11 pius augustine x ray basics

A detailed account of X-rays will be available on
my youtube videos
Please search:
xray diffraction series
SH Vision
Please go through
Lectures – 1 to 4
for better
understanding
Dr. Pius Augustine, S H College, Kochi

Röntgen radiation is a form of
electromagnetic radiation having
wavelength in the range of 10 to 0.01 nm
(frequency 3 × 1016 Hz to 3 × 1019 Hz) and
energies in the range 120 eV to 120 keV.
What are X-rays?

Serendipity – Accidental discovery
roentgen discovered X-rays – was a
serendipity
“ Dressed people might be made to appear
unclothed”, this was the speculation that
followed after the discovery of X-rays.

4.Some affects were noticed after covering
the discharge tube with opaque paper.
2.Photographic plate lying in a closed box near
the discharge tube was affected.
1.Glass wall glowed with brilliant greenish light
during discharge.
3.Florescent screen covered with Barium
Plateno Cyanide kept near the tube became
luminous.
OBSERVATIONS

Roentgen concluded, there
was some invisible radiation
penetrating out of the
discharge tube named as
x-ray

• In 1895
• Wilhelm Conrad Roentgen- German physicist
• Discovered X-rays while doing some
experiment with the discharge tube
• when cathode rays are allowed to fall on a
metal target called anticathode placed in there
path, a new rays are produced.
• Radiations are named as x-ray.

X-ray properties
•Electromagnetic waves
•wavelength - 0.01nm to 10nm
•Travel in straight line (v = c )
•Do not contain any charged particles
•Not deflected in electric and magnetic
field

X-ray properties
•Affect photographic plates.
•Ionizes the gas through which pass.
•Produce fluorescence.
•Cause photoelectric effect.
•Highly penetrating.

Coolidge tube Hot cathode ray tube

Lecture 2

When filament act as cathode - is
heated by passing suitable current
through it -emit thermions.
High p.d is maintained b/w the
cathode and a metallic target
accelerate the electrons.

Face of the target is at an angle about
450 relative to the electron beam and
X – rays leave the target pass through
the side of the tube.
Tube is highly evacuated to permit
electrons to get to the target
unimpeded.

An Electrostatic lens to focus the
beam onto a very small spot on the
anode usually made out of
tungsten or molybdenum.
The tube has a window designed for
escape of the generated X-ray
photons.

The anode is specially designed to
dissipate the heat
i. Mechanically spun to
increase the area heated by
the beam.
ii. Cooled by circulating coolantDr. Pius Augustine, S H College, Kochi

Types Of X-rays:
Soft X-rays: X-rays of long
wavelengths, low energy – used for
medical applications
Hard X-rays: X-rays of short
wavelengths, high energy, used for
industrial applications

Production of X-rays:Coolidge X-ray tube
Dr. Coolidge
In 1913 designed Coolidge X-ray tube
Electrons are produced by ionization
using a discharge tube.
Electrons are produced by heating a
filament in vacuum to about 2000ºC

i) A glass tube G: Nearly perfect vacuum of
about 10-5 mm of Hg.
ii) Tungsten Filament F: When F is
heated, electrons are emitted.
iii) Cylindric Shield S: Electrons are
focused by S to the target.

iv) Target T: Under high potential
difference, between F & T, the electrons
are accelerated to high velocity. T is
made of tungsten or molybdenum
having high atomic number & high
melting point.

v) Copper tube A: T is mounted on a
hollow copper tube A and is sloped at
45º to the electron beam.
vi) A part of Kinetic Energy is converted
to heat. 99.8% of energies of the
electron beam will be converted to
heat. Dr. Pius Augustine, S H College, Kochi

vii) The copper conduct heat to the
external cooling fins or water cooling
system.
viii) High voltage needed to operate the
X-ray tube is obtained from a rectifier
containing a suitable transformer.

X U
Unit to measure the
wavelength of X – rays
1XU = 10-3Å

Intensity of x – ray
Depends on the no. of electrons
striking the target, which in turn
depends on temperature of the
filament, which in turn depends on
current through the filament
ie. intensity proportional to current.

Quality of X – ray
Decides the penetrating power
or energy of X – rays.
Depends on p.d b/w filament
and anode
Decides the K.E of the striking
electrons.

How will you controll the intensity and quality of
X-rays?
Intensity – filament current (which
controlls the temperture of the filament
and number of electrons emitted)
Quality – penetrating power of electron
(depends on the p.d applied across anode
and cathode)

Hard and Soft X - rays
High frequency (low λ) x – rays
have high energy (E = hν) and
very high penetrating power
are known as hard X –rays
(industrial use)
Reverse – soft X rays (medical
use) Dr. Pius Augustine, S H College, Kochi

1) Energetic electrons directly collide
with innermost electron shells of the
atoms which are on the top surface of
the target
Shells get excited and the electrons in
the shells will produce hard X-ays.
Production of X rays

2) The energetic electrons which are
passing without colliding innermost shells
Slowdown by the collision with the outer
shells of the atoms (energy is lost)
Energy will be sufficient only to excite the
shells that are above of the innermost
electron shells of the atoms in the inner
region of the target. Dr. Pius Augustine, S H College, Kochi

Since these electrons are retarded by
the outer shells of the target atoms,
that outer shells will be excited and a
continuous spectrum (Bremsstrahlung)
of frequencies from infrared rays to soft
X-rays will be produced by the electrons
that present in the excited shells.Dr. Pius Augustine, S H College, Kochi

Characteristic X - rays
Sharp lines superimposed on
the continuous spectrum are
known as characteristic X
rays which are characteristic
of the target materials.Dr. Pius Augustine, S H College, Kochi

Characteristic spectrum arise,
when a bombarding electron
collides with a target atom has
sufficient energy to remove an
inner shell electron from the
atom.

thVacancy created is filled by electrons from
higher level drop down to it, which release
photon of energy equals difference in
energy b/w levels.
K series Kα - e from L – shell drops to K –
shell
Kβ - from M – shell drop down to K-
shell
L series Lα – from M to L
Lβ- from N to L

•X-rays are high-energy photons
that are produced when electrons
make transitions from one atomic
orbit to another. These transitions
can be generated via the
photoelectric effect.

If we send a photon into an atom with an
energy greater than the binding energy of
an electron in that atom, the photon can
knock that electron out of its orbit, leaving a
hole (or vacancy). This hole can then be
filled by another electron in the atom, giving
off an x-ray in the transition to conserve
energy. This process is fluorescence.

Many different atomic electrons
of different binding energies
can fill this hole, so we would
expect to see many energy
peaks in an x-ray spectrum.

The law states that the frequency of a
spectral line in the X-ray spectrum varies
as the square of the atomic number of
the element emitting it
√ν = K (Z – b).
Where ‘Z’ is the atomic number of the element ,b and K
are constants which depend upon the particular line (
Kα , Kβ …)
Mosley’s Law

b is the intercept on the atomic number axis
called the screening factor and k is the slope of
the line Dr. Pius Augustine, S H College, Kochi

•For Kα line (b=1) , √ν = a (Z – 1),
where a is calculated using
Bohr’s theory as a = 4.98 x
107 (Hz)1/2
•Or freq = (4.98 x 107 )2 ( z-1) 2.

i. Moseley could prove that, it is the
atomic number (and not atomic wt)
of an element determines the
physical and chemical properties .
ii. Helped in the discovery of new
elements - hafnium, illinium , etc
Importance of Mosley’s law

iii. Moseley’s law has helped to find
the atomic numbers of rare-earth
elements and to fix their
positions on the periodic table
correctly.
Importance of Mosley’s law

Mosley was killed in the First World
War at an early age of 27 yrs.

Rapidly moving electrons on colliding
with the target are suddenly brought
to rest (-ve accn)
Continuous X – rays produced in this
process is expressed by German term
‘ bremsstrahlung ‘(breaking radiation)
Continuous X – rays

X – rays produced at a given
accelerating potential V,
vary in λ upto a minimum
(λmin) or Emax.
Continuous X – rays

eV = hνmax Or eV = hc/λmin
This is known as Duane-Hunt law
λmin = hc /eV = 12375 / V
λmin -cut off or threshold wavelength.
As accelerating potential increases λmin
decreases or νmax increases

Energy of X-ray radiation
hv = ½ mv2 – ½ mv1
2
v - initial velocity v 1 - Final velocity
When electrons are completely stopped
v1 = 0
X- ray photon emitted will have maximum
frequency (νmax) Dr. Pius Augustine, S H College, Kochi

↑
νmax
Accelerating voltage →

“The continuous X-rays are called
‘bremstrahlung’. “ Why?
Because they are produce by the
slowing or braking down of the
incident charged particles.

Show that the minimum wavelength
of X-rays emitted from an X-ray
tube under an accelerating voltage
V is λm = (12400/V) Å.

Calculate the minimum
applied potential required to
produce X-rays of 1Å
wavelength.

The p.d across an X-ray tube is 105 V and a current
of 5 mA flows through it. Find the maximum speed
of electrons striking the target. If only 0.15% of the
incident energy is converted into X-radiations, find
the rate of production of heat.
Hint: eV = ½ mv2.
Quantity of charge in one second I = Q = ne,
5 x 10-3 = ne so, n is out.
Total energy of n electrons = neV
Now find 0.15% Dr. Pius Augustine, S H College, Kochi

An X-ray tube is operated with an anode potential
of 10 kV and an anode current of 15 mA. Calculate
i) the number of electrons hitting the anode per
second ii) the rate of production of heat at the
anode stating any assumptions made and iii) the
frequency of the emitting X-ray quantum of
maximum energy.
Hint: eV = ½ mv2.
Quantity of charge in one second I = Q = ne,
15 x 10-3 = ne so, n is out.
Total energy of n electrons = neV (assuming whole energy is converted in to
heat).
eV = hνmax. νmax. – can be determined. Dr. Pius Augustine, S H College, Kochi

The minimum wavelength emitted by
a X-ray tube is 20 pm. What is the
electric field between the cathode
and the target if the distance
between the two electrodes is 75 cm?
eV = hc/λmin. Find V, then E = V/d.

A pictorial representation of x-ray fluorescence using
Bohr model of atomic structure

ORIGINAL PATENT DIAGRAM OF
THE FIRST CAT- SCAN

• X-ray are used to study crystalline and atomic
structures
• X-rays are used in locating broken and diseased
bones and in detecting stones kidneys
• Barium either swallowed or injected as an
enema shows esophagus, stomach or intestine
on the X-ray plate and helps in visualization of
gastro-intestinal tract and its abnormalities

SCATTERING OF X-RAYS
Two types:
1) Thomson scattering or
Coherent scattering
2) Compton scattering

i . Thomson scattering
Incident and scattered x-rays have
same wavelength.
Hence the name Coherent scattering.
ii. Compton scattering
Scattered x-ray consists of two wavelengths
The incident wavelength and a radiation
of longer wavelength. Dr. Pius Augustine, S H College, Kochi

ARTHUR HOLLY COMPTON
(1892-
1962)

COMPTON EFFECT
i)The scattering of a photon by an
electron
ii) Compton effect –is an experimental
proof that photon is born as a particle
iii) Propagates as particle
iv) And dies as particle.

The scattered beam contained two
wavelengths:-
i) One of the beams - the same wavelength
as the incident beam or primary beam.
ii) Second beam - wavelength
longer than that of the primary beam.
Loss of energy is due to elastic interaction.

lines corresponding to
the longer wavelength
in the scattered beam
Compton lines

Elastic collision of x-ray photon with
electron:1. A photon of energy hv & mass
hv/c2 incident on a free electron
at rest.
2. Momentum of incident photon
= mass X velocity
=hvc/c2 = hv/c
Quantum theory explains
compton scattering

iii) Rest mass energy of
electron = moc2
mo- rest mass
iv) initial momentum of
electron is zero (rest).

V) The direction of scattered photons
making an angle φ with the initial
direction of the incident photon
Vi) Energy of scattered photons hv’
(smaller)
Vii.) During collision a part of the
energy is carried by electrons move
with a velocity v along a direction
making an angle θ with the
direction of the primary photon.Dr. Pius Augustine, S H College, Kochi

Viii) Hence ν’ is less than the incident
frequency v or the scattered photon
has a longer wavelength than the
incident photon.
ix) The momentum of the scattered
electron = mv. Its energy = mc2

x.) During collision, total energy &
momentum are conserved.
Total energy before collision =
Total energy after collision
hv + moc2 = hv΄+mc2……(i)

xi.) Momentum of the scattered photon &
scattered electron are resolved along x &
y direction and apply principle of
conservation of momentum.
Total mom before collision =Total mom after collision
hv/c = hvʹ cos φ + mv cos φ
(for x component)………..(ii)
0 = hv’/c sin φ - mv sin θ
(for y component)……….(iii)

mo → rest mass electron,
m- mass at velocity v, c- velocity of light
Relativistic variation of mass of electron
m=m0/√(1-v2/c2) ………..(iv)
∆ λ of the photon (Compton shift equation)
∆ λ= λ’ – λ = h (1-cos
φ)……….(v)
m0c
λ = wavelength of the incident photon
λ’ = wavelength of the scattered photon
h /moc = Compton wavelength of free

3. ∆ λ depends only on angle of
scattering φ .
Result:
1. λ̕ > λ , λ̕ = λ+ (h/moc) (1-cos
2. ∆ λ doesn’t depend on the
wavelength of the incident photon or
the nature of the scatterer.

b) When ф = 90º,
Δλ = h/moc
a) When ф = 0,
Δλ = 0

c) When ф = 180º,
Δλ= h/moc (1- cos 180º)
= 2h/moc
[cos 180º = -1]

d) When ф = 45º,
Δλ = h/moc (1- cos 45º)
= {h(1-1/√2)} / moc

e) When ф = 135º

Photon can get scattered in all directions
Electron get recoiled only in the onward direction
at angles less than 90º
θ = 90º , when ф = 0,
θ = 0º when ф =180º.
Scattering angle and the angle of recoil are
related by the equation
cot θ = (1+ h/λmoc) tan ф/2 ….(vi)

Give the importance of Compton Scattering ?
i. It verifies the particle nature of
electromagnetic waves.
ii. It supports quantum theory.
iii.It verifies Einstein’s relativistic
equation for mass and energy.

When a photon strikes a tightly bound electrons,
what is the effect on the Compton shift?
∆ λ= λ’ – λ = h (1-cos φ)
m0c
When the electron is tightly bound, the whole
atom will recoil and will represent the mass of
the atoms as a whole. ie. m0 is replaced with
mass of the atom, which is very large. So,
Compton shift will be negligibly small. – cannot

Calculate the Compton wavelength
for an electron. (Given h = 6.62 x 10-
34 Js, c = 3 x 108 m/s and m0 = 9.11 x
10-31 kg.

A cosmic-ray photon of energy hν is
scattered through 90o by an electron
initially at rest. The scattered photon
has a wavelength twice that of the
incident photon. Find the frequency of
the incident photon and the recoil angle
of electron.

Bragg ‘s Law.
X – ray diffraction by crystal planes.
nλ = 2d sinθ
d- interplanar spacing
λ- wavelength of X-rays
n- order of diffraction.
For monochromatic X-rays, λ is fixed.

Questions
1.Can X-rays be used for photoelectric effect?
2.Can X-rays be polarized?
3.X-rays and visible light travel at the same speed in
vacuum. Do they travel at the same speed in glass?
4.Can a hydrogen atom emit characteristic X rays?
Comment.
5.Why is exposure to X-ray injurious to health but
exposure to visible light is not, when both are
electromagnetic waves?

For my youtube videos: please visit -
SH vision youtube channel
or
xray diffraction series
SH Vision

92
Appeal: Please Contribute to Prime Minister’s or Chief
Minister’s fund in the fight against COVID-19
Dr. Pius Augustine, Dept of Physics, Sacred Heart College, Thevara
we will
overcome
Thank You
http://piusaugustine.shcollege.ac.in
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Dr. Pius Augustine, Asst. Professor, Sacred Heart College, Thevara, Kochi.

11 pius augustine x ray basics

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