We saw that a if you add this spin orbit interaction
in dot s term on harmonic oscillator single
particle nuclear potential, you will be produce
all the magic numbers. So, far as these shell
closures are concerned magic numbers are concerned,
it does not matter whether you take informative
square well potential or this harmonic oscillation
potential or a finite square well potential
with everything, you can get those magic numbers.
But the values of energy the separation between
different energy levels for example or sometimes
the order of energy levels inside a sub shell,
that is a function of a what kind of potential
you chose. So, for any realistic calculations
people choose more realistic potentials. So,
as compared to infinite square well potential
for harmonic oscillator potential.
Finite potential, finite square well potential
is more realistic, because as we discussed
earlier, you can take out a neutron or a proton
from the nucleus, whereas these infinite potentials
do not allow you to do that, then the sharp
change in potential as in infinite square
well potentials, suddenly it becomes 0, after
some or equal to capital R. So, then also
a little bit of unrealistic. So, people add
on it some kind of rounding of at the edge,
finite square well potential, but there is
some kind of rounding at the edge to make
it more realistic. The one which is which
suits every all these discussions is it is
would section potential right, it is would
section potential.
The shape is something of this type. So, more
sharp edges and no infinite potentials v is
a function of r is this would section potential,
that is taken in most in many of the calculations,
as the basic single particle neutral potential
and on this on add l dot s ten with appropriate
strength to get all those energy levels. So,
I will show you on the screen how this varies
how this differs from harmonic oscillator
energy levels or if you use those harmonic
oscillator energy levels and then apply L
dot S on it. You do get some energy levels.
And if you use this type of wood section,
normally this is around 50 m e v depth and
this r, becomes the size of the nucleus and
this how it tails of that is a right. So,
how these energies levels compare with the
harmonic oscillator energy levels.
Why is it called wood saxon?
Those are persons, those are scientists, who
had a come out with this kind of potentials.
So, the name is there to give them credit,
that these are the persons who had simple
say miller joules potentials. They were scientists
who came up with this potential that look
use these potential and you get the desired
results. We, I will just go to the screen
and you see how the these energy levels differ
from on how do they compare from this right.
So, these are the harmonic oscillator potentials
or with the harmonic oscillator potentials,
these are the energy levels, this you remember.
No sir.
1f then 1 p and 1 d 2 s together, we generate
1 f 2 p together 1 g 2 d 3 s together right,
that you can write easily. This is how the
wood saxon potential will look like. Now,
compare 1 s 1 s are 1 at the same height 1
p has come down all right. Then 1 d and 2
s both have come down, this is a general feature
from infinite potential, if you are coming
to finite potential normally, here energy
levels will come down in that that case. So,
1 d and 2 s are coming down by different amount.
So, and therefore, the degeneration is left
that. Similarly, 1 f and 2 p they have come down
by different amounts. So, 1 f has come down
more 2 p has come down less and that degeneration
is lifted then 1 g from that upper state has
come down by quite some larger amount. So,
that it is at that almost at the same level,
where the original 1 f 2 p was there. Similarly,
where 2 d 3 s, we have come down and the last
line that, we are seen at the top is coming
from 1 h. 1 h.
Next level was 1 h and then this 2 f 3 p those
things were there that 1 h has come down insignificantly,
and it is there. Now, if you add this spin
orbit potential on this these levels see what
happens. So, you have these wood saxon with
l dot s term. So, 1 s becomes 1 s half, what
you get half, the j value l and s, they have
to combined and they have given as j. So,
s state l is equal to 0 and spend is half
j is half. So, it is l s half and this 1 p
had split into parts p j equal to 3 by 2 that,
we call P 3 half that has come down and 1
p half has gone up. So, this 1 p is split
degeneration lift up.
Next one 2 s will not change, because if s
is take that l dot s term is 0, that does
not change. That does not split 1 d will split
in 2 parts d 5 by 2 coming down and d 3 by
2 going up. Then here is 1 f in 2 p, so 1f
split in 2 parts. The lowest 1 is f 7 by 2
and f 5 by 2 have gone up in 2 p also split
in 2 parts 3 by 2 and half p 3 by 2 is coming
between the 2 f states. So, 1 f has split
in 2 parts 1 f 7 by 2 and 1 f 5by 2, 5 by
2 going up 7 by 2 coming down 2 p has also
split in 2 parts p 3 half and p half and the
structure, that we have drawn is the 2 p 3
half is coming between f 7 half and f 5 by
2, then you have 1 g.
So, 1 g will split and 1 g splits in 2 parts
g 9 by 2 g 7 by 2 and 2 d on top of that,
you have 2 d that will also slip split 3 s
will not split and 2 d goes like this and
then you have this 3 s coming. And then you
have this 1 h coming from the top right. 1
h 11 by 2 coming from the top, these are the
energy levels, if you use this woods saxon
potential. The magic numbers are same shell
closers are same place s half will have 2
occupancies 2 neutrons or 2 protons can go
there. So, 2 similarly, that p 3 half and p half
that combined will give you the 6 and the
next shell closer is 8 then 21 f 7 by 2 is
stand alone. So, 8 8 nucleon can go there
8 nucleon of the same type 8 newtronce or
8 protons can go there. So, that shall closes
28 in after that the shall will close that
g 9 by 2 that will be 50 and at h 11 by 2,
that will be 82 and similarly, you can go
up you will get that 126.
Now I will like you 2 practices writing these
energy levels atleast up to say 50 nucleons
that magic number 50. So, you will just to
try to do that on bold, you can do in on your
notebook. So, that whenever need rises, you
can write this order of energy levels, these
sub levels quickly. So, let us do this manually
on the board. So, you can start with in the
same fashion.
1 f and then 1 p like harmonicoslator and
next one is 1 d and 2 s. So, 1 d will come
down and 2 s will go will be slightly up they
are split. Then 1 d and 2 s will be 1 f and
f will be somewhere, 1 f and 2 p will be,
slightly up 2 p somewhere here. So, you can
write quickly, if it is harmonicoslator it
is 1 s 1 p and then 1 d and 2 s together.
So, you split down little bit and then next
1 will be 2 p and 1 f, so you split down.
Then putting that l dot s tern, this p, what
will you do this p, you put in 2 parts p 3
by 2 down behalf of right. This p goes here
then this d, you can this shell is closes
at 2 and this is 6 of this closer at 8. next
closer at 20 2 and 8 10 for 10 more with this
d, what will happen d 3 by 2 will come down
d 5 by 2 right, d 5 by 2 will come down and
then d 3 by 2 will go up and this 2 s will
be remain there. So, this 2 s will remain
there, if you wish you can also write that
n.  Up to this part you can do that easily and
how many what is the number of newtronce or protoconce that can be accommodated 6 plus
2 8 8 plus 4 12. So, 8 plus 12 20 then f 7 by 2 will come down
and that will be alone 1 f 7 by 2. So, that
is next shell closer. And f by 2 will go up
f by 2 f 5 by 2 go up and then this 2 p this
2 p 3 by 2 and 5 by 2 and how is that written,
f comes between the 2 p levels. So, this 2
p let me write it slightly lower. So, that
this 2 p when it is bits 1 level goes here
and another level goes here. This is 2 p,
this is 2 p 1 by 2 and this is 2 p 3 by 2
and this 1 is 1 f 5 by 2 in between. How many
4 plus 6 10 and plus 2 12 and this is how
much 12 and 10 will come from 1 g, next 1
is 1 f 2 p 1 is 1 g 1 g will be there then
2 d will be there. And 3 s will be there and
this 1 g will give you 1 g will come down
and join here. 1 g 9 by 2 that makes it 12
plus 10 22 and 28 plus 20 is 50, you do not
know to climb.
These levels that to first it is the half
then p 3 or then p half and then d 5 by 2
then is a is almost impossible then crime
all those things to remember all that author.
But you construct them, if you remember the
basics of how it is spitting n, how it is
ordering from that basic construct basic knowledge,
you can construct 8 in how much time, you
will take from it is clean slate. If you have
to write all these things, how much time will
you take. Yes, how much time you will suppose,
I will graph everything and ask you to write
the levels, starting from scratch, how much
time will you take.
5 minutes.
5 minutes.
2 minutes.
2 minutes.
Maximum 2 minutes.
Maximum 2 minutes.
What you have to do, you have to first you
imagine your harmonic oscillator potential
1 s 1 p 1 d 2 s 1 f 2 p and then the 1 g and
2 d and 3 s harmony that should take you some
30 seconds. Then write them separate them
little bit in harmonic potential, you would
have written these thing at 1 level. These
2 at 1 level, these 3 at 1 level, separate
little bit and remember higher l value or
lower n value that will come down.
So, separate that, writing this, this, this,
this, this, this, and this 1 g at least 1
g should take about 30 seconds. And then you
have to split 1 s will remain at such this
will 3 by 2 coming down 1 by 2 going up d
5 by 2 coming down 3 by 2 going up and this
s is in between where f comes down. And this
is a you remember that, this is this close
is up to next magic number 8. Next is 20 next
is 28. So, this 8 will be alone and it is
corresponding side by 2 will go up and this
p will split into 2 parts and this f is signed
with between t half and t 3 half and 1 g comes
from here. And you reach this 50 magic number.
So, now tell me how much time, you should
will take 2 minutes or 1 minute.
1 minute
Yes it should be able to do it in 1 minute,
whenever you need as is you can, you should
be able to write it in 1 minute alright good.
Now, there are certain things about, this
ordering, 1 is this nuclear potential depends
little bit on the size of the nucleus. Because,
more is the bigger is the size this potential
that you have written will be wider. So, the
basic features remains the same, but the values
of energy will change. Values of energy will
change and therefore, the separation between
different energy levels that separation will
also change, that depend on how big is the
nucleus. And if the separation changes some
times, it may also happen that 2 adjacent
levels cross over. It may happen suppose,
this separation is changing as a is increasing
and so this separation is decreasing.
So, at certain instant it can cross over,
it can interchange, that happens first here.
This 2 s half and this 2 d half here, we have
written 2 s half between the 2 d levels. This
d level is splitting in 1 part here and another
part here, 2 s is remaining here. Now around
a equal to 50 or so this features. So, for
this is for lower values A less than fifty,
if a is larger and these, this section on
this things, the order will remain the same
this separation can change. But, order will
remain the same.
But, this portion will become d 5 by 2 and
then d 3 by 2 1 d 1 d and then 2 s half here.
So, this order can change similarly. Here
this order can change this order can change
somewhere a greater than 60 70 and so on.
So, just keep this in mind it is not a script
order, but if you normally shell model is
a used is quite successful for lighter nuclei
and then in the middle heavy range also there
are sections.
So, for lower lighter nuclei this is very
good and if you are more than 50 more than
60 more than 100 at some places, you may get
some cross overs. But the group will remain
as it is it will not no level will leave it
is own group and come join another group.
So, the magic numbers will remain the same.
So, that is 1, second thing is every separation
between the energy levels, that decreases
as this nucleon number, mass number A increases
right. It comes from experiment essential.
That last filled energy level, you can all
it nuclear fermi level, last fill energy levels.
That is a that does not vary much lighter
nuclei, heavy nuclei, muddle weight nuclei
that does not vary much varies within say
5 written middies of saw, similarly that depth
of the potential.
That also remains almost the same. So, this
total occupied energy difference the bottom
and top that is of the same order of all nuclei
and then if a is large then those who energy
level should be there, if A is small then
less number of energy levels are there, I
am only talking of occupied energy levels.
So, the highest occupied energy levels, if
that remains almost at the similar, place
and the lowest energy level, that remains
roughly at the same place and then you are
pushing 10 nucleons. So, you should have roughly
5 levels each one or yeah. In fact, nucleon
means of protons and neutrons, when energy
level will take 4 2 protons and 2 neutrons.
So, you will have, but if u have say 200 nuclei
nucleons and all those 200 are to be commutated.
So, you should have 50 levels in that same
range. So, that every separation between the
energy levels, that is you are seen here.
That is a function of capital A and as capital
A will increase this whole thing will get
squeezed. So, in the same height of the black
board, you should be able to bring all those
1 26 and 184 all those things is compressed.
The separation average separation between
the energy levels decreases as capital A increases.
That is why in semi empirical mass formula,
when we are writing that a symmetry n minus
z, whole square divided by capital A. So,
that is a because of this, the third thing
is this proton, this these energy levels or
this potential itself is some, what different
for proton and neutron. Because, of limbic
interaction, if you are writing for neutron
is a single practical potential for all nucleons,
we are saying that this is the potential and
these are the energy levels. But, then that
single particle, if it is neutron it is not
experiencing any calamine direction, where
as if it proton it is interacting with the
rest of the nucleus through, coulomb potential
also coulomb interacting also.
So, the potential it some would different
for proton and for neutron in lighter nuclei,
it does not matter much right. Light nuclei
it does not matter much and the levels are
same almost the same on the, but for middle
weight or heavy weight. This coulomb interaction
may become significant as compared to this
nuclear potential depth, you remember the
formula for a coulomb potential.
If you assume it will be a spherically charged
system spherically, symmetric charge distribution
z e z minus 1, you can say when the a flume
potential can be written as or coulomb potential
energy, can be written as some z e square
4 pie epsilon not capital R. Capital R is
this and then 3 by 2 minus 1 by 2 r by capital
R square right. We have derived this, when
we where considering the radius of the nucleus
am initial lecture and how this electrostatic
interaction between the nucleus and electron
changes the energy and x-ray energy, alpha
energy all those thing we discussed. So, we
have derived this equation that time is class
eleventh. So, columbic potential is like this.
So, for low z if the you have light nuclei,
z is low it does not matter, because this
is quite high, but for middle n heavier nuclei
it will matter. This is for r than r and r
greater than r also, you have no electrostatic
potential. Because it is long range force
and that goes by 1 r, if you have z e or q
charge in this sphere outside the potential
will just go as 1 by r. So, your potential
first thing it is lifted up, it is positive
right. Proton and the rest of the nucleus.
Both are positive charge. So, it is positive.
So, it will lift up at r equal to 0 small
r equal to 0, it is 3 by 2 times of this much
is added at r equal to capital R. This is
3 by 2 minus half is 1. So, this into 1 lifted.
So, at the center it is lifted more, here
it is lifted less. So, over all it is lifted
and then you have 1 by r type fall. So, this
is how the potential will modify for neutron
potential is like this for proton potential
is like this. So, if I drop both side by side.
Suppose this is the potential for neutron
and then you can draw a similar thing here.
This is 0 and this goes up, this was the original level 
and now this has been shifted up  it goes up and it falls 1 by 1 type. The energy
level starts here, if these are the energy
level correspondingly, you will have energy
levels here. But, these energy levels are
missing am for neutron energy levels will
starts here and here it will start here. So,
that is one simple effect another effect,
because of this lowering of this depth here
spacing is little bit larger than this. And
if the nucleons are to fill this these energy
levels for bottom to keep the total energy
of the nucleus minimum possible. So, neutrons
and if neutrons and protons are allowed to
interchange be technically.
So, the energy will be filled up to some level
right. If that is the case you will have more
neutrons and less number of protons, because
top the energy should be same, if we are filling
the higher levels then it should change through
beta decay. And then fill that lower energy
levels, if you have to keep the total energy
of the nucleus minimum possible. So, protons
and neutrons both should fill up to almost
the same level poly exclusive for example,
has to be obeyed up to same. Otherwise, if
you have much more things there they can interchange
between neutrons can go into proton or proton
can go into neutron and this empty shells
will be filled up.
And if is to be filled at the same level energy,
same energy then the number of energy levels
appear in here is less number of energy levels
appearing here is more. That means, you will
have more neutrons than protons n greater
than z. So, this is 1 way to understand way
in heavier nuclei, you have more neutrons
than protons, for light nuclei coulomb interaction
is not very effective. And therefore, the
2 potential and 2 schemes of energy levels
are almost the same right.
And therefore, n is equal to z to keep that
energy minimum n is equal is z. But, once
this coulomb interaction becomes effective
then to keep that to total energy minimum
you must have larger number n neutrons than
protons. These are some of the things, which
i wanted to talk about that. Now, coming to
connecting this whole model, this whole scheme
to the miserable properties, the first thing,
which the shell model can quite satisfactorily
explain is spin and parity of nuclei with
in ground state right. So, spin parity of
nuclei in ground state.
A spin when I say spin of a nucleus, what
to I mean, this is total angular momentum
remember. This is the terminology used in
nuclear physics not in atomic physics, in
atomic physics, we spin we by spin we understand
spin angular momentum. But, in nuclear physics
if i say spin of this nucleus, it is the total
angular momentum not the spin angular momentum.
So, that is the terminology used. So, when
I say spin it is j, total j all orbital spin
everything combine for the whole nucleus,
this is the final j am and then parity pi,
whether the space part of vary function is
symmetric or any symmetric and this pie depends
on l. So, if l is even, we say that parity is positive,
your space part of v function depend on this
l and if l is odd, we see that parity is odd,
we have discussed little bit of this while
doing dethrones parity negative. So, as super
script it is written plus or minus, if it
is written plus, that means parity is positive.
If it is written minus, parity is negative
and j is the total spin.
So, first even nuclei when I write even even
nuclei, what do I mean z is even and n is
even, z is even and also n is even. These
are called even even nuclei, now one assumption
or consistence with the observation is that
in particular energy level, where you have
2 neutrons and 2 protons going 1 with spin
up and 1 with spin down. Because, all these
levels which are drawn here. So, each will
contain 2 contemn number n contemn number
l contemn number j and n j right.
So, those 2 neutrons or 2 protons in that
same level, they will pair up. And pair up
means, if these 2 neutrons for example, p
half in p half, we have 2 neutrons on s half,
we have to neutrons or 2 protons or p 3 half,
we have 4 counter states there, then 2 will
pair up and 2 will pair up and when they pair
up. They pair up to angular momentum 0 and
magnetic moment 0 everything 0. So, we call
it extreme single particle model well we say
that all neutrons and protons, wherever possible.
They will pair up pairing decreases there
total energy.
So, some kind of an addition to this potential.
This potential tells that this is the energy,
this is the energy, this is the energy on
tomb of that we say that if neutrons 2 neutrons
pair up. Pair up means, if they combine there
angular momentum to 0 of these 2 0, j is 0
and m j is anyway 0. Those 2 only not of the
hole minus then we say that, they have paired
up and this pairing takes the energy down.
So, as many pairs can form the energy of the
nucleus is going down.
So, in extreme, what we call extreme single
particle mortal, we say that neutrons and
protons will pair up to the extent possible
greatest extent possible right. If you have
even even nucleus all the protons will pair
up and all the neutrons will pair up. So,
all the neutrons will pair up, all the protons
have pair up. then the total angular momentum
of the nucleus is also 0. Because, pair wise
it is 0 0 plus 0 plus 0 plus 0. The whole
thing is 0, it should be 0 15 should be 0
and since there are 2 of them.
So, whether it is l is equal to even or l
equal to odd for each of the 2. The parity
will become positive, if there are in say
d phi by 2 state, so then for this neutron
d for this neutron. So, plus parity combined
parity is also plus. If they run 2 3 by 2
p l is equal to 1, but then. So, negative
parity. So, negative parity of the first particle,
negative parity of the second particle, total
in the symmetric and anti-symmetric multiplied
becomes symmetric. So, parity is positive.
So, independent of what levels these pairs
are occupying pair wise, the parity is positive.
So, the entire nucleus that also will have
positive parity 0 plus all even-even nuclei
in there ground state remember  will have is j totally. J is yeah. That is terminology here guys terminology
here, you must remember and the word nuclear spin is ment for the total angular momentum
of the nucleus, so that the term the definition of the term nucleus spin right. Why that nomenclature
has come that could be some historical reasons,
that today we only knew our d s, if the nucleus
is talking of nucleus spin, he must have talked
about the total angular momentum right. So,
all even-even nuclei in the ground state should
have 0 plus spin parity according to the extreme
single particle shell model and this comes
out to be true in all experiments. So, that
was about even-even, now let us come to what
we call odd A.
Odd A nuclei, either z is odd together with
an even of l is odd together with z even.
So, 1 is even 1 is odd. So, the total will
be odd. And what are the third variety possible,
even-even then you had odd and the other possibilities
odd odd. Both are odd the total is even both
are odd, but very few odd odd nuclei stable
nuclei exists right. Some 4 of them in the
lighter 1’s nitrogen 7 7 14, nitrogen 14
7 7 7 that is a stable element, and so on.
So, bigger 1 bigger 1 itself 1 1
Oxygen
Oxygen is 8 8 protons it is not it is not
odd odd, oxygen means 8 protons, you can take
different isotopes 16 oxygen 17 oxygen 15
oxygen. But that 8, you cannot change that
8 oxygen is always z equal to 8. So, z is
even. So, odd A if you are discussed even-even
and odd A, you have discussed almost all the
nuclei barring, some 4 or 5 in odd A now here
that you would not get extreme single particle,
that is relevant here. Because, odd A means
there is 1 neutron or 1 proton which does
not have a chance for pairing, the total number
is odd and for pair you need 2.
So, if there are 9 of them only 8 can pair
up that 9 one has no chance of pairing up.
So, the problem in the extreme single point
of a model, we assume that the property of
this nucleus is decided by that last nucleon
only holding other, which had paired up will
give you 0 0 spin 0, parity 0 angular momentum
everything 0, 0 magnetic moment. Only that
last unpaired nucleon will decide all those
properties, that is extreme single particle
model to some extent even this simplified
model works and that we will see. Now, tell
me some light odd A nucleus and we will look
at the spin parity of that. So, give me 1
anyone.
Nitrogen Nitrogen odd A 7, you have to give 7 here
and then you can get 8 here or you can give
odd A 6, you can give 6 or you can give 8
or you can give 10 15. This will be odd A
nucleus is it stable do not remember, I also
do not remember, but let us just work on it
if this is the nucleus for whatever time it
is there. What would be the spin and parity
in it is own ground state. So, let us draw
those energy levels. So, tell me what is the
lowest 1, 1 s half, next 1.
1 2. 3 by 2 next 1,next 1,1 d,next next all right
this is enough. 2 parts are done, 1 for the
proton, 1 for neutron. So, 8 protons 8 neutrons
here, this 8 neutrons will go here right.
Lowest energy of the nucleus, 7 protons where
should they go 4 here p 3 by 2 4 here, so
2 plus 4 6 and 2 8, so 8 neutron are going
here and then 7 protons 2 here, 4 here and
1 here. Alright according our model this gibbs
is 0 spin 0 parity. This gibbs is 0 spin 0
parity, this gibb’s is 0 spin and 0 parity
2 pairs.
And similarly this and similarly this, so
the spin of the nuclease is same as the angular
momentum of this part and that is half j equal
to half. So, if capital J should be half in
ground state capital J should be half. And
since it is p state l is equal to 1, so the
parity negative right. That is how we calculate
take another example oxygen. Oxygen 15, oxygen
15, what do you expect tell me spin parity
neutron proton different, now protons 8.
So, all these you will have a proton here
and you will have a 1 neutron here. So, same
half minus right, what about 17 oxygen. It
protons 2 plus 4 6 plus 2 8 protons will be
there, now 9 9 neutrons. So, 8 neutrons and
9th neutron must go here. So, 5 by 2 plus
correct, it should be 5 by 2 plus the ground
state spin parity should be by 2 plus 13 carbon.
Carbon is z is equal to what, so n is equal
to 7. So, z is equal to 6 4 plus 6 correct
6 here and this 1 is 7. So, 1 neutron will
go here, correct 2 plus 4 plus 1 7 right.
7 neutrons, 7 neutrons and 6 protons. So,
what should be the spin parity, half negative
yes that is how you can calculate, in most
of the cases, it comes out to be correct right.
Most of the cases it comes out to be correct,
you can take for example, 27 silicon, what
is z for silicon. Silicon is very important
element 14 14 13 right. So, 14 what 14 z z
14 2 plus 4 6 plus 2 8 and 6 14 n 13 so that
means, what should be spin parity.
5 by 2 plus.
5 by 2 plus correct. So, it is 5 by 2 plus
most of this odd a nuclei, you will find that
single by article model prediction is correct.
I gave you 1 exception also and that is nickel
61 nick for nickel z is 28. Iron, cobalt,
nickel iron is z is equal to 26 then 27 and
28. So, nickel is 28 how much remains 33.
Total is 61 correct. So, now, lets work it
out let me draw the fresh diagram yeah, dictate
me 1 s half tell me.
1 p 3 by 2.
1 p 3 by 2.
1 p 3 by 2.
1 p 3 by 2.
1 p half.
1 p half.
1 d 5 by 2.
1 d 5 by 2.
1 d 5 by 2.
1 d 5 by 2.
2 s half.
2 s half.
1 d 3 by 2.
1 d 3 by 2, 1 f 7 by 2,1 f seen by 2, next
is,2 p 3 by 2. That is enough, 28 28 will
come up to here right. So, this is which side
I take proton which side neutron lets take
this side neutron and this side proton. So,
28 up to here, this will be 28 remember 20
and this makes this 28, so everything up to
here. So, fill proton everything is all this
is fill that will make it 28.
And similarly 28 means all 3 how do I write
8 of them, 8 of them let me write 8 p all
8 protons are here, all 4 protons are here,
all 2 protons are here, all 6 protons are
here that will make it 28 correct up to her
it is 28. So, up to here it is 28 neutrons,
how many more are there. So, 1 2 3 4 5 ok,
it cannot contain 5th. So, n here right. So,
this the ground state should be the ground
state, should be according to this calculation
j pi, should be 5 by 2 minus. Now and the
observed value is 3 by 2 minus, observed value
is 3 by 2 minus ground state spin parity of
61 nickel is measured to be 3 by 2 minus.
No no . So, that will be here, that will be
here, know at since you have 61. This nucleons
therefore, the order will change I told you,
but that was here that was here not here.
This order remains the same how can I get
this 3 by 2 minus as the ground state can
you see a 3 by 2 minus. In this scheme, where
a and odd neutron should be there in this
scheme to give you 3 by 2 minus. 2 p 3 by
2 here, 4 neutrons here, if you have 5 neutrons
here, it will give you 3 by 2 minus.
But, 5 neutrons cannot common this is 3 by
2 only. If not 5 neutron 3 neutrons will also
give you 3 by 2 minus. So, if this neutron
is not here, and this neutron is here, and
you have only 3 of them, that can give you
3 by 2 minus, but why should it go to a higher
energy, when the lower energy is available.
And the answer is that, if it is here it will
be it will pair with this neutron, if it is
here it will pair with this neutron. Now when
2 neutrons or 2 protons pair up energy of
the nucleolus goes down, but by how much amount
it is not fixed.
It depends on l. So, if 2 neutrons are getting
paired up in p level. The pairing energy will
be different, if they pair up in f level the
pairing energy will be different and that
pairing energy the amount by which the energy
of nucleolus is reduced due to pairing is
larger, if the pairing is made in a larger
l state. So, turns out that if the pair is
made here. The energy comes down by larger
amount, then if the pair is made here, this
is f l is larger here it is p l is smaller
l is 1 here, l is 3 here.
So, if pairing is done in a higher l level
that pairing energy is larger, so it. So,
happens that difference in pairing energy
compensates, more than compensates for this
rise here. The energy is because of pairing
energy is reduced by certain amount, here
it is reduced by much larger amount. So, even
if it has to go to a higher energy level that
saving due to paring is more than, this going
up and hence in ground a state it is n n n
here, and n n here. So, that extreme single
particle model is still working, we are only
bases on that only, we can also explain this
type of situation nickel 61 type of situation
by taking proper care of pairing energy.
Why we called only ground state.
We discuss ground state excited state also,
you can discuss with this extreme single particle
model for example, I give you the excited
the state of let us say oxygen 17.
Oxygen 17 if ground state, I am telling you
the observed values 5 by 2 plus then the next
one is half plus this is the first excited
state. And the next one is half minus and
then you have 5 by 2 minus and so on. Now,
look at this 17 O structure your 8 protons
and 9 neutrons. So, protons will anyway form
a close, shell we have minus half then you
have 1 p 3 by 2 1 p half what is next. 1 d
5 by t, 1 d 5 by 2,2 s half,2 s half,1 d 3
by 2,1 d 3 by 2 enough, enough, enough protons
we do not have to see look at the neutrons
only. So, you have all these will be filled
up to here it is 8 right. So, you will have
protons and then protons and then protons.
So, protons is anyway fill neutron talk of
neutrons all right protons is also. So, neutrons
up to here are all filled and a next 1 is
here, 5 by 2 plus is ground state. Now if
energy has to go upward this neutron should
go here.
So, the first excited state will be obtained,
if this neutron goes here and if this neutron
goes here. It should be half plus and it is
half plus. Next way is half minus, how can
I get the second excited the state half minus.
Next one is 3 by 2, if this neutron from here
it goes here it will it should be 3 by 2 plus
it is not it is half minus. The same story
this can give you half minus. So, if this
neutron one of these neutron goes here and
pairs up.
Then you will get 1 by 2 minus. So, that is
the second excited state for hence, this pair
is broken and here, the pair is made this
is p this is d. So, some energy saved, but
this energy difference is larger say it has
gone up. So, even excited the states also
many of the excited the states also you can
understand on this. So, we discussed ground
state first and then excited states also many
of the excited. The states of these at least
this light nuclei can also be understood using
single particle shell model right. So, the
next property that we will discuss with this
model is magnetic moment of the nucleus.
