Phy 310 chapter 6

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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

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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

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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


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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


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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.


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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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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


8

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


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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.


10

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 :


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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


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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


13

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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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)


L line Electron transition from M shell (n =3)


M lines series is defined as the lines spectra produced

due to electron transition from outer shell to M shell

vacancy.


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.


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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)


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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


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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


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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


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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


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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 :


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Estimate the K wavelength for molybdenum (Z =42).



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


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


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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 :


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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


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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


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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.



Example 4 :

Shell Energy (eV) 103

K 8.5

L 1.0

M 0.5

Table 6.3


30

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


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State Moseley’s Law and explain its impact on the

periodic table.

Learning Outcome:

6.2 Moseley’s law (½ hour)


32

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


33

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


34

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 :


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


36


Derive with the aid of a diagram the Bragg’s equation.

Use

Learning Outcome:

6.3 X-ray diffraction (½ hour)

nd sin2


37

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


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


39

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.


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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


41

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 :


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


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


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


45

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)


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


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


48

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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Phy 310 chapter 6

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