ELECTROMAGNETIC SPECTRUM
X-rays are electromagnetic radiations of very short
wavelength ranging from 0.1 Å (0.01 nm) to 100 Å (10 nm).
X-rays can be produced when kinetic energy of fast moving
electrons is transformed into energy of electromagnetic
waves.
Frequencies: 3 x 1016 Hz upward
Wavelengths: 10 nm downward
Quantum energies: 124 eV upward
There are FIVE important processes that may take place
when the fast-moving electrons hit the anode and interact with
the target atoms,
(i) Scattering of electrons by the target,
(ii) excitation of an outer orbital electron,
(iii) ionization of an outer orbital electron,
(iv) ionization followed by the emission of a characteristic
X-rays
(v) Continuous X-rays or bremsstrahlung ("braking
radiation") production.
The first two of these processes lead to the production of heat.
In an X-ray tube 95% to 99% of the energy from decelerating
electrons goes to heat via excitation and ionization of outer
orbital electrons.
The third and fourth of these processes lead to the
production of X-ray photons. Between 1% and 5% goes to
X-ray energy, mostly Bremsstrahlung.
Pyrex glass
Electron envelope
beam
Filament

Cathode
Anode
Tungsten
target
X  rays

Typical X-ray tubes
Production of X-rays
X-rays are produced when electrons with high velocity
(energy) interact with atoms. When high energetic electron
beam incident on a solid target material, most of the energy of
the electrons will be dissipated as heat and a very small
amount of energy is emitted in the form of X-rays.
Production of X-rays continued
When high energetic electrons strike the target, X-rays are
produced.
The wavelength of X-rays emitted will depend on various
factors including,
i)the initial and final energy of the incident electron
ii)nature of the target material
iii)the nature of the interaction.
Hard X-rays and uses
Hard X-rays: possessing higher penetration power (high
applied voltage, lower wavelengths) – materials crystal
structure analysis. (X-ray diffraction experiment, X-ray
scattering experiment, X-ray absorption experiment etc.)
Some X-ray diffraction equipments for material analysis
Debye-Scherrer
camera
attached X-ray
machine
Rigaku-Japan
X-ray diffractometer
Bruker-Germany
Soft X-rays and uses
Soft X-rays: possessing lower penetration power (low applied
voltage, higher wavelengths) – medical imaging.
Diagnostic x-rays:  = 1 to 0.1 Å; Therapeutic:  = 0.1 to 104Å
Typical medical X-ray tube
INTENSITY AND FREQUENCY OF X-RAYS
Intensity of X-rays: depends on the number of electrons
striking the target per second  filament temperature  varies
with filament current.
The Frequency of X-rays: depends on the voltage between
the anode (target) and cathode. Directly proportional to the
applied voltage between the cathode and anode.
PROPERTIES OF X-RAYS
1. They are not charged particles
2. They affect photographic plates
3. They ionize the gas
4. They produce fluorescence in many substances
5. They are highly penetrating. Lead (Pb) is practically
opaque to X-rays
6. They travel in straight lines with the velocity of light
7. They undergo reflection, refraction, interference,
diffraction and polarization like light waves
8. They produce photoelectric effect and thereby exhibit
corpuscular nature also.
X-ray spectra – radiation from X-ray tube
Characteri stic X  rays
I
K
K
Continuous X  rays
Bremsstrahlung
Braking
V  50 kV
V  40 kV
V  30 kV
V  20 kV
 min
 min  min


 min 
max max max

X-ray spectra – radiation from X-ray tube, continued
Characteri stic X  rays
I
K
K
Continuous X  rays
Bremsstrahlung
Braking
V  50 kV
V  40 kV
V  30 kV
V  20 kV
 min
 min  min
 max  max
 min 

max
Based on the characteristics & the origin of X-rays,
X-ray spectra may be classified as,
a)Continuous X-ray spectrum [bremsstrahlung] :
a background of continuous radiation.
b)Characteristic X-ray spectrum: the superimposed
lines on the continuous background, characteristic of
the material of the target, that occurs only when the
applied voltage is greater than a particular value.
Origin of – Continuous X-ray spectrum
– bremsstrahlung – “braking” radiations
Same t arg et
Applied voltage different
I
V  40 kV
V  30 kV
V  20 kV
 min  min
 min
 max max 

max
Result of Interaction (collision) between high speed
electron and nucleus.
As the electron moves past the nucleus of an atom, it
slows down or brakes. Electron is deflected and loses part of
its energy, which is emitted as radiation.
Each electron may have one or more bremsstrahlung
interactions resulting in loss of part or all its energy, therefore
photon may have any energy up to initial electron energy.
"Bremsstrahlung" means "braking radiation" and is retained
from the original German to describe the radiation which is
emitted when electrons are decelerated or "braked" when
they are fired at a metal target. Accelerated charges give off
electromagnetic radiation, and when the energy of the
bombarding electrons is high enough, that radiation is in the
X-ray region of the electromagnetic spectrum. It is
characterized by a continuous distribution of radiation which
becomes more intense and shifts toward higher frequencies
when the energy of the bombarding electrons is increased.
The bombarding electrons can also eject electrons from the
inner shells of the atoms of the metal target, and the quick
filling of those vacancies by electrons dropping down from
higher levels gives rise to sharply defined Characteristic Xrays.
Ei 
1
2
mv 2
Ei  Ef  h
Ef 
1
2
mv 2f
1
Let Ei  mv 2 be the initial energy of the electron and
2
1
Let E f  mv 2f be the final energy of the electron
2
after the interaction with target atom.
Then the energy of the photon emitted is given by
Ei  E f  h 


hc 1
 m v2  v2f ......(1)
 2
m  mass of electron
The short wavelength limit corresponds to the case when the
electron loses all its kinetic energy. That is, v f  0.
Ei  Ef  h max
hc
1

 mv 2 ......(2)
 min 2
Ei  Ef  h max
hc
1

 mv 2 ......(2)
 min 2
Since the kinetic energy of the electron is acquired due to
the application of an electric potential V, we have,
1
mv 2  eV ......(3)
2
Comparing Eqs. (2) and (3) we get,
hc
hc
 eV   min 
........(4)
 min
eV
12400 ( is in angstroms & V is in volts)
  min 
V
Equation (4) is known as Duane-Hunt law (limit) and this
equation can be used for the experimental determination of
Planck’s constant.
Properties of continuous X-ray spectrum
1. Continuous X-ray spectrum is that portion of the
spectrum in which the intensity of X rays varies
continuously & smoothly over a wide range of
wavelength.
2. For each applied voltage that accelerates the
impinging electrons, there is a certain minimum cut off
wavelength called the Duane-Hunt limit. Below this cut
off wavelength no X ray is emitted.
3. The minimum cut off wavelength is independent of the
target element but is dependent on the applied voltage.
4. The cut off wavelength decreases with the increasing
applied voltage & shifts towards the shorter
wavelength.
5. The maximum intensity peak increases with the
increasing applied voltage & shifts towards the shorter
wavelength region.
ORIGIN OF CHARACTERISTIC X-RAY SPECTRUM
V  60 kV
V  50 kV
V  40 kV
Electron transitions to lower atomic levels in heavy atoms
have quantum energies which place them in the X-ray
region of the electromagnetic spectrum. The x-ray
emissions associated with these transitions are called
characteristic X-rays.
During the transition of electron from outer shell n2 to inner
shell n1,the energy difference En2  En1 appears as X-ray
photon of frequency
En2  En1

h
A K series of lines results from the transition of electron
from the higher shell to K shell.
Example: L  K transition  K M  K transition  K
M  L transition  L N  L transition  L
n3M
E0
EO
EN
EM
n2L
EL
n
n5O
n 4N
K
K
n  1 K
L  L L 
EK
I
 min
 K  K
 L  L  L

Summary of continuous and characteristic X-rays
MOSELEY’S LAW
“The frequency of the characteristic x-rays emitted by
different target elements varies directly as the square of
the effective atomic number of the elements.”
Mathematically, Moseley’s law may be stated as
 = a2(Z – b)2   = a(Z – b)
where,  is the frequency of the x-ray emitted, Z is the atomic
number of the element. b is called the screening constant
whose value is different for different series of the x-ray
spectrum. For K series, b=1; for L series, b=7.4. (Z-b) is
called the effective atomic number.
 = a(Z – b)
 = a(Z – b)
The above relation can be obtained by considering Bohr’s
theory with modification for the screening effect of electrons
as follows.
From Bohr’s theory of hydrogen atom, the energy of the
electron in the nth shell is given by,
R  chz 2
En  
n2
R = Rydberg constant = 1.1  107/m
c = speed of light in vacuum = 3  108 m/s
h = Planck’s constant = 6.62  1034J.s
n = principal quantum number, n = 1, 2, 3,…..
2
R
chz
For hydrogen atom, En   
n2
For hydrogen atom, only one electron moves in the field of
positive nucleus, but for the other atoms, the electron is under
the action of two fields – nucleus and other electrons.
Hence the nucleus is screened by electrons surrounding it.
This effect can be involved in the above equation by replacing
Z by (Z – ), where  represents the nuclear screening
constant.
R  chZ   2
 for many electron atom, En  
n2
Hence the energy corresponding to the transition of electron
from level n2 to level n1 is,
1
2 1
h  En2  En1  R  chZ     2  2 
n1 n2 
En2  En1
1
2 1
 
 R  c Z     2  2 
h
n1 n2 
En2  En1
1
2 1

 R  c Z     2  2 
h
n1 n2 
1
1
  R  c  2  2  Z   
n1 n2 

  a z  b 
1
1
where a  R  c  2  2   a cons tan t & b  
n1 n2 
The electrons in the K shell of an atom of an element with
atomic number Z will experience the influence of the entire
nuclear charge Ze. But for an electron in a L shell, the two
1S electrons in the K shell screen the positive charge on the
nucleus & the effective charge experienced by the electron in
the L shell is therefore (Z – 2)e & not Ze.
1
1
  R  c  2  2  Z   
n1 n2 
When an impinging electron removes an electron from the K
shell, the electron in the L shell will see only (Z-1)e as the
effective charge on the nucleus. This is due to the screening
effect of the one K electron. So, for the K X ray, the
screening constant is 1 & the Zeff is (Z – 1).
En2  En1
1
2 1

 R  c Z     2  2 
h
n1 n2 
c
1
1
1
2 1
2 1
 R  cZ  1  2  2  
 R  Z  1  2  2 
K
K
n1 n2 
n1 n2 
Therefore for K line   1 & n1  1
1
1
2
 R  Z  1 1  2 
where, n2  2, 3, 4, .....
K
 n2 
1
2
K  R  cZ  1 1  2 
where, n2  2, 3, 4, ....
 n2 
1
1
1
2
2
 R  Z  1 1  2  & K  R  cZ  1 1  2 
 K
 2 
 2 
1
 K
2
1
 R  Z  1 1  2 
 3 
&
K
2
1
 R  cZ  1 1  2 
 3 
En2  En1
1
2 1

 R  c Z     2  2 
h
n1 n2 
c
1
1
1
2 1
2 1
 R  cZ  1  2  2  
 R  Z  1  2  2 
K
K
n1 n2 
n1 n2 
Similarly for L line, n1  2 & b    7.4
1
1
2 1
 R   Z  7 .4   2  2 
where, n2  3, 4, .....
L
n2 
2
1
2 1
 L  R  c  Z  7 .4   2  2 
where, n2  3, 4, ....
n2 
2
1
1
1
2 1
2 1
 R  Z  7.4   2  2  & L  R  cZ  7.4   2  2 
 L
2
2
3 
3 
1
1
1
1
1
 R  Z  7.4 2  2  2  & L  R  cZ  7.4 2  2  2 
 L
2
2
4 
4 
Applications of Moseley’s Law
1.Periodic table was previously an arbitrary scheme of
classification of elements. Moseley’s law gave a strong
link between the periodic table & the atomic theory. It
provided a systematic way of arranging the elements in
the periodic table with atomic number as the basis.
2. It removed the discrepancies in the arrangement of
certain elements such as,
Argon (Z=18, A=40)
&
Cobalt (Z=27, A=58.9)
&
Tellurium (Z=52, A=127.6) &
Potassium (Z=19, A=39)
Nickel (Z=28, A=58.7)
Iodine (Z=53, A=126.9)
By assigning proper places in the periodic table.
3. It provided a simple, direct & powerful method of
determining the atomic number of elements.
4. It predicted & hence led to the discovery of many
new elements for which gaps were provided in the
periodic table. Elements such as technetium (z=43),
cerium (z=58), hafnium (z=72), Promethium (z=61) etc
were discovered this way.
5. It ruled out the possibility of any new element which
would occupy the periodic table in between the
existing elements.
The potential difference across an X-ray tube is 50kV and the current
through it is 2.5 mA. Calculate (a) the number of electrons striking the
anode per second, (b) the speed with which they strike it and (c) the
approximate rate of production of heat in the anode.
(a) One ampere current is a flow of charge of one coulomb per
second. Thus 2.5 mA is equivalent to,
2.5  10  3
16

1
.
56

10
electrons / sec ond
19
1.6  10
1
2eV
2  1.6  10 19  50  103
2
8


1
.
33

10
m/ s
(b) mv  eV  v 
31
2
m
9.1 10
(c)
watt  volt  ampere
 50  103  2.5  10  3  125 W
Which element (what is the atomic number? Identify the
element) has a K X-ray line whose wavelength is 0.180 nm?
Given: Rydberg constant R = 1.1  107/m
We have,
for K 
2
  a z       a 2  z   
1
1
where, a  R  c  2  2 
n1 n2 
X  rays,   1 & n1  1; n2  2
3
3
R c  a2  R c
4
4
c
3
2
2
2
but
,


 K  a z     R  cz  1
K
 K
4
c
3

 R  c(z  1)2  z  12  676.60 OR z  27
 K 4
(Cobalt)
 a
When electrons bombard a molybdenum target, they produce
both continuous and characteristic X-rays. If the accelerating
potential is 50 keV, determine (a) min (b) the wavelength of
K line and (c) the wavelength of the K line. Given: Atomic
number of molybdenum = 42; Rydberg constant = 1.1 
107m1.
Given : V  50 keV; z  42; R   1.1 107 m1
e  1.60  10 19 C;
c  3  108 m / s
h  6.626  10  34 Js
a)  min
hc

 0.0248 nm
eV
Given : V  50 keV; z  42; R   1.1 107 m1
2
c
1
1
but ,  
we have ,   R  cZ     2  2 

n1 n2 
c
1
1
1
2 1
2 1
 R  c Z     2  2  
 R  Z     2  2 


n1 n2 
n1 n2 
a)

b)

for K  X  rays,   1 & n1  1;
1
 K
1 1 3
 R  Z  12  2  2   R  z  12
1 2  4
for K X  rays,   1 & n1  1;
1
 K
n2  2
 K  0.072 nm
( Answer )
n2  3
1 1 8
 R  Z  12  2  2   R  z  12
1 3  9
 K  0.061 nm
( Answer )
In a characteristic x-ray spectrum of Co, the wavelengths of
K line are 178.9pm for cobalt and 143.5pm for a second
faint line due to impurity. What is the impurity element?
Given :  K(Co )  178.9 pm; z Co  27
 K(Imp)  143.5 pm; z (Imp)  ?
we have ,
  a z   
&
c


for K  X  rays,   1

zImp  1
 Co

 zImp  30 (zinc)
 Imp z Co  1
HRK-Sample Problem
48-1:
Calculate
the
cutoff
wavelength for the continuous spectrum of x-rays
emitted
when
35-keV
electrons
molybdenum target.
Solution:
MIN 
hc
e V
MIN  3.55 x10
 35 .5 pm
11
m
fall
on
a
HRK-Exercise 48.1: Show that the short-wavelength
cutoff in the continuous x-ray spectrum is given by
MIN
1240 pm

where ΔV is the applied potential
V
difference in kilovolts.
Solution: The highest energy x-ray photon will have an
energy equal to the bombarding electrons,
hc
MIN 
e V
1240 pm

V
HRK-Exercise 48.9: X-rays are produced in an x-ray
tube by a target potential of 50.0 keV. If an electron
makes three collisions in the target before coming to
rest and loses one-half of its remaining kinetic energy
on each of the first two collisions, determine the
wavelengths of the resulting photons. Neglect the
recoil of the heavy target atoms.
Solution
Eo  50 KeV (incident electron )
50 KeV
c
E1Photon 
h
2
1
 1 
hc
E1Photon
6.625 x1034 x3 x108
12


49
.
68
x
10
m
3
19
25 x10 x1.6 x10
Energy of electron before the sec ond collision  25KeV
25KeV
c
E2 Photon 
h
2
2
34
8
hc
6.625 x10 x3 x10
12
 2 


99
.
375
x
10
m
3
19
E2 Photon 12.5 x10 x 1.6 x10
Energy of electron before the third collision  12.5KeV
c
E3 Photon  12.5Kev  h
3
34
8
hc
6.625 x10 x3x10
12
 3 


99
.
375
x
10
m
3
19
E3 Photon 12.5 x10 x 1.6 x10
HRK-Exercise 48.12: The binding energies of K-shell
and L-shell electrons in copper are 8.979 keV and
0.951 keV, respectively. If a K x-ray from copper is
incident on a sodium chloride crystal and gives a firstorder Bragg reflection at 15.9 when reflected from the
alternating planes of the sodium atoms, what is the
spacing between these planes ?
Solution:
Ln  2 
BE2  0.951 keV
K
BE1  8.979 keV
Kn  1
Ln  2 
BE2  0.951 keV
K
BE1  8.979 keV
E2  E1  h K
 K
Kn  1
hc

K
34
8
hc
6.625 x10 x3x10


E2  E1 (8.979  0.951) x103 x1.6 x10 19
K  0.154 nm
2d sin   n ,

for first order , n  1

0.154 10 9 m
d

 282 pm.

2 sin  2  sin( 15.9 )
HRK-Exercise 48.5:
Electrons bombard a molybdenum
target, producing both continuous and characteristic xrays.
If the accelerating potential applied to the x-ray
tube is 50.0 kV, what values of (a) λMIN (b) λKβ (c) λK
result ?
The energies of the K-shell, L-shell and M-shell
in the molybdenum atom are –20.0 keV, –2.6 keV and -0.4
keV, respectively.
MIN
1240 pm 1240 pm


 24.8 pm
V
50
E2  E1  h K
hc

K
hc
6.625 x10 34 x3x108
 K 

E2  E1 (20  2.6) x103 x1.6 x10 19
K  71.39 pm
E2  E1  h K
hc

 K
hc
6.625 x10 34 x3x108
  K 

3
19
E2  E1 (20  0.4) x10 x1.6 x10
K  63.37 pm
X-RAYS AND THE NUMBERING OF THE ELEMENTS
Moseley’s observation on the characteristic K x-rays shows
a relation between the frequency (f) of the K
x-rays and
the atomic number (Z) of the target element in the x-ray
tube:
f  C Z  1
C is a constant.
Based on this observation,
the elements are arranged
according to their atomic
numbers in the periodic table
MOSELEY PLOT OF
THE K X-RAYS
Bohr theory and the Moseley plot: Bohr’s formula
for
the
frequency
of
radiation
corresponding
to
a
transition in a one-electron atom between any two
atomic levels differing in energy by ΔE is
E
mZ e
f 

h
8 o2h3
2
In
a
many-electron
atom,
4
 1
1 
 2  2 
ni 
 nf
for
a
K transition,
the
effective nuclear charge felt by an L-electron can be
thought of as equal to +(Z–b)e instead of +Ze, where
b is the screening constant due to the screening effect
of the of the only K-electron.
 Frequency of the K x-ray is
MOSELEY PLOT OF
THE K X-RAYS
m Z  b e4  1
1 
 2  2 
f 
2 3
8 oh
2 
1
2
1
2
and
or
 3 m e4 
 Z  b
f  
2 3 
 32 oh 
f  C Z  1
since b  1
HRK-Sample problem 48-2: Calculate the value of
the constant C in the Moseley’s relation for x-ray
frequency and compare it with the measured slope of
the straight line in Moseley plot.
SOLUTION:
 3me 

c  
2 3 
 32 o h 
4
1
2
2
3m e
c
32  o h 3 / 2
c  4.95x107 Hz1/ 2
fromgraph
c  4.96x107 Hz1/ 2
HRK-Sample Problem 48-3: A cobalt target is bombarded
with electrons, and the wavelengths of its characteristic xray spectrum are measured. A second, fainter characteristic
spectrum is also found, due to an impurity in the target.
The wavelengths of the K lines are 178.9 pm (cobalt) and
143.5 pm (impurity). What is the impurity ?
f  C Z 1
c
co
 C Z co  1
f 
and
Co z X  1


 X zCo  1
Z X  30.0
c

c
X
 C Z X  1
178.9 pm z X  1

143.5 pm 27  1
(Zinc )
What is the screening constant ?
• It is 1 , 9 etc for Hα, Hß lines of Lymen series,
respectively, what it is for other series?
MIT-MANIPAL
BE-PHYSICS-QUANTUM
MECHANICS-2010-2011
59
X-RAY DIFFRACTION
diffraction of X-rays by the crystals
The wave nature of X-rays is depicted by the diffraction
phenomenon. When a beam of monoenergetic X-rays is
made to incident on a sample of a single crystal [say, ZnS,
Calcite (CaCo3) etc,], diffraction occurs resulting in a pattern
consisting of an array of symmetrically arranged diffraction
spots, called Laue’s spots.
When X-ray wavefront falls on the atoms of a crystal, the
atoms act as sources of secondary wavelets. These
secondary wavelets from different sets of points (atoms) form
different wavefronts of X-rays. That is, the arrangement of
atoms in a crystal results in the formation of planes with high
density of atoms which different X-rays preferentially along
specific directions. Hence a study of diffraction pattern helps
in the analysis of the crystal parameters.
Bragg Planes: According to Bragg, in every crystal, several
sets of parallel planes called the Bragg planes could be
identified. Each of these planes had an identical & a definite
arrangement of atoms. Different sets of Bragg planes are
oriented at different angles & are characterized by different
inter planar distances.
Different sets of Bragg planes in a cubic crystal
d
d
A crystal of NaCl showing two sets of Bragg planes
BRAGG’S LAW
D


X
C
N

iB i

d
P
Y
F



P
Q
2d sin 
X/
d
Y/
E
d
d
Z/
Z
  glancing angle;

i    90 ;
PE  EQ  d sin ;
i  angle of incidence
Reflection planes

A
CBP  angle of deviation,    2
Path difference  PE  EQ  2d sin 
The beam is scattered in all the directions by the atoms. The
secondary wavelets from the atoms reinforce only in one
direction for which the angle of reflection is equal to angle
of incidence.
A
C
N

X

iB i

d
P
Y
F



P
Q
2d sin 
X/
d
Y/
E
d
d
Z/
Z
  glancing angle;

i    90 ;
PE  EQ  d sin ;
i  angle of incidence
Reflection planes
D
CBP  angle of deviation,    2
Path difference  PE  EQ  2d sin 
The condition for two wavelets to be in the same phase is,
The path difference between the two reflected rays is
equal to integral multiple of wavelength of X-rays.
A
C
N

X

iB i

d
P
Y
F



P
Q
2d sin 
X/
d
Y/
E
d
d
Z/
Z
  glancing angle;

i    90 ;
PE  EQ  d sin ;
i  angle of incidence
Reflection planes
D
CBP  angle of deviation,    2
Path difference  PE  EQ  2d sin 
That is, the condition for constructive interference is,
PE + EQ = n
n is an integer
d sin + d sin = n
2d sin = n
(Bragg’s law)
A
C
N

X

iB i

d
P
Y
F



P
Q
2d sin 
X/
d
Y/
E
d
d
Z/
Z
  glancing angle;

i    90 ;
PE  EQ  d sin ;
i  angle of incidence
Reflection planes
D
CBP  angle of deviation,    2
Path difference  PE  EQ  2d sin 
2d sin = n
(Bragg’s law)
The above equation together with the requirement that the
angles of incidence and reflection must be equal,
constitutes Bragg’s law of X-ray diffraction.
Bragg’s X-ray spectrometer – verification of Bragg’s law.
3 parts
a) Source of Xrays
b) A crystal on a
turn table
c) A detector
(ionization
chamber)
In practice, the crystal table is geared to the ionization
chamber so that the chamber turns through 2 when the
crystal is turned through , with the direction of incident
beam.
Monochromatic X rays collimated by two fine slits S1 & S2 are
allowed to fall on the reflecting plane of a crystal, mounted on
the turntable at a glancing angle . Starting from a small
angle, the turntable is continuously rotated. The electrometer
readings are noted for different values of . A graph of the
ionization current I versus  is drawn. The peaks
corresponding to 1, 2, 3 are the 1st, 2nd and 3rd order
maxima. It will be observed that
sin1: sin2 : sin3…….: : 1: 2: 3…….
which will verify Bragg’s law.
Bragg’s law is,
2d sin = n
2d sin1 : 2d sin2 : 2d sin3 = 1 : 2 : 3
sin1 : sin2 : sin3 =
1: 2: 3
Applications of Bragg’s law
1.
2.
From the Bragg’s law  = 2dsin  / n. By
measuring  for a given order of reflection from
a given plane of a crystal of known inter planar
spacing, the wavelength  of the monochromatic
X rays may be determined.
The inter planar spacing d for a given crystal
can also be calculated by knowing the
wavelength of the X rays used.
Note: While explaining the above, you need to explain
how to get  for various n.
What is the wavelength of the x rays that emerged as
the first order scattered beam from a crystal of rock salt
that has a spacing of 2.81Å between its principal Bragg
planes, at an angle of 10 relative to the incident beam?
N
Given : d  2.81A
n1
D

X

iB i

d
P
Y
F



P
Q
2d sin 
X
d
/
Y/
E
d
d
Z/
Z
  glancing angle;

i    90 ;
PE  EQ  d sin ;
i  angle of incidence
CBP  angle of deviation,    2
Path difference  PE  EQ  2d sin 
n  1,
Reflection planes
A

C
deviation,
  2


2
10

 5
2
2d sin   n
   0.49 nm
The binding energies of K shell and L shell electrons on
copper are 8.979 and 0.951 keV respectively. If a K X-ray
from copper is incident on a sodium chloride crystal and gives
a first order Bragg reflection at an angle of 74.1 measured
relative to parallel planes of sodium atoms, what is the
spacing between these parallel planes?
E2  0.951keV
Ln  2 
K
Kn  1
E1  8.979keV
hc
E2  E1  hK 
 8.979  0.951keV   K  0.154nm
 K
2d sin   n , for first order , n  1

0.154  10  9 m
 d

 80 pm.

2 sin  2  sin(74.1 )