Lesson 9 X-rays Roentgen s Early Experiment How are X-rays Produced Hard and Soft X-RAYS X-ray Spectra X-ray spectra according to Bohr s Theory Properties of X Rays Uses of X Rays Dangers of X Rays Tutorials Photoelectron Effect Energy Levels
Roentgen s Early Experiment Wilhelm K. Roentgen, who in 1901 was the first man to receive the Nobel Prize in physics, first observed x rays in 1895. He was studying the light produced when electricity was passed through a gas in a tube at low pressure. He noted that a paper screen coated with a fluorescent material glowed when it was in the vicinity of the tube under operation. We now know that x rays are produced if electrons are accelerated through a potential of the order of 10 4 or more volts and then allowed to strike a metal target. Classical electromagnetic theory indicates that the deceleration of an electric charge causes it to radiate energy. In this case the form of electromagnetic radiation is called x rays. Early observations showed that this newly detected radiation had greater penetration power than any other electromagnetic radiation known at that time. It was also observed that these x rays affected a photographic film and would ionize atoms. These effects are utilized in detecting devices for x rays. Since x rays are electromagnetic waves, they are not deflected by either an electric or a magnetic field.
Roentgen s Early Experiment
How are X-rays Produced? Filament X-Ray Tube Invented by Coolidge in 1913. The most widely-used laboratory X-Ray source. Major components are a water-cooled target (anode) and a tungsten filament (cathode) that emits electrons. A high potential (up to 60 kv) is maintained between the filament and the anode, accelerating the electrons into the anode and generating X-Rays. Cooling water is circulated through the anode to keep it from melting (>99% of input power generates heat). Interior of the tube is evacuated for the electron beam; thin beryllium windows transmit the X-Rays Ceramic Diffraction X-Ray Tube Schematic View
How are X-rays Produced? Filament is heated, releasing electrons via thermionic emission (V f ~ 10V, If ~ 4A, resulting in T>000 o C) X rays are produced by high-speed electrons bombarding the target Typically < 1% of energy is converted to x rays; the rest is heat Schematic diagram of x-ray tube and circuit
How are X-rays Produced? The electron beam produced and controlled by the current that is passed through the filament. Stable high voltage and filament current power supplies are needed (old-style transformers high frequency supplies). Power rating: applied potential electron beam current (example: 50 kv and 40 ma kw). Maximum power determined by the rate of heat removal (without water, a tube can be destroyed in seconds flow interlocks). The anode is electrically grounded, while the filament is kept at negative kv s (the water-cooled anode won t short out, and the filament is protected by glass insulation).
How are X-rays Produced?
Hard and Soft X-RAYS The wavelength of the x rays is controlled by the applied voltage between the cathode and anode. For the higher potential differences (short wavelengths) the term hard x rays is used and for the lower potential differences (long wavelengths) the term soft x rays is used to describe the quality of the radiation. Since the tube is highly evacuated, the electron beam current (usually in the 10 milliampere range) is determined by the filament current. When these electrons strike the metal anode target some of them generate x rays as they are abruptly brought to rest. This deceleration radiation is called Bremsstrahlung (braking radiation) which appears as the continuous x- ray background shown in Figure
X-ray Spectra The Bremsstrahlung Spectrum
X-ray Spectra Much of the electron energy goes into heating up the anode. Some of the electrons in the beam interact with the innermost, most tightly, bound electrons in the target and "knock" them into excited states. When these excited atoms return to their ground state, photons are emitted. Since the target material is a metal with many electrons, the innermost electron energy levels are of the order of thousands of electron volts (kev). The photons emitted due to this excitation are characteristic of the target material and are called the characteristic spectra. The maximum energy x rays correspond to the conversion of the maximum electron beam energy into a photon, electron beam energy = photon energy maximum ev = hf max = hc/ λ min (30.1) where V is the accelerating voltage for the tube, fmax is maximum x-ray frequency, h is Planck's constant, and e is the change of the electron.
X-ray Spectra The x-ray photons interact with matter through the photoelectric effect and Compton scattering. The photoelectric effect increases rapidly with Z, the atomic number of the target material. The photoelectric interaction probability is proportional to Z and is the greatest for low-energy photons. The Compton effect is relatively independent of energy and the atomic number of the absorber. These two processes determine the absorption coefficient of the material for x rays. The intensity x rays passing through a material of thickness x can be expressed as an exponential function, I is the x-ray intensity at a distance x in the material, µ is absorption coefficient of the material, Δx is the thickness of the material, and IO is the incident intensity of the x rays in joules per unit area per second.
Properties of X Rays i. Like radio waves and other electromagnetic radiation, X rays have a wide range of wavelengths (Obey the relationship. Strong, deeply penetrating, and highly destructive rays with short wavelengths are called hard X rays. Those with longer wavelengths and less penetrating power the type used in medical and dental diagnosis are known as soft X rays. ii. iii. iv. X rays can penetrate some substances more easily than others. For example, they penetrate flesh more easily than bone, and bone more easily than lead. Thus they make it possible to see bones within flesh and a bullet embedded in bone. The ability of X rays to penetrate depends not only on their wavelength, but also on the density and thickness of the substance. X rays affect photographic film in the same way as light rays do. An X-ray photograph is made by passing a beam of X rays through the subject onto photographic film. In a more recent technique, called xeroradiography, an electrostatic ally charged metal plate is substituted for photographic film. When X rays pass through an object and strike the plate, they discharge it in proportion to the density of the object. To bring out the electrostatic image thus formed, the plate is sprayed with a powder that adheres to the charged area. The powder gathers more thickly in heavily charged areas than in lightly charged spots, producing a detailed picture. X rays cause certain substances to glow, or fluoresce.
Properties of X Rays Bragg's Law According to W. L. Bragg, X-ray diffraction can be viewed as a process that is similar to reflection from planes of atoms in the crystal. In Bragg's construct, the planes in the crystal are exposed to a radiation source at a glancing angle θ and X rays are scattered with an angle of reflection also equal to θ. The incident and diffracted rays are in the same plane as the normal to the crystal planes. Constructive interference occurs only when the path length difference between rays scattered from parallel crystal planes would be an integer number of wavelengths of the radiation. When the crystal planes are separated by a distance d, the path length difference would be dsin θ.
Properties of X Rays Thus, for constructive interference to occur the following relation must hold true. n λ = d sin θ This relation is known as Bragg's Law. Thus for a given d spacing and wavelength, the first order peak (n = 1) will occur at a particular θ value. Similarly, the θ values for the second (n = ) and higher order (n > ) peaks can be predicted. From v then 1 ev m mv d sin ev and n m And p h ev m mv P h mv, nh so evm where h mv where p h evm Broglie wavelength The above derivation assumes that phase differences between wavelets scattered at different points depend only on path length differences. It is assumed that there is no intrinsic phase change between the incident and scattered beams or that this phase change is constant for all scattering events. However this is not always the case (see anomalous scattering elsewhere). h is the de
Properties of X Rays
Uses of X Rays In Medicine X rays help dentists detect diseases of the teeth. Doctors use X rays to locate bullets and other foreign objects within the body; to guide them in setting broken bones; and to detect cancer, ulcers, kidney stones, and other abnormalities. Various types of X-ray scanners have been developed that allow highly detailed views of a particular section of the body. One type, known as a CT (computerized tomography) scanner, sends narrow beams of X rays at various angles through a patient's body. The information obtained from the X rays is processed by a computer to produce an image of a crosssection of the body. The image shows much more detail than an ordinary X-ray picture. A section of the body can be studied in three dimensions by producing a series of adjacent cross-sectional images. X rays can halt the growth of cells and even destroy them altogether. They are therefore used to destroy benign and malignant tumors. X rays have also been used in the treatment of leukemia and bursitis.
Uses of X Rays In Industry X rays are used to inspect canned goods and other packaged products. A conveyor carries the goods past a beam of X rays. If a container is improperly filled, or if it contains a foreign substance, the X rays set off an alarm or set into action a device that removes the container from the conveyor. X rays are similarly used to separate beryl from granite and to inspect airplane and automobile parts, rubber goods, plastics, metal castings, and a variety of other products. When used as a target in an X-ray tube, every element gives off X rays of specific wavelengths. These characteristic rays are used to analyze metal alloys, paint pigments, and other substances.
Uses of X Rays In Science (crystallography) Scientists have learned much about the structure of matter by means of X rays. Among other things, they have learned how atoms are arranged in crystals. The average wavelength of X- rays is about equal to the distance between the atoms in crystals. Crystals therefore act as diffraction gratings for X rays. That is, they scatter X- rays in a pattern that shows the positions of their atoms. When the patterns of specific crystalline substances are known, technicians can use X- rays to analyze substances of which they are a part. Petroleum products, metal alloys, and other substances are thus analyzed.
Uses of X Rays Other Uses Airport security and other personnel use X- rays in examining luggage and packages to check for weapons or smuggled articles. X-rays show whether pearls are natural or cultured, and whether gemstones are natural or synthetic. X-rays have also been used to learn whether paintings attributed to noted painters are authentic. Sometimes they have revealed changes made in the original work, or an earlier painting under the one that appears on the surface. In geology labs, the most common targets used are copper and cobalt.
Dangers of X- Rays Because X rays can kill living cells, they must be used with extreme care. When improperly used they can cause severe burns, cancer, leukemia, and cataracts. They can speed aging, reduce immunity to disease, and bring about disastrous changes in the reproductive cells. Lead screens, sheets of lead-impregnated rubber, and leaded glass are used to shield patients and technicians from undesired radiation. The effect of X radiation is cumulative. That is, a number of minor doses over a number of years is equivalent to a large dose at one time.
Tutorials 1. Electrons are accelerated from rest through a p.d of 10Kv in an x ray tube. Calculate: (i) The resultant energy of the electrons in ev. (10 4 ev) (ii) The wavelength of the associated electron waves. (1.3x10-11 m) (iii) The maximum energy and the minimum wavelength of the x ray radiation generated (assume, m e,e, h=6.6x10-34 Js, c=3x10 8 m/s): (1.6x10-15 J, 1.4x10-10 m) Hint:use the de Broglie equation and h P where p h evm evm hc min 1.4x10 ev. The energy of an x ray photon is hf. X rays are emitted from a target bombarded by electrons which have been accelerated from rest through 10 5 V. Calculate the minimum possible wavelength of the X rays assuming that the corresponding energy is equal to the whole of the kinetic energy of one electron. ( assume h, e, c ) ans, 1.43x10-11 m e 10 m (use: hf = ev),
Tutorials 3. a) State the de Broglie relationship in words. b) Show that the speed of electrons which have been accelerated from rest through a p.d of Kv is.6x10 7 m/s and calculate the wavelength with a beam of these electrons. (.7x10-11 m). c) Suggest why electron diffraction is a useful tool for studying the arrangement of atoms in a crystals where the separation of atoms is typically 0.3nm (λ is less than 0.03nm so diffraction is observed) 4. Electrons are accelerated through a p.d of 50V. Calculate, (a) The speed acquired.(4.13x10 6 m/s) (b) Their momentum (P = mv)(37.5x10-5 kgm/s) (c) The de Broglie wavelength associated with them. (h/p = 0.17nm)
Tutorials 5. What is the k.e of an electron with a de Broglie wavelength of 0.1nm.through what p.d should it be accelerated to achieve this value? Assume e, m e, h. h p h m v e, v h m p mv but k. e 1 mv 1 m e h m e (use since v h m e ) also ev k. e, so V h em e, V 0.151Kv 6. An x-ray operates at 30Kv and the current through it is.0ma.calulate: (i) The electrical power output (ii) The number of electrons striking the target per second. (iii)the speed of the electrons when they hit the target (iv) The lower wavelength limit of the x-rays emitted.
Tutorials Solution. (i) ( i) ( ii) of n ( iii) min I e p I electrons x10 1.6x10 1 mv v hc ev VI ne where 3 19 ev m 30x10 60w ev 0.41x10 n stricking 1.3x10 10 3 x1.6 x10 31 9x10 m x.0x10 16 is 16 3 the the 1x10 8 m s number t arg et persecond 7. Monochromatic X-rays of wavelength 1.x10-10 m are incident on a crystal. The ist order diffraction maximum is observed at when the angle between the incident beam and the atomic plane is 1 0 (i) What is the separation of the atomic planes responsible for the diffraction? (d =.89x10-10 m) (ii) What is the highest order Bragg diffraction observable? ( max n = 4)
Tutorials 8. An x-ray machine can accelerate electrons of energies 4.8x10-15J. The shortest wavelength of the x- rays produced by the machine is found to be 4.1x10-11m. Use this information to estimate the value of the plank constant. ( use, hc E hf, h 6.56x10 min 34 Js 9. Molybdemumk a x-rays have wavelength 7x10-11m. Find: (i) The minimum x ray potential difference that can produce these x rays. (17.7Kv) (ii) Their photon energy in e V (17.7 Kv x e = 17.7KeV) 10. The spacing between Principal planes of Nacl crystal is.8å. It is found that the first order Bragg diffraction occurs at an angle of 10 0. What is the wavelength of the x rays? d sin n d sin / n 9.8x10 11 m
Tutorials 11. Bragg s spectrometer is set for the first order reflection to be received by the dectector at a glancing angle of 9.8 0. calculate the angle through which the detector is rotated to receive nd order reflection from the same face of the crystal. d sin 1 glamcing d sin9.3 and d sin angles for and d sin the where first 1 and and second are order This gives θ = 18.8 0 and therefore 0 1 18.8 9.3 9. 5
Photoelectron Effect The photoelectric effect relates to the following phenomena: if a metal surface is illuminated by visible or ultraviolet light radiation, electrons are released provided that the frequency of the radiation exceeds a critical threshold. In 190 Lernard found that the velocity or kinetic energy of the electron emitted from an illuminated metal was independent of the intensity of the particular incident monochromatic light. It appeared to vary only with the wavelength or the frequency of the incident light. Quantum Theory of Radiation, Planck s Constant In 190 Planck had shown that the experimental observations in black body heat radiation could be explained on the basis that the energy from the body was emitted in separate packets of energy. Each packet was called quantum of energy and the amount of energy E carried is given by E =hf Where h = 6.63 10-34 radiation Js called the Planck s constant and f is the frequency of
Photoelectron Effect Work Function The least or minimum amount of work or energy necessary to take free electron out of a metal against the attractive forces of surrounding positive ions is called the work function of the metal, symbol w. The work functions of cesium, sodium and beryllium are respectively about 1.9 ev,.0 ev and 3.9 ev. Electron volt (ev) is a unit of energy equal to 1 electron charge e 1 volt, which is 1.6 10-19 C 1V = 1.6 10-19 J Hence for example the work of sodium, w =ev = 1.6 10-19 J = 3. 10-19 J. Einstein s Particle (Photon) Theory In 1905 Einstein suggested that the experimental results in photoelectricity could be explained by applying a quantum theory of light. He assumed that light of frequency f contains packets or quanta of energy hf. On this basis, light consists of particles called photons. The number of photons per unit area of cross section of the beam of light per second is proportional to its intensity. But the energy of a photon is proportional to its frequency and is independent of the light intensity.
Photoelectron Effect Considering the case of sodium whose work function is ev or 3. 10-19 J. According to the particle theory, if the quantum energy in the incident light is 3. 10-19 J, then the electrons are just liberated from the metal. The particular frequency f is called the threshold frequency of the metal. The threshold wavelength c/ threshold frequency f and threshold for sodium is E=hf w, f. Hence the f w h 3. 10 6.6 10 19 34 14 4.8 10 Hz c f 8 3 10 4.8 10 14 7 6. 10 m So electrons are not liberated from sodium if the incident light has a frequency less than 4.8 10 14 Hz or a wavelength longer than 6. 10-7 m.
Photoelectron Effect Einstein s Photoelectric Equation If the quantum of energy hf in the light incident on a metal is say 4. 10-19 J, and the work function w o of the metal is 3. 10-19 J, the maximum energy of the liberated electrons is (4. 10-19 J - 3. 10-19 J) = 1.0 10-19 J, because w o is the least energy to liberate electrons from the 1 metal. The maximum kinetic energy of the liberated electrons is given by E or m v. max e m E hf max E hf max w w The above equations are called the Einstein s photoelectric equations.
Photoelectron Effect Example Calculate the maximum energy and kinetic energy of emitted electron when sodium is illuminated by radiation of wavelength 150 nm. (w o for sodium is.0ev) Photon energy is given by E = hf h c 6.63 10 3 10 9 150 10 34 8 19 13. 10 J So maximum kinetic energy = hf w 19 19-18 13. 10 1.6 10 J=10 J The threshold frequency is given by f w h 19 1.6 10 J -34 6.63 10 Js 14 4.8 10 Hz
Photoelectron Effect Stopping Potential Stopping potential is defined as the negative potential difference (p.d.) which can stop liberated electrons with maximum energy. Stopping potential is denoted by V s, hence E max =ev s hc hf w w Example Caesium has a work function of 1.9eV. Find i) its threshold wavelength; ii) maximum energy of liberated electrons when th metal is illuminated by light of wavelength 4.5 10-7 m; iii) stopping potential. (take 1eV = 1.6 10-19 J, c = 3 10 8 m/s, h = 6.63 10-34 Js). i) Threshold wavelength 3 10 6.63 10 19 1.9 1.6 10 8 34 c c ch f w / h w 7 6.5 10 m recall that E=hf w
Photoelectron Effect ii) Maximum energy of liberated electrons =hf-w o where f c / hence 34 8 hc 6.63 10 3 10 19 19 Max. Energy w 1.9 10 1.4 10 J 7 4.5 10 iii) the stopping potential V s is given by ev s = 1.4 10-19 J where e = 1.6 10-19 C recall E max =ev s -19 1.4 10 therefore V= 0.9V (approx) -19 1.6 10 Homework 1 Light of frequency 5.0 10 14 Hz liberates electrons with energy.31 10-19 J from a certain metallic surface. What is the wavelength of ultraviolet light which liberates electrons of energy 8.93 10-19 J from the same surface? Take c =3.0 10 8 m/s, h=6.63 10-34 Js. When light of frequency 5.4 10 14 Hz is shone on to a metal surface the maximum energy of the electrons emitted is 1. 10-19 J. If the same surface is illuminated with light of frequency 6.6 10 14 Hz the maximum energy of the electrons emitted is.0 10-19 J. Use this data to calculate a value for the Planck s constant.