So, we talked about neutron drip line, we
talked about alpha decay, we talked about
composition of nucleus number of protons and
number of neutrons for having a stable or
beta active nuclei and so on. So, today I
will take one more example or one more application
of this semiempirical mass formula, and that
is kind of extrapolation into a very big nucleus.
You had seen that for a particular value of
z, for a particular value of z, if you look
at how many neutrons can be absorbed in this.
Then you have some limit after which neutron
cannot be absorbed. But there is an object
there are objects in this universe where you
have neutrons, and neutrons, and neutrons
only and these are neutron stars. So, what
are neutron stars, what are stars? Stars are
big celestial bodies, in which the gravitation
plays a great role. Stars are formed when
all these hydrogen and little bit helium in
space that shrink because of the gravitational
attraction and then somehow the density of
core increases, and the same time the temperature
also increases. So, when this mass cloud that
collapses at the core, because of this gravitational
potential energy converted into thermal energy,
the temperature increases, the density increases,
and at a certain stage all the temperatures
and densities are so high that the nuclear
fusion starts and that it becomes a star.
So, you haveall those stars in the universe,
Sun is our own star from which we are getting
all this light and energy and everything to
survive.
Now, what happens if the star is very big,
much bigger than our Sun say, 10 times of
that, or 8 times of that, or 6 times of those
very heavy stars. So, there the cycle is,
it goes from hydrogen to helium, the first
reaction in stars is, hydrogen goes into helium,
4 hydrogen nuclei that is 4 protons. They
combine and form an alpha particle protons
converting into neutrons and all those things.
So, that is going on in our Sun. So, hydrogen
going to helium, I am not talking much about
that, I am only giving you a very small introduction
to reach this neutron star as quickly as possible.
So, hydrogen goes into helium and then next
phase is this, helium goes into carbon, then
the next phase is carbon, oxygen and neon
and other things, other things. Finally, it
comes to iron after which fusion as such is
not possible. So, depending on the mass of
the star the things can stop anywhere. In
our Sun the thing will stop after this carbon,
but if the mass of the star is much larger
than this then all this can happen and then
at the end nuclear fusion at the core stops,
then the gravitational collapse is too fast,
okay? Because when the nuclear fusion goes
on the energy, is radiated outside, outward,
and this energy makes some kind of equilibrium
between this energy coming out and trying
to expand the gas there.
The cloud there and the gravitational attraction
which tries to collapse there some kind of
equilibrium is maintained, but if the nuclear
fusion stops, then this gravitational collapse
is much faster and then you have very hard
dense core and the outer layers which collide
on this hard core. They rebound from there
and the star ends its life in what is called,
supernova explosion, all those outer layers
are thrown away into space. But then, that
dense core which is left, that keeps collapsing
because of the gravitational attraction and
the pressure builds up too high and the density
becomes too high. It is so, stage comes when
all those electrons are forced to combine
with protons, and then this proton plus electron
making neutrons that takes place.
So, if a most of the protons is converted
into neutrons by combining with electrons
then, you have a star made mostly of neutron.
So, almost the entire star barring some small
surface layer is made of neutrons only. Of
course, there are very interesting properties
this neutron star once it is in this format,
it rotates very at large spin velocity and
all those things, so that maybe we will talk
about that later.
But here the idea is to tell that there are
bodies in our universe were unstable, where
you have mostly neutrons only, so mostly you
have neutrons only. How many neutrons you
think it would be here once it is a neutron
star? The mass is say of the order of the
mass of the Sun, or little larger than that
one or two masses, solar masses. So, this
could be something like one to two mass of
the Sun and, the radius could be few tens
of kilometres, say 10 20 kilometres depending
on which kind of star it is?
So, in about 10 20 kilometres of radius, the
mass much more than solar mass that is combined
compacted, so how many neutrons will be there
if the mass is mass of this Sun, how many
neutrons you think would be here? That could
be a really very large number and this object
of so many neutrons is also stable. We had
talked about this stability. So, these many
neutrons compacted in a small volume, small
in the sense that the the number of neutrons
which is, which are here and the mass which
is here, in the comparison it is small. It
is still 10 20 kilometres radius. So, we had
been talking of nuclei of the radius, 4 femtometers
and 5 femtometers and 2 femtometers and and
so on. Number of nucleons 100, 200, 50, 60,
10, so these we were talking of.
But here we are talking of large, much larger
number of neutrons. If we calculate how many
neutrons will be there, if it is one solar
mass, then neutrons make one solar mass. So,
that number will come out to be something
like, 10 to the power 55 or so. So, on Earth,
we do not see a nucleus of 2 neutrons. Even
a nucleus of 2 neutrons will be unstable,
three neutrons, five neutrons will be unstable.
You need protons together with protons you
have neutrons and then you make the nucleus.
For a particular value of Z there is is a
fixed number of neutrons which can get into
it and after that it drips neutron drip line
that we talked yesterday. Here, with no proton
in the core barring some surface layer where
there can be something some some protons or
some other nuclei, the whole of this big mass
is made of neutrons only and it is stable.
So, how do I understand this? If two neutrons
cannot make a nucleus and you do need protons
to make a nucleus; what stops having too many
neutrons and too few protons at asymmetry
energy? N minus Z whole square that term that
increases the mass energy and the binding
energy will become negative, so but here you
have a case in our space you have such a big
nucleus and the clue is gravitation.
The neutron star is formed from gravitation
only. This kind of extra ordinary density
is resulting from gravitation only. These
many neutrons, some 10 to the power 55 56
neutrons, they are packed into a volume of
some few kilometres. Whereas, if you look
at the Sun, it is a thousand times much more
bigger size. That is because of gravitation,
so gravitation is playing a big role and if
gravitation is playing a big role, maybe we
have to consider. When we consider binding
energy, we have to consider gravitational
potential energy also, okay?
In our discussion, we took care of coulomb
potential energy. We are taking care of this
nuclear interactions potential energy coming
from there binding energy, coming from there,
that is a V times A and so on. So, all those
things we are taking care of, but we are not
looking at the gravitational attraction between
the protons and neutrons. That is because
at that smaller scale where we have nuclei
commonly found on E, the gravitational attraction
is much, much smaller absolutely negligible
as compared to the nuclear in femtometer size.
In femtometer distances nuclear forces are
much, more effective than gravitational force.
So, we altogether we did not see towards gravitational
attraction when we were discussing this nuclei.
But these kinds of objects neutron stars,
where the gravitation is so important. Then
gravitational potential energy should also
be taken into account that is 1.
Now, the question is, the values or this model
of semi empirical mass formula, that is derived
on the basis of observation of this terrestrial
nuclei, A V is equal to 15.5 M E V or A C
is equal to 0.72 M E V or all those numbers
and even thisstructure of semi empirical mass
formula that is derived from the observations
of nuclei, that we find on Earth or we find
in our laboratories. Now, can this be extended
to objects like neutron star after taking
account of gravitational potential energy
of course? Will the the structure of the formula
and will the values of those parameters which
are obtained from these nuclei commonly seen
nuclei, will the same structure and will the
same parameters be applicable for an object
where you have ten to the power 55 or 56 neutrons?
So, let us do it and see what happens, okay?
I write the semi empirical mass formula together
with gravitational potential energy. How much
will be gravitational potential energy? That
also we can work out. You must have done in
your school.
If you have a mass in a sphere an assume uniformly
dense sphere, total mass capital M and total
net radius R, so what will be gravitational
potential energy of this object? So, its simple
to derive you take the density to be rho and
rho will be M divided by pi R cube, this will
be the density. We are talking of just mass
density, mass per unit volume. Construct this
sphere layer by layer, right? So, from infinity
you bring masses, pieces of masses, small
pieces of masses and then construct this sphere.
That is how you calculate potential energy.
Similar is the story when you calculate potential
energy of an uniformly charged sphere.
So, when the radius is small r that time you
bring masses from infinity and put that mass
here, so that radius becomes r plus dr. See
how much is the work done in bringing that
mass from infinity and putting it here, right?
So, that is easy to calculate the potential
of this object, this sphere here at the surface
and then multiply that potential by the mass
that you are bringing that is the potential
energy of that particular layer. So, that
gravitational potential energy that you can
calculate, if the mass already accumulated
here is small m radius small r.
Then next layer that you are bringing will
have a mass 4 pi r square d r times rho where
the radius is increased by d r. Already accumulated
mass that will have a potential G m by r equal
to this, negative of that sign. We will take
care. So, the work done that you are doing,
bringing this layer from infinity to here
will be this potential times, this mass that
you are bringing, which will be G 4 by 4 pi
r square rho and times this 4 pi r square,
d r times rho. This is G, 4 by 3 pi here,
this 4 by 3 pi , r square into r square, will
be r 4 and then what is left? So, we have
taken this G, we have taken this 4 by 3 pi,
we have taken this r square, we have taken
this r square, this 4 pi, this 4 pi will be
there, correct? So, 4 pi and rho square d
r, that is G, take 3 here, this is4pi square
and rho square is total mass, rho is total
mass square and divided by total volume.
That is 4 by 3 pi capital R cube, this is
the total volume whole square d r and where
is that r 4? Here is that r 4, so this will
be equal to G by 3, this 4 pi square and this
4 pi square goes away and this 3 square here
will be 9 1 by 9 So, 1 by 3, so this 3 will
go up, so 3 goes up, so this 3 goes, so this
3 cancels and this 3 goes up. So, let me write
it 3 G. So, this we have taken this, we have
taken this, we have taken 4 pi square and
4 pi square will be cancel out and then what
is left?
M square, small r 5 by 5small r 5 by, I am
integrating now integrated r 5 by 5 from 0
to capital R and in the denominator R to the
power 6. Check, I have integrated, so this
r square d r 4 d r that has become r 5 by
5. So, r 4 d r is taken care of, m square
is taken care of and other things are all
taken care of, correct? So, this is 3 by 5
this is 5 capital G, M square by R. So, this
is that gravitational binding energy, right?
Gravitational because of this gravitational
attraction this is the kind of energy that
is resulting, if you are looking for the potential
energy. It is negative of that from all pieces
at infinity when you combine it into one mass.
Then because of this attraction the total
potential energy is going down, so total mass
is going down, binding energy is going up.
So, this term will be added to binding energy
. So, if I take this gravitational binding
energy, in the total energy of the nucleus,
we had not done so far because we were talking
of tiny nuclei carbon and oxygen and uranium
and iron and zinc. These are tiny nuclei,
the radius was in femtometers 3, 4 femtmeters.
So, at that length scale the gravitational
forces are totally negligible, as compared
to the nuclear forces. So, gravitational binding
energy, we had never considered, but now that
we are talking of neutron star and this and
that where it is the gravitational attraction
that is playing the key role. We will write
our mass formula once again, assuming that
it is, it can be applied to those big masses,
because this mass formula was derived on the
basis of observations of these commonly available
nuclei on Earth. I will just assuming that
this can be extended to these systems like
neutron stars and so on.
So, our mass formula should be modified to
binding energy, a v times A minus a s times
A to the power 2 by 3 minus you speak loudly
minus a c times z z minus 1 upon a power 1
by 3 minus asymmetric N minus Z, the whole
square by a plus delta. This was the these
five terms were there, so far we had considered
this and now you add this plus or minus? Plus,
because it is adding to binding energy it
is adding to binding its attraction it is
providing you binding energy, so plus 3 capital
G m square over 5 R, okay?
We can write this term interms of capital
A and so on this last term here 3 G M square
by 5 R. In fact let me do that. So, this will
be 3 G, M will be mass of neutron times A
and R will be R naught A power 1 by 3, we
are now talking of neutron stars, this is
for neutron star. So, all the nucleons are
neutrons, so mass of that neutron star is,
mass of all those neutrons, mass of 1 neutron
is this M n and there are capital A neutrons.
I am not going into the detailed structure
of neutron star, I am just taking it as just
neutrons making this whole body, once you
go for neutron star Physics. You will have
more detailed structures the core is neutrons,
but then you can have on the surface some
other nuclei floating here and there. But
take it as just neutron only neutron a star
of only neutrons, so there are capital a nucleons
all neutrons. Therefore, mass capital M is
M n times A so square of that radius, radius
of a nucleus is R naught A to the power 1
by 3. What is R naught? 1.2 femtometers.
Once again, this 1.2 femtometers times capital
A to the power 1 by 3 that is, that we obtained
from measurement of radii of commonly available
nuclei, right? The radius was measured using
electron scattering or other experiments on
those nuclei and from there we had derived
that radius goes as something like 1.2 femtometers
times capital A 1 by 3. So, I am just taking
this formula two neutron stars. We do not
know whether it is valid there or not but
just let us take it there whatever we have
seen for thiscommonly available small nuclei.
We just assume that let us apply it there
and see what happens, justification will come
if there is some agreement. So, I am writing
this as this, so this term will be, so let
me in the next line we will add that. So,
a v times A now look at this second term this
is proportional to a to the power 2 by 3 it
is surface term and you know asyou increase
the volumethe contribution from surface goes
down in proportion to the contribution from
the volume.
If you increase the total mass, total volume,
then surface also increases. If you think
of a sphere and you increase the volume then,
what will you do? You will increase the radius
and once you increase the radius surface area
also increases, so any contribution from surface
will also increase. But then volume increases
in proportion to radius cube. Whereas surface
increases, surface area increases in proportion
to r square, so as thisr gets larger and larger,
the increase in contribution from volume will
be much more than increase in the contribution
from the surface. So, the importance of surface
will go downas you increase the size of the
object, although the surface is increasing,
but in proportion to volume the contribution
from surface in proportion to volume that
importance will keep going down.
On other side, if you decrease the volume
of your object then the surface becomes more
and more important. That is why nano particles,
nano materials have every different properties
set than the bulk materials, because the particle
size is very small. So, as the particle size
gets smaller the contribution from those surface particles .
sitting on the surface that becomes dominant
So, in usual bulk material the properties
are decided more by those volume interactions.
The particles which are in the volume they
decide the properties, but in nano particles
the surface which decides the properties.
Therefore, you have a new class of materials
with new properties with which you can manufacture
newer things you can do Physics. So, this
surface term let us neglect as compared to
the volume term, because the object is likely
to be very, very large. So, this we will neglect.
This is coming from coulomb potential energy
interaction between protons if there are no
protons, so this also we neglect. This 1 is
minus a sym N minus Z square, there are no
protons this is capital N square, but then
capital N its same as capital A, right?
So if there are no protons then the number
of nucleons is same as number of neutrons,
so capital N its same as capital A or you
also remember we wrote it capital A minus
2 Z whole square. So, if Z is 0 capital A
square here and A here. This pairing is also
we are going to neglect, this is coming from
even number of nucleons and odd number of
nucleons. So, if you are likely to have some
ten to the power fifty fifty-four fifty-five
fifty-six nucleons, so one more or one less
will not, is not going to change the energy
of the system by any significant account.
So, this also we can neglect or you can also
think in terms of, it is some constant divided
by capital a to the power3 by 4 and capital
A is so large, so forget this also. Then comes
this plus, this one lets write from here this
is 3G M n square by 5R naught A to the power
5 by 3. Now, this is binding energy neglecting
those relevant terms and keeping these terms
and including the gravitation. Now, for a
system to be bound, binding energy should
be positive right because binding energy,
is difference between total mass energy when
the whole thing are separated, the parts are
separated from each other, all nucleons separated
from each other.
Then the mass is larger and the bound system
mass is small. That is why it is bound system
mass is smaller and that difference is binding
energy, so this binding energy has to be positive
then it is bound. So, you can have a nucleus
of only neutrons provided, this binding energy
is positive, right? We are writing these expressions
assuming that I have a nucleus with only neutrons,
and now we are, we have written an expression
for binding energy. If this binding energy
turns out to be positive yes I can have that
so let us see in what condition this will
become positive.
So, for this neutron to exist neutron only
nucleus of only neutron to exist, binding
energy should be greater than 0, or say equal
to 0, so that means a v minus a sym capital
A and plus 3 G M n square 5 R naught A, this
should be greater than equal to 0, I can divide
by capital A. So, it is a v minus a sym plus
3 G M n square 5 R naught a power 2 by 3 should
be greater than equal to 0. Let us put the
values, how much is a v 15.5 M e V and how
much is this a sym 23 M e V plus 3 into G?
Next line, I will write the values, let me
write G M n square by 5 R naught a 2 by 3
greater than 0. Now, from here itself you
can see, if you do not have this gravitation
coming here, if you do not have this gravitational
term coming here, this will not be greater
than equal to 0. It is 15.5 and minus 23,
right? This is negative.
So, without gravitation you would not have
that nucleus of only neutrons, right? But,
once this gravitation is there, positive term
is there it is possible, so this equation
is 3 G M n square by 5 R naught A power 2
by 3, should be greater than or equal to,
I am taking it to the other side. So, it is
23 minus 15.5, how much it is, 7.5. So, this
is 7.5, remember the units or A 2 by 3 should
be greater than equal to 7.5 M e V times 5
R naught divided by 3 G M n square. Right,
A 2 by 3 should be greater than or equal to
5 by 7.5 M e V multiplied by this 5 R naught
and divided by 3G M n square. Now, let me
put more numbers.
So, this side is 7.5, let me do everything
in SI units, may we have to put this capital
G which I know in SI units, better. So, let
us do everything in SI units. Mega electron
volts here is mega capital M 10 to the power
6 and electron volt. One electron volt is
how many joules? 19 joules, so I am writing
everything in SI units so 7.5 MeV I have written,
into 5 times R naught, capital R naught, how
much is capital R naught? 1.2 femtometer.
So, minus 15 divided by 3 into capital G,
6.67 into 10 to the power minus 11 into mass
of neutron in kilograms. How much is this?
1.67 minus 27 kg m square, and this should
beunitless. This is coming capital A to the
power 2 by 3 that we are doing, so if I have
written everything correctly in SI units.
All those units will cancel out, and you will
get just a number. Let me work out it little
bit further, 7.5, 1.65, 1.2 into 10 to the
power, numerator look at the exponent, how
much it is? 6 and minus 19, that is minus
13minus 13 and minus 15, minus 28.
And this side 3 into 6.67 into 1.67 into1.67
and then you have 10 to the power minus 11
from here and 54 from here, so minus 65, right?
This will be some number, it is not going
to affect much, this side, 7.5 into one point.
This will be around 10 50, so this side will
be around 50 or 60 or so, and this is also
like that. So, this will be some number around
say 1 to 10 or something like that ten to
the power, so this is of the order of 10 to
the power minus 28 and plus 65. How much it
is? 37. This is remember capital A to the
power 2 by 3.
So, capital A should be greater than 10 to
the power 37, I am writing the order of magnitudes,
right 10 to the power 37 to the power 3 by
2, this is 10 to the power 37 and multiplied
by 10 to the power how much, is a 9 19, this
is 10 power 37 multiplied by square root of
10 to the power 37 and something is coming
from this side, also and I can see its more
than 1, so it should have been 18.5. If I
take square root of this but, then this factor
also in order of magnitude calculation, so
it is this, how much it is? 10 to the power,
so you cannot have a nucleus of 2 neutrons,
you cannot have a nucleus of four neutrons,
you cannot have a nucleus of 10 neutrons,
you cannot have a nucleus of hundred neutrons.
But you can have a nucleus of 10 to the power
56 neutrons.
Now, let us see in terms of mass kilograms,
how much it is mass for this number? This
is, it should be greater than this, right
A should be greater than this, so if this
is the number of nucleons, number of neutrons,
what will be the mass? Mass will be equal
to 10 to the power 56 and mass of a neutron
kg, how much is this? 10 to the power 29,
the mass of the Sun is around ten to the power
thirty kg, mass of the sun is around 10 to
the power thirty kg. You see these given you
the right order, neutron stars are observed
or are known with masses somewhat larger than
mass of the Sun.
One solar mass or 1.5 solar mass or two solar
mass this is the order of neutron star masses
that are actually observed. It says that mass
should be greater than 10 to the power 29
kgs. It is hitting the right order, right,
so the semi empirical mass formula which we
worked with or which we wrote making observations
on tiny nuclei’s and taking all those models
and those things, that can be extended to
these astro objects. These celestial objects
where very, very large masses, are involved
and the calculations with the same number
that we derived for this common nuclei are
not too bad. They are giving me the right
kind of order of magnitude. It is not that
this is totally a failure, it is giving me
that a neutron star can exist with 10 to the
power 20 neutrons. It is not giving that,
or you need 10 to the power seventy neutrons
minimum, no. Its giving the right kind of
order. Now, this particular example has been
worked out in a book I have taken.
It is from this book it is called, basic Ideas
and Concepts 
in Nuclear Physics in Nuclear Physics by Heyde
and it is published to by Institute of Physics
Publishing, okay? So, that is all I wanted
to talk about, this semi empirical mass formula,
remember I recapitulate things, this we wrote
using two different varieties of modelling
one, where nucleus was taken as a liquid drop.
The kind of terminology we use for liquid
drop that we are using here the volume the
surface and Coulomb interaction, so it is
charged drop so in a liquid drop the interaction
between molecules is also short range.
Because molecules are electrically neutral,
so and the interaction between molecules is
coming from electromagnetic interaction only.
So, if it is neutral, if we have two neutral
objects then the there’s no coulomb force
as such. But then if they are very close to
each other, so that the there is not really
charge 0. It the total charge is 0. But the
charge is distributed plus and minus inside
the molecule if the two molecules are close
to each other, they see this structure and
from there the interaction occurs.
If, the molecules are slightly apart then,
they are neutral particles and they do not
see much of the charge distribution and the
force is 0. So, at that scale, the molecular
forces can also be treated as short range
and therefore, the binding energy is proportional
to volume and then surface and all those things.
The nucleus is treated as a liquid so that
is called, liquid drop model of the nucleus.
The first three terms in that semi empirical
mass formula are essentially treating nucleus
as a liquid drop whereas, the other two terms
there it is more like quantum mechanical energy
levels and poly exclusion principles. Two
neutrons should not have same quantum number
and this and that, all those things, so that
is those energy levels coming in discreet
energy levels and nucleons filling all those
things. Then the coupling, pairing, where
the angular moment are coupling to in a particular
way spin and all those things.
So, this is the model which we call, nuclear
shell model. The semi empirical mass formula
is kind of a combination of nuclear shell
model and liquid drop model and from there
we derive this mass energy, binding energy
and from there many of the observed properties
of nucleus can be explained rightly, as we
have seen the stability against beta decay,
the mass parabola, existence of one stable
isobar or two stable isobars, even A nucleus
odd A nucleus, alpha particle stability neutron
many things fission, right fission heavier
nuclei will be unstable against fission. So,
many of these energy related energy related
properties of nucleus can very nicely understood
with semi empirical mass formula.
Just now as I had said that it can even be
extended for us those astrophysical objects
where the numbers are very large there also
this formula with the same numbers is not
that bad so that is all about this. The next
topic that I will take next lecture will be
on, we are going towards nuclear shell model,
but first we will discuss the nuclear interactions
between two nucleons. Just two nucleons and
those two nucleons, a nucleus with two nucleons
is deuteron. We will study little bit about
deuterons just this two nucleon system and
then we will move to nuclear shell model.
