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# Phy 310 chapter 6

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### Phy 310 chapter 6

1. 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. 1
2. 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. 2
3. 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. 3
4. 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 4
5. 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.     5
6. 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.  6
7. 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 7
8. 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 8
9. 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. 9
10. 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). 10
11. 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. 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,  12
13. 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) 13
14. 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. 14
15. 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) 15
16. 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.  16
17. 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. 17
18. 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. 18
19. 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,  19
20. 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,  20
21. 21. DR.ATAR @ UiTM.NS 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,  21
22. 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 22
23. 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) 23
24. 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. 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 25
26. 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) 26
27. 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 27
28. 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      28
29. 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) 29
30. 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 30
31. 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. 31
32. 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. 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). 33
34. 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) 34
35. 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. 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 36
37. 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 37
38. 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. 38
39. 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. 39
40. 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). 40
41. 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. 41
42. 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        42
43. 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   43
44. 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. 44 0 1 2 3 4
45. 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) 45
46. 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 46
47. 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. 47 ANS. : 0.012 A; 1.33108 m s1
48. 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 48
49. 49. DR.ATAR @ UiTM.NS PHY310 X-RAY 49