1. Dr Ahmad Taufek Abdul Rahman
School of Physics & Material Studies, Faculty of Applied Sciences, Universiti Teknologi MARA Malaysia, Campus of Negeri Sembilan
is defined as an electromagnetic
radiation of shorter wavelength
than UV radiation produced by the
bombardment of atoms by high
energy electrons in x-ray tube.
CHAPTER 6: X-rays
(2 Hours)
discovered by
Wilhelm Konrad Rontgen
in 1895.
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2. DR.ATAR @ UiTM.NS
PHY310 X-RAY
Learning Outcome:
6.1 X-ray spectra (1 hour)
At the end of this chapter, students should be able to:

Explain with the aid of a diagram, the production of
X-rays from an X-ray tube.

Explain the production of continuous and characteristic
X-ray spectra.

Derive and use the formulae for minimum wavelength for
continuous X-ray spectra,
min

hc

eV
Identify the effects of the variation of current,
accelerating voltage and atomic number of the anode on
the continuous and characteristic X-ray spectra.
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3. DR.ATAR @ UiTM.NS
PHY310 X-RAY
6.1 X-ray spectra
6.1.1 Properties of x-rays

Its properties are
 x-rays travel in a straight lines at the speed of light.
 x-rays cannot be deflected by electric or magnetic fields.
(This is convincing evidence that they are uncharged or
neutral particles)
 x-rays can be diffracted by the crystal lattice if the spacing
between two consecutive planes of atoms approximately
equal to its wavelength.
 x-rays affect photographic film.
 x-rays can produce fluorescence and photoelectric
emission.
 x-rays penetrate matter. Penetration power is least in the
materials of high density.
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4. DR.ATAR @ UiTM.NS
PHY310 X-RAY
6.1.2 Production of x-rays

X-rays are produced in an x-ray tube. Figure 6.1 shows a
schematic diagram of an x-ray tube.
Tungsten target
(anode)
Cooling system

X-rays
Evacuated glass
tube
Heated filament
(cathode)
Power supply
High voltage source
for heater
Figure 6.1
An x-ray tube consists of
 an evacuated glass tube to allow the electrons strike the
target without collision with gas molecules.
Electrons
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5. DR.ATAR @ UiTM.NS
PHY310 X-RAY
a heated filament as a cathode and is made from the
material of lower ionization energy.
 a target (anode) made from a heavy metal of high
melting point such as tungsten and molybdenum.
 a cooling system that is used to prevent the target
(anode) from melting.
 a high voltage source that is used to set the anode at a
large positive potential compare to the filament.
When a filament (cathode) is heated by the current supplied to
it (filament current If), many electrons are emitted by
thermionic emission (is defined as the emission of electrons
from a heated conductor).
These electrons are accelerated towards a target, which is
maintained at a high positive voltage relative to cathode.
The high speed electrons strike the target and rapidly
decelerated on impact, suddenly the x-rays are emitted.




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6. DR.ATAR @ UiTM.NS
PHY310 X-RAY
X-rays emission can be considered as the reverse of the
photoelectric effect. In the photoelectric effect, EM radiation
incident on a target causes the emission of electrons but in
an x-ray tube, electrons incident on a target cause the
emission of EM radiation (x-rays).
 The radiation produced by the x-ray tube is created by two
completely difference physical mechanisms refer to:
 characteristic x-rays
 continuous x-rays (called bremsstrahlung in german
which is braking radiation).
Characteristic x-rays
 The electrons which bombard the target are very energetic
and are capable of knock out the inner shell electrons from
the target atom, creating the inner shell vacancies.
 When these are refilled by electrons from the outer shells,
the electrons making a transition from any one of the outer
shells (higher energy level) to the inner shell (lower energy
level) vacancies and emit the characteristic x-rays.

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7. DR.ATAR @ UiTM.NS

PHY310 X-RAY
The energy of the characteristic x-rays is given by
E  hf  Ef  Ei


(6.1)
Since the energy of characteristic x-rays equal to the difference
of the two energies level, thus its energy is discrete . Then its
frequency and wavelength also discrete.
Figure 6.2 shows the production of characteristic x-rays.
hc
M
L
K
E2  EL  EM  hf2 
2
hc
E1  EK  EL  hf1 
1
High speed electron
vacancy
Figure 6.2
Electron in the shell
Nucleus
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8. DR.ATAR @ UiTM.NS
PHY310 X-RAY
Note:

In the production of the x-rays, a target (anode) made from a heavy
metal of multielectron atom, thus the energy level for multielectron
atom is given by
Z  12 ; n  1,2,3,...
En  13.6 eV
2
(6.2)
n
where

En : energy level of n th state (orbit)
Z : atomic number
n : principal quantum number
Table 6.1 shows a shell designation for multielectron atom.
n
Number of electron
1
K
2
2
L
M
8
3
Table 6.1
Shell
18
4
N
32
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9. DR.ATAR @ UiTM.NS
PHY310 X-RAY
Continuous x-rays (Bremsstrahlung)
 Some of high speed electrons which bombard the target
undergo a rapid deceleration. This is braking.
 As the electrons suddenly come to rest in the target, a part
or all of their kinetic energies are converted into energy of
EM radiation immediately called Bresmsstrahlung, that is
kinetic energy of the electron
K  E energy of EM radiation
1 2
mv  hf
2

Note:

(6.3)
These x-rays cover a wide range of wavelengths or frequencies
and its energies are continuous.
The intensity of x-rays depends on
 the number of electrons hitting the target i.e. the filament
current.
 the voltage across the tube. If the voltage increases so the
energy of the bombarding electrons increases and therefore makes
more energy available for x-rays production.
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10. DR.ATAR @ UiTM.NS
PHY310 X-RAY
Example 1 :
Calculate the minimum energy (in joule) of a bombarding electron
must have to knock out a K shell electron of a tungsten atom
(Z =74).
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11. DR.ATAR @ UiTM.NS
PHY310 X-RAY
Example 1 :
Calculate the minimum energy (in joule) of a bombarding electron
must have to knock out a K shell electron of a tungsten atom
(Z =74).
Solution : ni  1; nf  
By applying the equation of the energy level for multielectron atom,
Z  12
En  13.6 eV
n2
74  12
For K shell, Ei  EK  13.6 eV
2
1
 7.25  10 eV
For n =, Ef  E  0
4
Therefore the minimum energy of the bombarding electron is given
4
by
E  Ef  Ei

E  0   7.25  10




 7.25 10 4 1.60 10 19
11
E  1.16  10 14 J

12. DR.ATAR @ UiTM.NS
PHY310 X-RAY
6.1.3 X-ray spectra

Since there are two types of x-rays are produced in the x-ray
tube, hence the x-ray spectra consist of line spectra (known as
characteristic lines) and continuous spectrum as shown in
Figure 6.3.
Kα
X-ray intensity
The area under the
graph = the total
intensity of x-rays
No x-rays is
produced if
Kγ
  min
0 min
Line spectra
(characteristic lines)
Kβ
1 2 3
Figure 6.3
Continuous
spectrum
Wavelength, 
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13. DR.ATAR @ UiTM.NS
PHY310 X-RAY
At low applied voltage across the tube, only a continuous
spectrum of radiation exists. As the applied voltage
increases, groups of sharp peaks superimposed on the
continuous radiation begin to appear. These peaks are lines
spectra (characteristic lines) where it is depend on the target
material.
Characteristic lines
 The characteristic lines are the result of electrons transition
within the atoms of the target material due to the production of
characteristic x-rays (section 6.1.2).
 There are several types of characteristic lines series:


K lines series is defined as the line spectra produced
due to electron transition from outer shell to K shell
vacancy.
K line
Electron transition from L shell (n =2) to
K shell vacancy (n =1)
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14. DR.ATAR @ UiTM.NS
PHY310 X-RAY
K line
K line

Electron transition from M shell (n =3)
to K shell vacancy (n =1)
Electron transition from N shell (n =4)
to K shell vacancy (n =1)
L lines series is defined as the lines spectra produced
due to electron transition from outer shell to L shell
vacancy.
L line
Electron transition from M shell (n =3)
to L shell vacancy (n =2)
L line
Electron transition from N shell (n =4)
to L shell vacancy (n =2)
L line
Electron transition from O shell (n =5)
to L shell vacancy (n =2)
 M lines series is defined as the lines spectra produced
due to electron transition from outer shell to M shell
vacancy.
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15. DR.ATAR @ UiTM.NS
PHY310 X-RAY
Electron transition from N shell (n =4)
to M shell vacancy (n =3)
M line
Electron transition from O shell (n =5)
to M shell vacancy (n =3)
M line
Electron transition from P shell (n =6) to
M shell vacancy (n =3)
These lines spectra can be illustrated by using the energy level
diagram as shown in Figure 6.4.
Mγ n
EP
6(P shell)
Lγ
M β 5 (O shell)
EO
Mα
Lβ
4 (N shell)
E
M line

N
EM
EL
Kγ
Lα
Kβ
3 (M shell)
2 (L shell)
Kα
EK
Figure 6.4
1 (K shell)
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16. DR.ATAR @ UiTM.NS
PHY310 X-RAY
These characteristic lines is the property of the target material
i.e. for difference material the wavelengths of the
characteristic lines are different.
 Note that the wavelengths of the characteristic lines does not
changes when the applied voltage across x-ray tube changes.
Continuous (background) spectrum
 The continuous spectrum is produced by electrons colliding with
the target and being decelerated due to the production of
continuous x-rays in section 6.1.2.
 According to the x-ray spectra (Figure 6.3), the continuous
spectrum has a minimum wavelength.
 The existence of the minimum wavelength is due to the
emission of the most energetic photon where the kinetic
energy of an electron accelerated through the x-ray tube is
completely converted into the photon energy . This happens
when the electron colliding with the target is decelerated and
stopped in a single collision.

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17. DR.ATAR @ UiTM.NS
PHY310 X-RAY
If the electron is accelerated through a voltage V, the kinetic
energy of the electron is
kinetic energy of the electron
K  U electric potential energy

K  eV

When the kinetic energy of the electron is completely converted
into the photon energy , thus the minimum wavelength min
of the x-rays is
eV  E
hc
eV 
min
hc
min 
eV

(6.4)
From the eq. (6.4), the minimum wavelength depends on the
applied voltage across the x-ray tube and independent of
target material.
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18. DR.ATAR @ UiTM.NS
PHY310 X-RAY
6.1.4 Penetrating power (quality) of x-rays



The strength of the x-rays are determined by their penetrating
power.
The penetrating power depends on the wavelength of the xrays where if their wavelength are short then the penetrating
power is high or vice versa.
By using the eq. (6.4) :
hc Penetrating E  hc

power




increases
decreases
eV
V  
E
P
t
P
X-rays of low penetrating power are called soft x-ray and
those of high penetrating power are called hard x-ray.
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19. DR.ATAR @ UiTM.NS
PHY310 X-RAY
6.1.5 Factors influence the x-ray spectra
X-rays intensity
Filament current
 When
it
is
increased, the
intensity of the
x-ray
spectra
also increased
as shown in
Figure 6.5.
Initial
Final
0 min
Figure 6.5
1 2 3
No change
Wavelength, 
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20. DR.ATAR @ UiTM.NS
PHY310 X-RAY
Applied voltage (p.d.) across x-ray tube
X-rays intensity
 When it is
increased, the
intensity of the xray spectra also
increased but the
minimum
wavelength is
decreased.
 The wavelengths
of the
characteristic
lines remain
unchanged as
shown in Figure
6.6.
0 f 
i
Figure 6.6
1 2 3
No change
Initial
Final
Wavelength, 
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PHY310 X-RAY
Target material
X-rays intensity
 When the target
material is
changed with
heavy material
(greater in atomic
number), the
intensity of the xray spectra
increased, the
wavelengths of
the characteristic
lines decreased.
 The minimum
wavelength
remains
unchanged as
shown in Figure
6.7.
0
min
Figure 6.7
Initial
Final
'
'

1122'3 3
No change
Wavelength, 
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22. DR.ATAR @ UiTM.NS
PHY310 X-RAY
6.1.6 Difference between x-ray emission spectra
and optical atomic emission spectra

is from the production aspect as shown in Table 6.2.
X-ray spectra


Optical atomic spectra
is produced when the
inner-most shell electron
knocked out and left
vacancy. This vacancy is
filled by electron from
outer shells.
The electron transition
from outer shells to inner
shell vacancy emits
energy of x-rays and
produced x-ray spectra.


is produced when the
electron from ground
state rises to the excited
state.
After that, the electron
return to the ground state
and emits energy of EM
radiation whose produced
the emission spectra.
Table 6.2
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23. DR.ATAR @ UiTM.NS
PHY310 X-RAY
Example 2 :
Estimate the K wavelength for molybdenum (Z =42).
(Given the speed of light in the vacuum, c =3.00108 m s1 and
Planck’s constant, h =6.631034 J s)
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24. DR.ATAR @ UiTM.NS
PHY310 X-RAY
Example 2 :
Estimate the K wavelength for molybdenum (Z =42).
(Given the speed of light in the vacuum, c =3.00108 m s1 and
Planck’s constant, h =6.631034 J s)
Solution : Z
 42
The energy level for K and L shells are
Z  12
En  13.6 eV
n2 2
42  1
E  13.6 eV
K
 22862 eV
and
12
42  12
EL  13.6 eV
2
 5715 eV
2
24

25. DR.ATAR @ UiTM.NS
Solution :
PHY310 X-RAY
Z  42
The difference between the energy level of K and L shells is
E  EK  EL
  22862    5715 

 17147  1.60 10 19
E  2.74  10 15 J

Therefore the wavelength corresponds to the E is given by
E 
2.74  10 15
hc

6.63 10 3.00 10 

34
8

  7.26  10 11 m
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26. DR.ATAR @ UiTM.NS
PHY310 X-RAY
Example 3 :
An x-ray tube has an applied voltage of 40 kV. Calculate
a. the maximum frequency and minimum wavelength of the emitted
x-rays,
b. the maximum speed of the electron to produce the x-rays of
maximum frequency.
(Given c =3.00108 m s1, h =6.631034 J s, me=9.111031 kg;
e=1.601019 C and k=9.00109 N m2 C2)
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27. DR.ATAR @ UiTM.NS
PHY310 X-RAY
Example 3 :
An x-ray tube has an applied voltage of 40 kV. Calculate
a. the maximum frequency and minimum wavelength of the emitted
x-rays,
b. the maximum speed of the electron to produce the x-rays of
maximum frequency.
(Given c =3.00108 m s1, h =6.631034 J s, me=9.111031 kg;
e=1.601019 C and k=9.00109 N m2 C2)
Solution : V  40  10 3 V
a. The maximum frequency of the x-rays is
hf max  eV
6.63 10  f
34
max


 1.60 10 19 40 103

f max  9.65 1018 Hz
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28. DR.ATAR @ UiTM.NS
PHY310 X-RAY
Solution : V  40  10 3 V
a. Since the frequency is maximum, thus the minimum wavelength
of x-rays is given by
min 
min
c
f max
3.00  108

9.65  1018
 3.11 10 11 m
b. The maximum speed of the electron is
1
2
mv max  hfmax
2
1
2
9.11 10 31 vmax  6.63 10 34 9.65 1018
2
vmax  1.19 108 m s 1





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29. DR.ATAR @ UiTM.NS
PHY310 X-RAY
Example 4 :
The energy of an electron in the various shells of the nickel atom is
given by Table 6.3.
Shell
Energy (eV)  103
K
8.5
L
1.0
M
0.5
Table 6.3
If the nickel is used as the target in an x-ray tube, calculate the
wavelength of the K line.
(Given the speed of light in the vacuum, c =3.00108 m s1 and
Planck’s constant, h =6.631034 J s)
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30. DR.ATAR @ UiTM.NS
PHY310 X-RAY
Solution :
The difference between the energy level of K and M shells is
E  EK  EM

 

 8.0 10 1.60 10 
  8.5  10 3   0.5  10 3
19
3
E  1.28  10 15 J
Therefore the wavelength corresponds to the E is given by
E 
1.28  10 15
hc

6.63 10 3.00 10 

34
8

  1.55  10 10 m
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31. DR.ATAR @ UiTM.NS
PHY310 X-RAY
Learning Outcome:
6.2 Moseley’s law (½ hour)
At the end of this chapter, students should be able to:

State Moseley’s Law and explain its impact on the
periodic table.
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32. DR.ATAR @ UiTM.NS
PHY310 X-RAY
6.2 Moseley’s law

In 1913, Henry G.J. Moseley studies on the characteristic x-ray
spectra for various target elements using the x-ray diffraction
technique.

He found that the K frequency line in the x-ray spectra from a
particular target element is varied smoothly with that element’s
atomic number Z as shown in Figure 6.8.
8
f K 10
24
Hz
1
2
Zr
16
Cl
8
Figure 6.8
Al
01
8
Y
Cu
Co
Zn
Cr Ni
Ti
Fe
V
K
Si
16
24
32
40
Z
32

33. DR.ATAR @ UiTM.NS

PHY310 X-RAY
From the Figure 6.8, Moseley states that the frequency of K
characteristic lines is proportional to the squared of atomic
number for the target element and could be expressed as
fK
 2.48 10
where

15
HzZ  1
2
(6.5)
f K : frequency of the K line;
Z : atomic number of the target element
Eq. (6.5) is known as Moseley’s law.
Moseley’s law is considerable importance in the development
of early quantum theory and the arrangement of modern
periodic table of element (Moseley suggested the
arrangement of the elements according to their atomic number,
Z).
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34. DR.ATAR @ UiTM.NS
PHY310 X-RAY
Example 5 :
For the K line of wavelength 0.0709 nm, determine the atomic
number of the target element.
(Given the speed of light in the vacuum, c =3.00108 m s1)
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35. DR.ATAR @ UiTM.NS
PHY310 X-RAY
Example 5 :
For the K line of wavelength 0.0709 nm, determine the atomic
number of the target element.
(Given the speed of light in the vacuum, c =3.00108 m s1)
9
Solution : K  0.0709 10 m
The frequency of the K line is given by
fK 
3.00 108
fK 
0.0709 10 9
 4.23 1018 Hz
c
K
By applying the Moseley’s law, thus the atomic number for element
is given by
2
15


 2.48 10 Z  1
f K  2.48 10 Hz Z  1
4.23 1018
Z  42
15
2
35

36. DR.ATAR @ UiTM.NS
PHY310 X-RAY
Learning Outcome:
6.3 X-ray diffraction (½ hour)
At the end of this chapter, students should be able to:

Derive with the aid of a diagram the Bragg’s equation.

Use
2d sin   n
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37. DR.ATAR @ UiTM.NS
PHY310 X-RAY
6.3 X-ray diffraction
6.3.1 Bragg’s law


X-rays being diffracted by the crystal lattice if their wavelength
approximately equal to the distance between two consecutive
atomic planes of the crystal.
The x-ray diffraction is shown by the diagram in Figure 6.9.
C
R
T

i

P
dsin

A
O
air
crystal
B
d
Q
d
dsin
Figure 6.9
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38. DR.ATAR @ UiTM.NS

PHY310 X-RAY
From the Figure 6.9, the path difference L between rays RAC
and TBO is given by ΔL  PB  BQ
ΔL  d sin θ  d sin θ
ΔL  2d sin θ


(6.6)
The path difference condition for constructive interference
(bright) is
(6.7)
ΔL  n ; n  1,2,3,...
By equating the eqs. (6.6) and (6.7), hence
2d sin   n
(6.8)
where

d : separationbetween atomicplanes
 : glancing angle (thecomplementof incident angle or
diffraction angle)
 : wavelength of x - rays
n : diffraction order  1,2,3,...
Eq. (6.8) is known as Bragg’s law and the angle  also known
as Bragg angle.
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39. DR.ATAR @ UiTM.NS
PHY310 X-RAY
Note:

The number of diffraction order n depends on the glancing angle
 where if  is increased then n also increased.

The number of diffraction order n is maximum when the glancing
angle  =90.

If n =1  1st order bright, the angle   1st order glancing angle

If n =2  2nd order bright, the angle   2nd order glancing angle
6.3.2 Uses of x-rays

In medicine, x-rays are used to diagnose illnesses and for
treatment.
 Soft x-rays of low penetrating power are used for x-rays
photography. X-rays penetrate easily soft tissues such as
the flesh, whereas the bones which are high density and
absorb more x-rays. Hence the image of the bones on the
photographic plate is less exposed compared to that of the
soft tissues as shown in Figure 6.10.
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40. DR.ATAR @ UiTM.NS


PHY310 X-RAY
Figure 6.10
 Hard x-rays are used in radio therapy for destroying
cancerous cells. It is found that cancerous cells are more
easily damaged by x-rays than stables ones.
In industry : x-rays are used to detect cracks in the interior of
a metal.
X-rays are used to study the structure of crystal by using xray spectrometry since they can be diffracted (Bragg’s law).
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41. DR.ATAR @ UiTM.NS
PHY310 X-RAY
Example 6 :
A beam of x-rays of wavelength 0.02 nm is incident on a crystal.
The separation of the atomic planes in the crystal is 3.601010 m.
Calculate
a. the glancing angle for first order,
b. the maximum number of orders observed.
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42. DR.ATAR @ UiTM.NS
PHY310 X-RAY
Example 6 :
A beam of x-rays of wavelength 0.02 nm is incident on a crystal.
The separation of the atomic planes in the crystal is 3.601010 m.
Calculate
a. the glancing angle for first order,
b. the maximum number of orders observed.
Solution :   0.02 10 9 m; d  3.60 10 10 m
a. Given n  1
By using the Bragg’s law equation, thus
2d sin   n
 n 
  sin 1  
 2d 
 1 0.02  10 9
 sin 1 
 2 3.60  10 10

  1.59 






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43. DR.ATAR @ UiTM.NS
Solution :
PHY310 X-RAY
  0.02 10 9 m; d  3.60 10 10 m
b. The number of order is maximum when =90, thus
2d sin   n
2d sin 90   nmax 
2d
nmax 

2 3.60  10 10

0.02  10 9
nmax  36


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44. DR.ATAR @ UiTM.NS
PHY310 X-RAY
Intensity
Example 7 :
A
B(25 kV)
5 6 7 8 9 (102 nm)
Figure 6.11
Curves A and B are two x-rays spectra obtained by using two
different voltage. Based on the Figure 6.11 , answer the following
questions.
a. Explain and give reason, whether curves A and B are obtained
by using the same x-ray tube.
b. If curve B is obtained by using a voltage of 25 kV, calculate the
voltage for curve A and obtained the Planck’s constant.
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0
1
2
3
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45. DR.ATAR @ UiTM.NS
PHY310 X-RAY
Solution :A  2.5 10 11 m; B  5.0  10 11 m; VB  25 10 3 V
a. For both curves, the characteristic lines spectra occurred at the
same value of wavelengths. That means the target material used
to obtain the curves A and B are the same but the applied
voltage is increased. Therefore the curves A and B are obtained
by using the same x-ray tube.
b. By applying the equation of minimum wavelength for continuous
x-ray,
hc
(1)
For curve A: A 
eVA
hc
For curve B: B 
eVB
(2)
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46. DR.ATAR @ UiTM.NS
PHY310 X-RAY
Solution :A  2.5 10 11 m; B  5.0  10 11 m; VB
b.
By dividing the eqs. (2) and (1) thus
 20 103 V
 hc 

 eV 
B  B 


A  hc 

 eV 

A 

B VA
VA
5.0  10 11


11
A VB
2.5  10
25  10 3
VA  50  10 3 V
By substituting the value of VA into the eq. (1) :
2.5  10
11


h 3.00  10 8

1.60  10 19 50  10 3



h  6.67  10 34 J s
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47. DR.ATAR @ UiTM.NS
PHY310 X-RAY
Exercise 6.1 :
Given c =3.00108 m s1, h =6.631034 J s, me=9.111031 kg
and e=1.601019 C
1.
Electrons are accelerated from rest through a potential
difference of 10 kV in an x-ray tube. Calculate
a. the resultant energy of the electrons in electron-volt,
b. the wavelength of the associated electron waves,
c. the maximum energy and the minimum wavelength of the xrays generated.
ANS. : 10 keV; 1.231011 m; 1.601015 J, 1.241010 m
2. An x-ray tube works at a DC potential difference of 50 kV.
Only 0.4 % of the energy of the cathode rays is converted into
x-rays and heat is generated in the target at a rate of 600 W.
Determine
a. the current passed into the tube,
b. the velocity of the electrons striking the target.
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ANS. : 0.012 A; 1.33108 m s1

48. DR.ATAR @ UiTM.NS
PHY310 X-RAY
Exercise 6.1 :
3.
Consider an x-ray tube that uses platinum (Z =78) as its
target.
a. Use the Bohr’s model to estimate the minimum kinetic
energy electrons ( in joule) must have in order for K xrays to just appear in the x-ray spectrum of the tube.
b. Assuming the electrons are accelerated from rest through
a voltage V, estimate the minimum voltage required to
produce the K x-rays.
(Physics, 3rd edition, James S. Walker, Q54, p.1069)
ANS. : 1.291014 J; 80.6103 V
4. A monochromatic x-rays are incident on a crystal for which the
spacing of the atomic planes is 0.440 nm. The first order
maximum in the Bragg reflection occurs when the angle
between the incident and reflected x-rays is 101.2. Calculate
the wavelength of the x-rays.
ANS. : 5.591010 m
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49. DR.ATAR @ UiTM.NS
PHY310 X-RAY
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