, where in this particular
example we have a deuterium and a tritium
getting fused together to produce a helium
nucleus, which is having a higher mass
number and therefore, it is expected to have
higher binding energy per nucleon.
Fusion is a more natural process because that
you can regularly find in extra terrestrial
bodies like the stars, but fission is a reaction
which is confined to heavy nucleus and the
chain reaction or fission reaction can be
controlled as per the present technology.
However, we are not in a position to control
the fusion reaction, one big problem with
fusion is, it requires very high temperature
and several other extreme parameters.
Research is on and hopefully shall be able
to capture the fusion reaction at physically
in
future.
In the fusion reactions substantial larger
amount of energy is also released compared
to
fission because associated elements are much
lighter in mass. Just compare the mass of
Uranium, which is having a mass number of
235 to that of a deuterium, or tritium which
are having mass number of 2 or 3 respectively
and therefore, fusion gives substantially
larger amount of energy release.
Here we are going in a example of typical
fission reaction. Don't need to go the
questions like why it is happening or how
it is happening, let us just observe the number
that you are having.
Here we can see one U 235 nucleus is being
stopped by 1 particular neutron, because of
this they get converted to and U 236, you
can think that the neutron has been swallowed
by this U 235, giving rise to a U 236 nucleus
and this particular reaction is associated
with release of energy amounting 5 MeV, this
U 236 is generally a very unstable
nucleus.
So, it further goes through a fission reaction,
where it gets broken into 2 smaller
components one is Krypton, having a mass number
of 92, other is Barium having a mass
number of 141. You can see the first stage
on the reaction corresponds to the mass deficit
or mass defect of 0.005365, you can calculate
this number just by seeing the amu values
that is present beside each of this nucleus.
Like for neutron 1.008665, whereas for the
U
235 it is 235.0493 and same way we can calculate
for others.
So, this mass defect of 0.05 amu, during the
first stage of the reaction corresponds to
approximately 5 MeV of energy. So, 5 MeV of
energy will be released during the first
stage. The same we can calculate the mass
defect for the second stage, where U 236 gets
broken into Krypton and Barium and corresponding
merge defective value can be
calculate as 0.2154 U, which is associated
with an energy, that means, if you multiply
this one with 931, we get an energy release
amounting 201 MeV.
Commonly the fission of one U 235, is associated
with release of about 212 MeV of
energy. Well, combining the figure shown here,
we are getting only 206, but the final
product that are shown here are krypton and
barium they are actually not the final
products because you can see particularly
for barium, its mass member is 140, which
is
far off from this optimum value of 60. Therefore,
it is likely that, the product here
Krypton and Barium may go through some further
nuclear reactions, there by releasing
some more amount of energy and finally getting
converted to even lighter nucleus.
So, combining all this energy release from
the direct reaction or later decay by the
products here, we generally get approximately
212 MeV of energy by fission reaction of
one U 235 molecule.
Typically, fission of one U 235 releases 212
MeV of energy, which are just mentioned. If
we divide this value by 931, we get corresponding
mass defect as 0.228 amu. Therefore,
fractional change in mass is 0.228 divided
by the atomic mass of Uranium and it comes
out to be of the order of 10 to the power
of minus 3.
Now, we compare this with a conventional chemical
reaction. Carbon is getting oxidized
and resulting in CO2 or carbon dioxide. Concerned
enthalpy of formation value is shown
here, if this enthalpy of formation probably
you have heard this term in thermodynamics.
This is associated with amount of energy released
or amount of chemical energy that gets
converted with thermal energy because of the
combustion reaction of 1 kilo mole of
reactants, that is the combine number of moles
of carbon and oxygen together has to be 1
kilo mole, then we are going to get this amount
of energy release.
Corresponding mass defect hence can be calculated
by dividing this figure with C square
and it comes out to be order of 10 to the
power minus 9 kg, but associated mass of
reactants that is 1 mole of reactants, will
be associated with approximately 12 for oxygen
and 32 for carbon to therefore, 42 kgs of
reactants and hence the fractional change
in
mass comes out to be of the order of 10 to
the power minus 10.
This is a very, very important figure or this
numbers here are very important to
understand the importance of nuclear reaction.
There are two points to note. Firstly, with
chemical reaction we are seeing a mass defect
of 10 to the power minus 10 amu. In
therefore, we can understand that whenever
we are having any kind of chemical reaction,
there is some amount of mass defect that goes
on between reactants and products and
that amount of mass defect gets converted
to energy, which we can see at the
manifestation during an exothermic chemical
reaction, but corresponding value of this
mass defect is coming out to be order of 10
to the power minus 10, which is extremely
negligible compare to the total mass of the
reactants participating in the reaction and
hence is often neglected.
So, in conventional chemical reaction during
analysis I always keep mass and energy as
two separate quantities and hardly considered
their equivalence, but when you consider
this figure with the figure presented above
for nuclear the ratio of energy release for
nuclear and chemical reaction, there is nuclear
reaction involving U 235 and chemical
reaction involving C 12, that is of the order
of 10 to the power 7.
That means, the amount of energy that you
are going to get by the combustion of 1 kg
of
carbon plus oxygen, we are going to get 10
to the power 7 times that energy by allowing
a nuclear reaction for 1 kg of U 235. That
is the most significant factor for nuclear
power
generation, you probably can remember that
during previous lecture while detailing the
merits and demerits of nuclear reactions or
nuclear power generation I mentioned about
a
point of higher energy density, this is the
point that I was referring to that time.
Energy density refers to the amount of energy
that you can harness from some unit
quantity or unit mass of the fuel and as you
can see here that the energy density of a
nuclear reactor will be 10 to the power 7
times of that with a conventional chemical
reactor or conventional thermal boiler. That
is the principle reason of people focusing
on
nuclear reaction as the next source of energy
generation and because of such higher
energy density we can get reactor cores to
be of much smaller in volume, that way saving
more space and sometime of fabrication cost
as well.
Here, let us try one numerical example, I
have taken a sample reaction, where Uranium
235 is being stuck by a neutron, leading to
the formation of Cesium and Rubidium and 4
further neutrons are released during the reaction.
So, let us try to solve this, the atomic mass
of Cesium and Rubidium are already
available here, we know that the mass of U
235 it was present in 1 of the earliest slides,
it
is 235.0493 amu. So, the mass on the left-hand
side, U will be equal to the mass of the
Uranium 235.0493 plus the mass of the neutron,
which is 1.008665 all in amus. Mass on
the right-hand side will be equal to mass
of Cesium which is given above 139.91711 plus
mass of Rubidium 91.91914 plus mass of 4 neutrons
1.008665.
So, if we compare these 2 figures, you will
find that the mass of left-hand side minus
mass of right-hand side, which we are calling
the mass defect is equal to I have chopped
out this number, it is equal to 0.187055 amu.
Correspondingly amount of energy released
will be delta m into c square which is equivalent
174 MeV.
So, we can see here that, this particular
reaction leads to the formation of 174 MeV
of
energy, there are several kinds of nuclear
reaction possible and ensure just a following
this mass balance you will be able to calculate
the amount of mass defect that we are
encountering and where remembering the relation
the conversion of mass to energy that
is E equal to m c square or by remembering
the equivalence 1 amu mass is equal to 931
MeV of energy. Here, this particular mass
value can just be multiplied with 931 to arrive
at this particular figure, we do not need
to multiply this m by c square and get the
conversion done.
So, we can always solve any such kind of problem,
yes. So, this prepares the background
for this course and this leads us to the end
of this particular module. I am going to
summarize the key point that we have learnt
here, you are after getting introduced to
the
topic of nuclear power generation, we have
learned about the concepts of isotopes and
we have seen that most of the natural elements
can have several isotopes because of the
presence of the different number of neutrons
and the nucleus.
Well, they have the same kind of chemical
properties, their nuclear properties strongly
vary. The mass of a nucleus is generally lesser
than the combined mass of his
constituency, which leads to the mass defect.
This concerned mass defect corresponds to
the binding energy; binding energy can be
thought about or can be viewed as the glue
which binds the nucleons together in the very
small volume of the nucleus.
Any nucleus attempts to have a higher value
of this binding energy per nucleon and as
we have seen from the earlier graph the highest
possible binding energy per nucleon is
around 8.8 MeV and therefore, any nucleus
attempts to participate in any kind of nuclear
reaction, which leads it to that particular
value of higher binding energy per nucleon
accordingly they can participate in fission
or fusion reaction.
And finally, the amount of energy released
during a nuclear reaction can be 10 to the
power 7 times larger than an equivalent chemical
relation, which substantiates the
importance of nuclear reaction as a source
of power generation. So, that is the end of
your module 1. This will be followed by an
assignment, which I am sure that you will
be
able to solve, if whenever you have any query
you please write to us and thanks for your
attention. In the next one, we are going to
start the very important topic of radio activity
in module 2.
Thank you.
