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1
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.
Dr Ahmad Taufek Abdul RahmanSchool of Physics & Material Studies, Faculty of Applied Sciences, Universiti Teknologi MARA Malaysia, Campus of Negeri Sembilan
2
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,
Identify the effects of the variation of current,
accelerating voltage and atomic number of the anode on
the continuous and characteristic X-ray spectra.
Learning Outcome:
6.1 X-ray spectra (1 hour)
eV
hcmin
DR.ATAR @ UiTM.NS PHY310 X-RAY
3
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.
6.1 X-ray spectra
DR.ATAR @ UiTM.NS PHY310 X-RAY
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X-rays are produced in an x-ray tube. Figure 6.1 shows a schematic diagram of an x-ray tube.
An x-ray tube consists of
an evacuated glass tube to allow the electrons strike the target without collision with gas molecules.
6.1.2 Production of x-rays
Figure 6.1
X-rays
Heated filament
(cathode)
Tungsten target
(anode)
Electrons
High voltage source
Cooling system
Evacuated glass
tube
Power supply
for heater
DR.ATAR @ UiTM.NS PHY310 X-RAY
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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.
DR.ATAR @ UiTM.NS PHY310 X-RAY
6
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.
DR.ATAR @ UiTM.NS PHY310 X-RAY
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The energy of the characteristic x-rays is given by
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.
if EEhfE (6.1)
KLM
vacancyHigh speed electron
Electron in the shell
Nucleus
1
1LK1
hchfEEE
Figure 6.2
2
2ML2
hchfEEE
DR.ATAR @ UiTM.NS PHY310 X-RAY
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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
Table 6.1 shows a shell designation for multielectron atom.
1,2,3,... ; 1
eV 6.132
2
nn
ZEn (6.2)
where (orbit) state of levelenergy : thnEn
number atomic :Znumber quantum principal :n
n Shell Number of electron
1 K 2
2 L 8
3 M 18
4 N 32Table 6.1
DR.ATAR @ UiTM.NS PHY310 X-RAY
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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
These x-rays cover a wide range of wavelengths or frequencies and its energies are continuous.
hfmv 2
2
1
EK
(6.3)
energy of EM radiationkinetic energy of the electron
Note:
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.
DR.ATAR @ UiTM.NS PHY310 X-RAY
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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).
Example 1 :
DR.ATAR @ UiTM.NS PHY310 X-RAY
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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 :
By applying the equation of the energy level for multielectron atom,
For K shell,
For n =,
Therefore the minimum energy of the bombarding electron is given
by
Example 1 :
fi ;1 nn
2
2
n
1eV 6.13
n
ZE
2
2
Ki1
174eV 6.13
EE
eV 1025.7 40f EE
if EEE 41025.70 E
J 1016.1 14E
194 1060.11025.7
DR.ATAR @ UiTM.NS PHY310 X-RAY
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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.
6.1.3 X-ray spectra
Figure 6.3
Line spectra
(characteristic lines)
Continuous
spectrum
X-ray intensity
Wavelength, 0
min31 2
αK
γKβK
min
No x-rays is
produced if
The area under the
graph = the total
intensity of x-rays
DR.ATAR @ UiTM.NS PHY310 X-RAY
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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)
DR.ATAR @ UiTM.NS PHY310 X-RAY
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L lines series is defined as the lines spectra produced
due to electron transition from outer shell to L shell
vacancy.
K line Electron transition from M shell (n =3)
to K shell vacancy (n =1)
K line Electron transition from N shell (n =4)
to K shell vacancy (n =1)
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)
L line Electron transition from M shell (n =3)
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.
DR.ATAR @ UiTM.NS PHY310 X-RAY
15KE
LE
MENEOEPE
1
2
3
456
n
(K shell)
(L shell)
(M shell)
(N shell)(O shell)(P shell)
These lines spectra can be illustrated by using the energy level diagram as shown in Figure 6.4.
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)
M line Electron transition from N shell (n =4)
to M shell vacancy (n =3)
αK
γK
βKαL
γL
βL αM
γM
βM
Figure 6.4
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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.
DR.ATAR @ UiTM.NS PHY310 X-RAY
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If the electron is accelerated through a voltage V, the kinetic
energy of the electron is
When the kinetic energy of the electron is completely converted
into the photon energy , thus the minimum wavelength min
of the x-rays is
From the eq. (6.4), the minimum wavelength depends on the
applied voltage across the x-ray tube and independent of
target material.
UK eVK
electric potential energykinetic energy of the electron
EeV
min
hceV
eV
hcmin (6.4)
DR.ATAR @ UiTM.NS PHY310 X-RAY
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The strength of the x-rays are determined by their penetrating
power.
The penetrating power depends on the wavelength of the x-
rays where if their wavelength are short then the penetrating
power is high or vice versa.
By using the eq. (6.4) :
X-rays of low penetrating power are called soft x-ray and
those of high penetrating power are called hard x-ray.
6.1.4 Penetrating power (quality) of x-rays
eV
hc
V t
EP
hcE
P
Penetrating
power
increases
decreases
DR.ATAR @ UiTM.NS PHY310 X-RAY
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Filament current
When it is
increased, the
intensity of the
x-ray spectra
also increased
as shown in
Figure 6.5.
6.1.5 Factors influence the x-ray spectra
min31 2
Initial
Final
X-rays intensity
Wavelength, 0
No changeFigure 6.5
DR.ATAR @ UiTM.NS PHY310 X-RAY
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Applied voltage (p.d.) across x-ray tube
Initial
Final
When it is increased, the intensity of the x-ray spectra also increased but the minimum wavelength is decreased.
The wavelengths of the characteristic lines remain unchanged as shown in Figure 6.6.
31 2i
X-rays intensity
Wavelength, 0f
No changeFigure 6.6
DR.ATAR @ UiTM.NS PHY310 X-RAY
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Target material
Initial
Final
Figure 6.7
When the target material is changed with heavy material(greater in atomic number), the intensity of the x-ray spectra increased, the wavelengths of the characteristic lines decreased.
The minimum wavelength remains unchanged as shown in Figure 6.7.
min31 2'
1'2
'3
No change
X-rays intensity
Wavelength, 0
DR.ATAR @ UiTM.NS PHY310 X-RAY
22
is from the production aspect as shown in Table 6.2.
6.1.6 Difference between x-ray emission spectra
and optical atomic emission spectra
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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Estimate the K wavelength for molybdenum (Z =42).
(Given the speed of light in the vacuum, c =3.00108 m s1 and
Planck’s constant, h =6.631034 J s)
Example 2 :
DR.ATAR @ UiTM.NS PHY310 X-RAY
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Estimate the K wavelength for molybdenum (Z =42).
(Given the speed of light in the vacuum, c =3.00108 m s1 and
Planck’s constant, h =6.631034 J s)
Solution :
The energy level for K and L shells are
and
Example 2 :
42Z
2
21
eV 6.13n
ZEn
2
2
K1
142eV 6.13
E
eV 22862
2
2
L2
142eV 6.13
E
eV 5715
DR.ATAR @ UiTM.NS PHY310 X-RAY
25
Solution :
The difference between the energy level of K and L shells is
Therefore the wavelength corresponds to the E is given by
LK EEE
571522862
191060.117147
J 1074.2 15E
hcE
83415 1000.31063.6
1074.2
m 1026.7 11
42Z
DR.ATAR @ UiTM.NS PHY310 X-RAY
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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 =3.00108 m s1, h =6.631034 J s, me=9.111031 kg;
e=1.601019 C and k=9.00109 N m2 C2)
Example 3 :
DR.ATAR @ UiTM.NS PHY310 X-RAY
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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 =3.00108 m s1, h =6.631034 J s, me=9.111031 kg;
e=1.601019 C and k=9.00109 N m2 C2)
Solution :
a. The maximum frequency of the x-rays is
Example 3 :
V 1040 3V
eVhf max
319max
34 10401060.11063.6 f
Hz 1065.9 18max f
DR.ATAR @ UiTM.NS PHY310 X-RAY
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Solution :
a. Since the frequency is maximum, thus the minimum wavelength
of x-rays is given by
b. The maximum speed of the electron is
18
8
1065.9
1000.3
m 1011.3 11min
V 1040 3V
max
minf
c
18342
max31 1065.91063.61011.9
2
1 v
18max s m 1019.1 v
max
2max
2
1hfmv
DR.ATAR @ UiTM.NS PHY310 X-RAY
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The energy of an electron in the various shells of the nickel atom is
given by 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 =3.00108 m s1 and
Planck’s constant, h =6.631034 J s)
Example 4 :
Shell Energy (eV) 103
K 8.5
L 1.0
M 0.5
Table 6.3
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Solution :
The difference between the energy level of K and M shells is
Therefore the wavelength corresponds to the E is given by
MK EEE
33 105.0105.8
193 1060.1100.8
J 1028.1 15E
hcE
83415 1000.31063.6
1028.1
m 1055.1 10
DR.ATAR @ UiTM.NS PHY310 X-RAY
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At the end of this chapter, students should be able to:
State Moseley’s Law and explain its impact on the
periodic table.
Learning Outcome:
6.2 Moseley’s law (½ hour)
DR.ATAR @ UiTM.NS PHY310 X-RAY
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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.
6.2 Moseley’s law
21
Hz10 8K f
Z0 168 3224 40
16
8
24
AlSi
ClK
TiV
CrFe
CoNi
CuZn
ZrY
1Figure 6.8
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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
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).
215K 1Hz 1048.2 Zf (6.5)
where line;K theoffrequency : Kf
element target theofnumber atomic: Z
DR.ATAR @ UiTM.NS PHY310 X-RAY
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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 =3.00108 m s1)
Example 5 :
DR.ATAR @ UiTM.NS PHY310 X-RAY
35
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 =3.00108 m s1)
Solution :
The frequency of the K line is given by
By applying the Moseley’s law, thus the atomic number for element
is given by
Example 5 :
m 100709.0 9
K
K
K
cf
9
8
K100709.0
1000.3
f
Hz 1023.4 18
215
K 1Hz 1048.2 Zf
21518 11048.21023.4 Z
42Z
DR.ATAR @ UiTM.NS PHY310 X-RAY
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At the end of this chapter, students should be able to:
Derive with the aid of a diagram the Bragg’s equation.
Use
Learning Outcome:
6.3 X-ray diffraction (½ hour)
nd sin2
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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.
6.3 X-ray diffraction
Figure 6.9
R
T
A
C
O
QB
P
dsin dsin
i
d
air
crystal
d
DR.ATAR @ UiTM.NS PHY310 X-RAY
38
From the Figure 6.9, the path difference L between rays RAC and TBO is given by
The path difference condition for constructive interference (bright) is
By equating the eqs. (6.6) and (6.7), hence
Eq. (6.8) is known as Bragg’s law and the angle also known as Bragg angle.
BQPBΔ L
θdθdL sinsinΔ
θdL sin2Δ
,...3,2,1 ; Δ nnL
(6.6)
(6.7)
nd sin2 (6.8)
where
planes atomicbetween separation: d
rays- xof wavelength: ,...,, n 321order n diffractio:
or angleincident of complement (the angle glancing: angle)n diffractio
DR.ATAR @ UiTM.NS PHY310 X-RAY
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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 andabsorb 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.
DR.ATAR @ UiTM.NS PHY310 X-RAY
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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 x-ray spectrometry since they can be diffracted (Bragg’s law).
Figure 6.10
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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 3.601010 m. Calculate
a. the glancing angle for first order,
b. the maximum number of orders observed.
Example 6 :
DR.ATAR @ UiTM.NS PHY310 X-RAY
42
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 3.601010 m. Calculate
a. the glancing angle for first order,
b. the maximum number of orders observed.
Solution :
a. Given
By using the Bragg’s law equation, thus
Example 6 :
m 1060.3 m; 1002.0 109 d1n
nd sin2
d
n
2sin 1
10
91
1060.32
1002.01sin
59.1
DR.ATAR @ UiTM.NS PHY310 X-RAY
43
Solution :
b. The number of order is maximum when =90, thus
max90sin2 nd
dn
2max
m 1060.3 m; 1002.0 109 d
nd sin2
9
10
1002.0
1060.32
36max n
DR.ATAR @ UiTM.NS PHY310 X-RAY
44
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.
Example 7 :
0 1 2 3 4 5 6 7 8 9
Inte
nsity
A
B(25 kV)
(102 nm)Figure 6.11
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Solution :
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,
For curve A:
For curve B:
V 1025 m; 100.5 m; 105.2 3B
11B
11A V
B
BeV
hc
A
AeV
hc (1)
(2)
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46
Solution :
b. By dividing the eqs. (2) and (1) thus
By substituting the value of VA into the eq. (1) :
V 1020 m; 100.5 m; 105.2 3B
11B
11A V
A
B
A
B
eV
hc
eV
hc
B
A
A
B
V
V
3
A
11
11
1025105.2
100.5
V
V 1050 3A V
319
811
10501060.1
1000.3105.2
h
s J 1067.6 34h
DR.ATAR @ UiTM.NS PHY310 X-RAY
47
Exercise 6.1 :Given c =3.00108 m s1, h =6.631034 J s, me=9.111031 kg
and e=1.601019 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 x-
rays generated.
ANS. : 10 keV; 1.231011 m; 1.601015 J, 1.241010 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.
ANS. : 0.012 A; 1.33108 m s1
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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 x-
rays 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. : 1.291014 J; 80.6103 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. : 5.591010 m
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DR.ATAR @ UiTM.NS PHY310 X-RAY