Cross-Section of Nuclear Scattering
A Project Report Presented to
The Faculty of the Centre for High Energy Physics
University of the Punjab
in the Partial Fulfillment of Requirement for the Degree of M.sc (2-Years).
Supervised By :
Dr. Talib Hussain
Submitted By :



Muzammil Asghar (1019MM)
Ali Raza (2319MM)
Arslan Anwar (2519MM)
July , 2021
Certificate
It is certified that the work contained in this project report has been done by Muzammal
Asghar Roll No. 1019MM, Ali Raza Roll No. 2319MM and Arslan Anwar Roll No. 2519MM in
Centre for High Energy Physics (CHEP), University of the Punjab, Lahore, under our guidance
and supervision. They took keen interested in research work and completed their work
satisfactorily.
Supervisors:
Dr. Talab Hussain
Assistant Professor
Centre for High Energy Physics,
University of the Punjab, Lahore.
Director
Centre for High Energy Physics,
University of the Punjab, Lahore.
Acknowledgement
We pay our humble gratitude to ALLAH Almighty, The
most Beneficent, The most Merciful, The most Gracious,
Benevolent and the Compassionate whose bounteous
blessing and exaltation brandished our thoughts. We also
are
deeply
thankful
to
Holy
Prophet
Hazrat
Muhammad (S.A.W) who's guidance encouraged us to
seek knowledge.
It is great pleasure to acknowledge our thank to Dr.
Talab Hussain (Assistant Professor) for his kind
supervision and the project relevant teaching. We would
like to express our gratitude to who helped us to seek
knowledge in the field of Cross-Section of Nuclear
Scattering. We are appreciative to for his creative and
comprehensive guidance, and continuous support in the
completion of this project.
An extreme thanks to our family, especially our kind
parents to support us by all means throughout writing
this project report.
Contents
Scattering………………………………………….1
introduction
Scattering
introduction
In mathematics and physics scattering theory is a work for studying and
understanding the scattering of wave and particles. Wave scattering corresponds to
collision of a wave with some material object. For example, sunlight scattered by
raindrop to from rainbows.
Whenever a beam of particles of any kind is directed at matter (incident on any
matter), the particles will be deflected from their original path as a result of
collision with the particles of matter.
Examples:
i. Interaction of Billiard Balls with table
ii. Rutherford scattering of alpha-particles by Gold-foil (nuclei)
Definition
“When a beam of particles collides with a target material then the particles will be
defalcated from their original paths and this process is called scattering”
What is a Beam?
“ The Group of same particles moving with same velocity, same energy in one
particular direction ”
1.4 Conditions For a Beam:
Let suppose if we are studying the electron beam , Then
 All the particles in the beam should be electron.
 If v is the velocity of electrons then all the electron will move with the same
velocity v.
 Because velocity of all electrons is same so momentum will also be same (i.e.
P = mv)
 If momentum is some & masses of all electrons are also same then the energy
of all the electron will also be same (i.e. E =
𝑃2
2𝑚
)
 All the particles should move in only one particular direction
“ These are the conditions for Beam But it is not necessary that all the
particles interact with the target some of them may not make an
interaction with the target ”
Some important terms about scattering :
Scattered Angle ( 𝜽 )
“ it is the angle that the scattered particles making with incident beam of
particles (take along z-direction) ”
Solid Angle ( dΩ )
Definition:
“ A three dimensional angle such as subtended by a cone is called a solid angle
(measured in steradians ) ”
We are considering scattering in 3-D
therefore scattered particles are
moving in all direction x y z, which
are forming a sphere of Radius R.
𝜃 = Scattered particle making with Z
𝜑 = angle is xy-plane
In 2D the angle is defined as
Angle =
𝑎𝑟𝑐
𝑟𝑎𝑑𝑖𝑢𝑠
(radian)
Because circle is in 2D-Plane there only
simple angle is discussed in 2D.
Here in the present case we are dealing
with sphere in 3D, so then angle is also in
3D.
Let we have a sphere consider an element
of the sphere the element will represent the
area.
Arc = (radius) (angle)
AB=r d𝜃
Angle =
𝑎𝑟𝑐
𝑟𝑎𝑑𝑖𝑢𝑠
In 3D:
Angle  solid angle
Arc  area
Solid Angle =
(Area of ABCD)
𝑅2
Radius  (radius) square
To find AD we shall consider projection
So Area of ABCD = AB*AD
= r d𝜃 × rsin 𝜃 d𝜑
Area
= 𝑟 2 sin 𝜃 d𝜃 d𝜑
So,
dΩ =
𝐴𝑟𝑒𝑎
𝑟2
=
r sin 𝜃 d𝜃 d𝜑
𝑟2
dΩ = sin 𝜃 d𝜃 d𝜑
This is the expression for solid angle
Types of Scattering
Depending upon whether there is energy loss or not we can classify scattering
events into two types of scattering, given bellow
1. Elastic Scattering
2. In-Elastic Scattering
Elastic Scattering
Elastic scattering is a form of particle scattering and scattering theory
nuclear physics and particle physics. In this process the K.E of a particle is
conserved in the center-of -mass frame but its direction of propagation is modified.
Further-more while the particle’s K.E in the center of mass frame is constant its
energy in the lab frame is not constant. Generally, elastic scattering describes a
process where the total K.E of the system is conserved.
During elastic scattering of high-Energy subatomic particles “linear
energy transformation takes place until the incident particle’s energy and
speed has been reduced to the some as its surroundings.”
The Origin of Elastic Scattering
The elastic scattering is caused by the incident electron beam interacting with
the nuclei in the specimen. In this atomic model shown below , if the electron
beam gets close to the nucleus of the atom due to the Columbic Force. The path
will be altered and the electron will called the Forward-Scattered electron, which
will be used in TEM. If the incident beam of electrons gets really really close to the
nucleus then it can be swung back and it is termed as Back-Scattered electron or
BSES, which are widely used in SEM.
The
scattering
angle
in
elastic
scattering is
usually
large
as
(theta 1) the
angle from
the
in
elastic
scattering is usually small as (theta 2)
The elastic scatterings are not absolutely elastic in many cases the nuclei in the
specimen slightly slow down the incident electron beam taking a small tae under
energy. The loss of energy is converted into X-rays called Bremsstrahlung Xrays.
In Germen it means Break in, The nuclei in the specimen is slowing down the
incident electron beams if it hitting the break.
All the diffraction patterns we see in the microscope or on the film are actually
caused by the elastic Scattering
Inelastic Scattering
In this scattering process the K.E of the incident particle does not remain
conserved. In an in-elastic process some of the energy of the incident particle is
lost or increased. The internal states of the scattered particles changed which
results in scattering atom to excite some of the electrons of the atom in the
specimen.
Origin Of Inelastic Scattering
It is caused by the incident electron beam interacting with the electron in the
specimen, Some of the energy from the incident electrons will be lost in this type
of scattering. The lost energy will be used to generate different signals. For
example:





Characteristics X-Rays
Cathodolumiescence (CL)
Auger Electrons
Plasmons
Phonons
The loss of energy from the incident electron beam can also cause Beam Damage
of TEM Specimen.
The Figure depicts the probability of signal generation from the electron beam
specimen interaction. The cross section for Plasma is the hieghest followed by the
elastically scattered electrons. Then L & K characteristics x-rays and lastly the
secondary electrons. The signals generated by inelastically scattering will focus on
the characteristic x-rays and Cathodolumiescence (CL)
Cross-Section (𝝈(𝜽𝝋))
Introduction
In nuclear physics one of the ways to study the nucleus of an atom is by nuclear
reactions and one of the very common ways to inducing nuclear reactions is to
bombard incident particles on to some sort of a target nucleus. This is a very
common way of inducing nuclear reactions
However, we have already know that the nucleus is very tiny object, so vast the
majority of space inside in any atom is usually empty space.So in this kind of
experiment when incident particles are bombarded onto some sort of a target
nucleus.
In this case vast majority of incident particles usually penetrate through the empty
space of the material and only few amount of incident particles actually interact
with the target nucleus inside the material and lead to some kind of nuclear
reactions.
Now is it possible to calculate how many of these incident particles will
actually interact to the target nucleus and lead to some sort of nuclear
reactions and how many incident particles will actually penetrate through?
To that kind of calculation we need to understand something know as nuclear
cross-section.
The nuclear cross-section is simply defined as the area around the nucleus on
which if incident particles are incident on it then it will most definitely lead to an
interaction with the nucleus and all the other incident particles outside this area
will penetrate through without interacting with the nucleus.
Dependence of Nuclear Cross-Section
1. The nuclear cross-section depend upon variety of factors i.e. It depends upon the
kind of incident particles (e.g. if the incident particles are Electron, Protons or
Neutrons)
1. If incident particles are Protons
 If we have protons as incident particles then we probability know that the
protons are rebel by the nucleus because proton is positively charged so is a
nucleus.
 So, protons (as incident particles) will most likely get repel by the nucleus
and move away
 It is very difficult to induce a reaction between proton and some sort of a
nucleus
 Therefore, the cross-section area corresponding to incident protons will be
very less
2. If incident particles are Neutrons
 If the incident particles have neutrons in them
 As neutrons are unchanged so they are not repel by the nucleus
 Therefore vast majority of the neutrons which come near the nucleus
will interact with the nucleus
 So, the area of cross-section corresponding to neutrons will be greater
3. The nuclear cross-section also depends upon the energy of incident
particles e.g. in the case of neutrons
 If the velocity of the incident neutrons is very high then that neutron is much
more likely to penetrate through the material without interaction/interacting
to the nucleus
 If the incident neutrons have low velocity or we take thermal neutron will
have a higher de-Broglie’s wavelength and it will be more like for that
particle to interact with the nucleus
By depending upon some factors (kind of incident particles and energy of
incident particles) we can define the nuclear cross-section also known as
interaction cross-section
“The nuclear cross-section/interaction cross-section is that area around the
nucleus facing the incident particles within which the incident particles will
lead to an interaction with nucleus and outside which the incident particles
will not interact with the nucleus”
Greater is the nuclear cross-section more is the like route of interaction.
This area can be greater than the actual size of the nucleus it could be less than the
actual size of the nucleus it could be equal to the actual size of the nucleus Based
upon the nature of the nuclear reaction
Unit
The quantity nuclear cross-section has a unit given by
1barn = 10−28 𝑚2 = 100f𝑚2
If for a particular nuclear reaction we know what is the nuclear crosssection of that reaction, is it possible to calculate how many of the incident
particles will actually interact with the nucleus.
For this we will do the simple derivation of calculating how many the
incident particles actually lead to a nuclear reaction.
Calculation
We have certain kind of incident particles which are incident on to some
sort of material, let suppose we are taking a rectangular slab of that material which
consists the target nucleus
Let the area of cross-section facing the incident
particles is A. let’s take a very small section of this particular rectangular slab
having thickness dx. Now we want to study how many nuclear interactions happen
inside this particular slab.
Now, we will define two quantities N(x) & N(x) + dN
N(x) ≡ No.
particles at distance x
of
incident
dN ≡ (Decrease in number of incident particles) or (
𝑁𝑜.𝑜𝑓 𝑖𝑛𝑡𝑒𝑟𝑎𝑐𝑡𝑖𝑜𝑛
𝑛𝑢𝑐𝑙𝑒𝑎𝑟 𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛
)
To calculate how many interactions will take place we need to know how
much target nuclei exist within this small section,
Let suppose the number density of the target nucleus within this small
section is basically given by
(
Number of atoms
number density inside
)=n=(
)
per unit volume
this kind of material
N = number density
So,
No. of target
no. of density (n) no. of density (n) cross − section thickness
(
) = [(
)(
)(
)(
)]
of Area A
dx
nucleus within 𝐝𝐱
of the material
of the material
No. of target nucleus within dx = nx . A . dx = (n A dx)
Where,
(nAdx) is the number of nuclei which exist inside the small section of slab
However, we also need to know the amount of area which is corresponding to this
no. of nuclei (nAdx) within which incident particles will actually lead to nuclear
interaction, so the amount of nuclear cross-section corresponding to (nAdx ) no. of
nuclei is simply given as
nuclear cross
Area of cross-section = (nAdx) (
)
section 𝜎
∴ 𝜎 = 𝜋𝑅2
Area of cross-section = nAdx𝜎
∴ (𝜎 = Area corresponding to 1 nucleus)
Now, the actual area facing the incident particles is A , But the amount of
area available for interaction is nAdx𝜎, so we can say that
(𝐴𝑟𝑒𝑎 𝑜𝑓 𝑛𝑢𝑐𝑙𝑒𝑎𝑟 𝑐𝑟𝑜𝑠𝑠−𝑠𝑒𝑐𝑡𝑖𝑜𝑛)
(𝑇𝑜𝑡𝑎𝑙 𝑎𝑚𝑜𝑢𝑛𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑎𝑚𝑜𝑢𝑛𝑡 𝑠𝑙𝑎𝑏)
𝑛𝜎𝐴𝑑𝑥
𝐴
𝑑𝑁
𝑁
=
=
(𝑁𝑜. 𝑜𝑓 𝑖𝑛𝑡𝑒𝑟𝑎𝑐𝑡𝑖𝑜𝑛𝑠)
(𝑇𝑜𝑡𝑎𝑙 𝑖𝑛𝑐𝑖𝑑𝑒𝑛𝑡 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠)
𝑑𝑁
𝑁
= n
𝑑𝑁
𝜎 = ( 𝑁 ) / ndx
It is the imaginary area perpendicular to the direction of incident beam
where the probability that the collision will take place between incident
particles & target atom is maximum and outside this area the probability
zero.
Differential Cross-Section
After scattering those particles that do not interact with the target get their motion
in the forward direction (undistributed), but those that interact with the target get
scattered (deflected) at some angle as depicted in the Fig. The number of particles
coming out varies from one direction to other. The number of particles scattered
into an element of solid angle dΩ is proportional to that quantity that plays a
central role in the physics of scattering : the differential cross-section. It is
denoted by
d𝜎(𝜃,𝜑)
dΩ
and is defined as:
“ The number of incident particles scattered into an element of solid angle
𝐝𝛀 in the direction of (𝜽, 𝝋) per unit time per incident flux ”
Given as
d𝜎(𝜃,𝜑)
dΩ
=
1
𝑑𝑁
𝐽𝑖𝑛 dΩdt
where
𝐽𝑖𝑛 = incident flux (incident density)
Incident Flux
No. of incident particles per unit area per unit time
Expression
Consider the incident beam of mono-energetic particles moving
with velocity 𝒗𝒐 , momentum p along z-direction as shown in fig above.
The incident beam may be represented by using the wave function given as 𝒆𝒊𝒌𝒐 𝒛
Where
Z = direction of incident beam
𝑘𝑜 = propagation of the beam
𝑝𝑜
𝑘𝑜 =
ℏ
=
𝑚𝑣𝑜
ℏ
This is function is sufficient to describe the incident beam.
Now this incident beam is scattered by the target atom and the amplitude of the
scattered wave is given by
𝒆𝒊𝒌𝒓
𝒓
𝒇(𝜽) or
𝒆𝒊𝒌𝒓
𝒓
𝒇(𝜽, 𝝋)
Where
K = propagation constant for scattered wave
k.r = (𝑘𝑥 𝑥 + 𝑘𝑦 𝑦 + 𝑘𝑧 𝑧) because beam is scattering in all direction x , y
& z.
𝜃 = the angle measured w.r.t z-direction more precisely; 𝜃 is the minimum angle
made by the scattered wave to the direction of incident beam Z-direction.
1
Also a factor of
appears which represent that scattering will be large at short
𝑟
distance and vice versa (e.g R.F scattering experiment)
Now,
The total probability of scattered beam.
P = | Amplitude |2 = |
𝑒 𝑖𝑘𝑟
𝑟
𝑓(𝜃, 𝜑) |2
2
𝑒 𝑖𝑘𝑧 𝑓(𝜃,𝜑)2
P =
∴ | 𝑒 𝑖𝑘𝑧 |2 = 𝑒 𝑖𝑘𝑧 𝑒 −𝑖𝑘𝑧 = 𝑒 0
𝑟2
P =
|𝑓(𝜃,𝜑)|2
𝑟2
Probability in the Area A
|𝑓(𝜃,𝜑)|2
𝑟2
|𝑓(𝜃,𝜑)|2
𝑟2
dA =
|𝑓(𝜃,𝜑)|2
𝑟2
𝑟 2 dΩ
dΩ =
𝑑𝐴
𝑟2
dA = |𝑓(𝜃, 𝜑)|2 dΩ …………………………..( 1)
Where dΩ is the solid angle subtended by area dA at the centre.
Let  represents the wave function of scattered wave
𝑒 𝑖𝑘𝑟
𝚿𝑠𝑐 =
𝑟
𝑓(𝜃, 𝜑)
The scattered flux 𝐽𝑖𝑛 is given by
𝑖ℏ
𝐽𝑠𝑐 =
(Ψ∇Ψ ∗ − Ψ ∗ ∇Ψ)
2𝜇
𝑖ℏ
𝐽𝑠𝑐 =
(Ψ𝑠𝑐 ∇Ψ∗ − Ψ𝑠𝑐 ∗ ∇Ψ𝑠𝑐 )
2𝜇
Putting value of 𝚿𝑠𝑐
𝐽𝑠𝑐 =
𝑒 𝑖𝑘𝑟
𝑖ℏ
(
2𝜇
𝐽𝑠𝑐 =
𝑖ℏ
2𝜇
𝑒 −𝑖𝑘𝑟
𝑟2
𝐽𝑠𝑐 =
𝑒 𝑖𝑘𝑟
(
𝑟
𝑓(𝜃, 𝜑){
𝑒 𝑖𝑘𝑟
𝑟
𝑓(𝜃, 𝜑)∗ ) −
𝑒 −𝑖𝑘𝑟
𝑟
𝑖ℏ
(
2𝜇
𝑒 𝑖𝑘𝑟
2𝜇
𝑒 𝑖𝑘𝑟 𝑒 −𝑖𝑘𝑟
(−
𝑒 −𝑖𝑘𝑟
𝑓(𝜃, 𝜑)∗ −
𝑒 −𝑖𝑘𝑟
𝑟2
𝑟
𝑒 𝑖𝑘𝑟
𝑟
𝑓(𝜃, 𝜑)
𝑓(𝜃, 𝜑){𝑖𝑘
𝑟
𝑟
2𝜇
(−
𝑖ℏ
𝐽𝑠𝑐 =
𝐽𝑠𝑐 =
2𝜇
2(−
−𝑖 2 ℏk
𝜇𝑟 2
ℏk
𝐽𝑠𝑐 =
𝐽𝑠𝑐 ∝
𝑖𝑘 |𝑓(𝜃,𝜑)|2
𝑟2
𝜇𝑟 2
−
𝑒 𝑖𝑘𝑟 𝑒 −𝑖𝑘𝑟
𝑟3
𝑒 𝑖𝑘𝑟
𝑟
𝑓(𝜃, 𝜑) −
𝑟2
𝑟3
𝑟3
−
)
|𝑓(𝜃, 𝜑)|2
……………………… (2)
Thus, equation (1) represents the scattered flux.
Similarly if 𝚿𝑖𝑛 = 𝑒 𝑖𝑘.𝑟 =𝑒 𝑖𝑘𝑜 𝑧
𝐽𝑖𝑛 =
ℏ𝑘𝑜
𝜇
𝑖𝑘 |𝑓(𝜃,𝜑)|2
|𝑓(𝜃, 𝜑)|2
𝑟2
𝑖𝑘 𝑓(𝜃, 𝜑)𝑓(𝜃, 𝜑) ∗ −
𝑖𝑘 𝑓(𝜃, 𝜑)𝑓(𝜃, 𝜑) ∗ )
𝑖𝑘 |𝑓(𝜃,𝜑)|2
𝑖𝑘 |𝑓(𝜃,𝜑)|2
ℏk |𝑓(𝜃,𝜑)|2
𝜇
𝑒 𝑖𝑘𝑟 𝑒 −𝑖𝑘𝑟
𝑖𝑘 𝑓(𝜃, 𝜑)𝑓(𝜃, 𝜑)∗ −
𝑖𝑘 𝑓(𝜃, 𝜑)𝑓(𝜃, 𝜑) +
𝑖ℏ
Now, let
𝑟
𝑓(𝜃, 𝜑) ∗ ∇
𝑓(𝜃, 𝜑) ∗ )
𝑒 −𝑖𝑘𝑟 𝑒 𝑖𝑘𝑟
𝐽𝑠𝑐 =
𝑟
𝑖ℏ
𝑟
𝑓(𝜃, 𝜑)∇
……………………………… (3)
𝑟2
+
𝑖𝑘 |𝑓(𝜃,𝜑)|2
𝑟3
)
no of particles scattered into
an element of solid angle dΩ
dN(𝜃, 𝜑) = in the direction (𝜃, 𝜑) passing
through a surface
( area = 𝑟 2 dΩ per unit time )
then,
𝑑𝑁 (𝜃, 𝜑)
𝑟 2 dΩdt
𝐽𝑠𝑐 =
dN = 𝐽𝑠𝑐 𝑟 2 𝑑𝑡 ……(4)
or,
𝑑𝑁
dΩ
=
𝑑𝜎
dΩ
𝑑𝜎
dΩ
ℏk
𝜇𝑟 2
=
=
𝑑𝜎
dΩ
𝑟 2 𝑑𝑡 |𝑓(𝜃, 𝜑)|2……….(5)
1 𝑑𝑁 1
𝐽𝑖𝑛 dΩ 𝑑𝑡
ℏk 𝑟 2 𝑑𝑡
𝜇
ℏk𝑜
=
𝜇𝑟 2
𝑘
𝑘𝑜
𝑑𝑡
|𝑓(𝜃, 𝜑)|2
|𝑓(𝜃, 𝜑)|2
This is the expression for the differential cross section of scattering.
Now we shall calculate total cross sectio
Total cross section
The total cross section is the integral of differential cross section over all
the solid angle
𝜎𝑇𝑜𝑡𝑎𝑙
= ∫ 𝑑𝜎
𝜎𝑇𝑜𝑡𝑎𝑙
=
𝑘
𝑘𝑜
∫ |𝑓(𝜃, 𝜑)|2 dΩ
Electron Scattering
A Pictorial View
Before electron beam enters the thin specimen in TEM (Transmission Electron
Microscope). We have uniform intensity of electron spatially. The uniform
intensity is depicted by the horizontal lines. After leaving the specimen the
intensity of the electron beam is not uniform any more and this gives rise to the
contrast we see in TEM as shown blew (Fig. a)
FIGURE
(a)
a
uniform intensity of electrons represented by the horizontal lines, falls on a thin specimen .Scattering within the
specimen changes both the spatial and angular distribution of emerging electrons. The spatial distribution is
indicated by the wave line
If we narrow down to a small region highlighted by the box here and look at the
scattering event again we can have an angular distribution of the forward scattered
beams. Although we start with only one incident beam were in describing the
scattering events there as shown in (Fig. b)
FIGURE (b) the changes in angular distribution is shown by an incident beam of electrons being transformed into
several forward scattered beams
Some terminologies of Scattering
There are some terminologies looking at the scattering direction we have backscattered electron which are mainly used is SEM. we also have the forward
scattered electron many are used in TEM.
This Fig is taken form the Williams and the carter Book and it, nicely
summarizes the scattering events inside the electron microscopes
Phase change / not
Looking at whether there is a phase change or not, this is really concerning
the wave nature of electron beam we have the coherent wave scattered electron
which remain step as well as the incoherently scattered electron which have no
phase relationship with the incident beam or the transmitted beam we
 Coherent -------remain in step
 In-coherent------no-phase relationship
Energy Loss
We can also look at whether we have energy loss as or not
during the scattering process. If there is no energy loss then it is called elastic
scattering, elastically scattered electron are usually coherent but not always. If
there is energy loss during the scattering process then it is termed as inelastic
scattering, in-elastically scattered electron are always incoherent
Elastic-----no energy loss, usually coherent
Inelastic------energy loss, always incoherent
Scattering once or more
We can also look at how many times the scattering event happened within the
specimen, single scattering as the name suggests there is only one scattering event,
scattering more than once is called plural scattering and more than 20-times is
called multiple scattering
Considering scattering there are four questions we need to address
Probability of Scattering
The probability of scattering is described by a concept called
cross-section and the probability of scattering is given as
𝑁𝑜 𝜎𝑎𝑡𝑜𝑚(𝜎𝑡)
𝜎𝑡𝑜𝑡𝑎𝑙 𝑡 =
𝐴
Where,
𝜎𝑡𝑜𝑡𝑎𝑙 = Total cross-section
,
t = spacemen thickness
𝜎𝑡𝑜𝑡𝑎𝑙 𝑡 = 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑜𝑓 𝑠𝑐𝑎𝑡𝑡𝑒𝑟𝑖𝑛𝑔 , 𝑁𝑜 = 𝐴𝑣𝑜𝑔𝑎𝑑𝑟𝑜 ′ 𝑠 𝑛𝑢𝑚𝑏𝑒𝑟
𝜌 = 𝑑𝑒𝑛𝑠𝑖𝑡𝑦
(𝑁𝑜 𝜌)
𝐴
,
A=Atomic weight
= Number of atoms per unit volume
𝜎𝑎𝑡𝑜𝑚 = Cross-section of a single atom
It is the integral of all the scattering events from scattering angle
𝜃 𝑡𝑜 𝜋 moving to what kind of angles we can get from scattering.
𝜋
𝜋 𝑑𝜎
𝜎𝑎𝑡𝑜𝑚 = ∫𝜃 𝑑𝜎 = 2𝜋 ∫0
𝑑Ω
sin 𝜃 𝑑𝜃
Angle of Scattering
For the elastically scattered electron most of them, they usually scatters by
one to the large deviations these electron are not coherent. There are also electron
getting scattered greater than ten degrees and due to the large deviations these
electron are not coherent any more. In coherent the in-elastically scattered
electrons usually have by small angles like less than 1 degree and diseased electron
are always incoherent.
Elastically Scattered Electron:
Usually 1-10° largely coherent
Can be >10° , become incoherent
In-elastically Scattered Electron:
Usually < 1° , always incoherent
Average distance of Scattering
Mean free path: average distance travelled by an electron between scattering
events, it is mathematically expressed as
Mean free path =
1
𝜎𝑡𝑜𝑡𝑎𝑙
=𝑁
𝐴
𝑜 𝜎𝑎𝑡𝑜𝑚 𝜌
From this relationship we can tell that lower the probability of scattering
longer will be the mean free path
The concept of man free path and cross-section are used to beyond TEM.
Energy Loss
Energy loss or not during scattering If there is energy loss, it is inelastic but
if there is no energy loss then it is elastic.
Effect of atomic Number (Z)
Higher the atomic number (z) higher will be the probability of
scattering this lower will be the mean free path e.g. with the some thickness of gold
and copper we see a lot more scattering events from gold then from copper. Also,
higher the atomic number (z) larger will be the scattering angle, let’s take example
of gold and copper again there is a higher probability for electron to be scattered at
a large angle from gold then from copper, this is the origin of Z constant.
Effect of Acceleration Voltage
To look at the effect of the acceleration voltage from TEM,
higher the acceleration voltage less likely for the electron to scatter inside the
material therefore we have a lot more scattering events in the 100kv TEM
compared to a 300kv TEM.
To sound more scientifically we can say that the 100kv TEM
gives a larger cross-section than the 300kv TEM.
Neutron Scattering
Electrons are charged and experienced Strong,Long-Range Coulumb interactions
in a solid. Therefore they typically only penetrate a few atomic layers in the solids.
Electron sCattering is therefore a Surface-Sensitive research. Neutrons are
uncharged and do not experience Coulumb interaction. The Strong Force
interaction is naturally strong but very short-range, and the magnetic interaction is
long range but weak. Neutrons therefore penetrate into most of the aterials, so that
neutron scattering is a bulk exploration.
In an elastic scattering reaction between the neutron and a target nucleus there is
no energy transformation into nuclear excitation, But an elastic scattering plays an
important role in slowing down the neutrons specially at the high energies and by
heavy nuclei.
Generally, a nuetron scattering takes place when a target nucleus emitts a single
neutron after a neutron-nucleus interaction. In an inelastic scattering reaction of
neutron and a target nucleus some of the energy of the incident neutron is absorbed
by the recoiling nucleus and the nucleas remains in excited state. Therefore the
momentum is conserved in an inelastic scattering and th K.E of the system is not
conserved.
Neutron Inelastic Scattering Experiment
The energy of neutron and the internal state of the sample is modified in this
process.The neutron source emit high energy poly-chromatic neutron beam, this
beam travels towards mono-chromator. The monochromatic (name itself says that
this mono-chromator) is going to send high energy beam towards the sample. The
neutron beam collides with the sample and it further travel towards the analyzer.
After striking the analyzer the bam is sent to the detector, the measurement are
done at the detector . If the mono-chromator and the analyzer are showing different
energies i.e
Energy at mono-chromator ≠ Energy at analyzer
The experimental arrangement is shown below
Therefore, we can say that this is an inelastic scattering
𝐸𝑖 =Energy of incident state
𝐸𝑖 = Energy of final state
Since kinetic energy is given as
K.E =
𝑃2
2𝑀𝑛
( kinetic of energy of incident neutron )
𝑀𝑛 = Mass of the neutron
Now,
As
P=ℏK
𝑃2 = ℏ2 𝑘 2
∴K=Wave vector of neutron
K.E =
ℏ2 𝑘 2
2𝑀𝑛
For scattered neutron, we have
K.E =
ℏ2 𝑘 ′
2
2𝑀𝑛
Using law of conservation of energy
ℏ2 𝑘 ′
2
2𝑀𝑛
=
ℏ2 𝑘 2
2𝑀𝑛
± ℏ𝜔 Energy of the photon create (+) or Absorbed (-) in the process
Application
It is used to measure the details of atomic and the molecular motion (we can study
about the atomic and molecular motion of the sample)
i.
ii.
It is used to study the composition of the sample
It is used in condensed matter research to study the atomic and molecular
properties.