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ﻓﻴﺰﻳﺎء ﻃﺒﻴﺔ Medical Physics “For Medicine College” ﺩﻜﺘﻭﺭ ﻤﺤﻤﺩ ﺨﻠﻴل ﺴﻌﻴﺩ Email: [email protected] www.nu.edu.sa/mksaleh.aspx numksaleh Course objectives • 1- Define the terminology of medical physics; • 2- Develop basic understanding of physics concepts; 3- Understanding the possibility of radiation hazards (Ionizing and non ionizing radiation) in medicine; • 4Describe application of physics in medicine such as an imaging system and break it down into its components and physical principles, for each of the imaging modalities covered (x-ray, fluoroscopy, and US); Course objectives • 5Develop basic understanding of radiotherapy • 6- Develop basic understanding of nuclear Medicine • 7- Develop basic understanding of magnatic resonance imaging. • 8- Enhance the background knowledge of the principles of medical instruments; Syllabus 1. Physics Quantities 2.Energy, Work, Power (Kinetic Energy, Unit of Work and Power, Potential Energy, Rest Energy, & Conservation of Energy) 3.Fluids (Density, Specific Gravity, Presssure, Because Pressure, Gauge Pressure, & Archimedes Principle) 4. Heat (Internal Energy, Temperature, Temperature Scales, Celsius and Fahrenheit Scale , Specific Heat Capacity, Change of State, Pressure and Boiling Point) Syllabus 5. Thermodynamics (Heat Engines, First Law of Thermodynamics, Second Law of Thermodynamics, Engine Efficiency, Refrigeration, Heat Transfer: Conduction, Convection, Radiation) 6. Principles of Thermography 7. Optics (Focal Length, Ray Tracing, Lens Equation, Magnification, Lens Systems, ) Syllabus 8. Basic of radiation physics (Atoms, Isotopes, Binding energy, Radiation types and production, Radiation decay and half life, Biological effects of radiation , Radiation units, Limits of radiation doses, Principles of radiation protection) 9. Introduction to Nuclear Medicine (Cyclotron-Produced Radionuclides, Radionuclide Generator, radiopharmaceuticals, Gamma Camera and Photomultiplier Tubes) Syllabus 10.Introduction to Diagnostic Radiology (Xray, Fluoroscopy, Ultrasound ) 11.Magnatisium in medicine (Magnetization Properties, Magnetic Characteristics of the Elements, Treatment of Nervous Diseases and Mesmerism, Magnetic Resonance Imaging). 12.Physics of radiotherapy (External and internal therapy, Co-60 units and medical linear accelerators, Percentage depth dose and tissue air ratio). Assessment Course total mark = 100 marks as follows: a) Mid term ( 40 marks) as following: 1- Practical (10 marks) 2- Essay & attendance (10 marks) 3- Midterm exam (20 marks) b) Course final exam ( 60 marks) as follows: 1- written exam ( 40 marks) 2- practical exam ( 20 marks) Or References References References References General Physics Physics Quantities + Math Calculations using power of ten The rules for finding powers and roots of powers of 10 are: Calculations using power of ten Example: Some common British and SI (metric) units of length and time are: Units Some common British and SI (metric) units of length and time are: Units Subdivisions and multiples of metric units are: Solved Problems Solved Problems Solved Problems Solved Problems Solved Problems Energy, Work, Power • WORK is a measure of the amount of change (in a general sense) a force produces when it arts upon a body. • The change may be in the velocity of the body, in its position, in its size or shape, and so forth. • By definition, the work done by a force acting on a body is equal to the product of the force and the distance through which the force arts, provided that F and s are in the same direction. Thus W = Fs Work = (force)(distance) • If F is perpendicular to s, no work is done. Work is a scalar quantity Energy, Work, Power UNITS OF WORK • In the SI system the unit of work is the joule (J). SI units: 1 joule (J) = 1 newton-meter = 0.738 ft.lb British units: 1 foot-pound (ft.lb) = 1.36 J POWER • Power is the rate at which work is done by a force. P= W / t Power = work done / time UNITS OF POWER • Two special units of power are in wide use, the watt and the horsepower Energy, Work, Power where 1 watt (W) = 1 joule/second = 1.34 x 10-3 hp 1 horsepower (hp) = 550 ft.lb/s = 746 W • A kilowatt (kW) is equal to 103 W or 1.34 hp. • A kilowatthour is the work done in 1 h by an agency whose power output is 1 kW; hence 1 kW . h = 3.6 x 106 J. ENERGY • Energy is the property something has which enables it to do work. • The more energy something has, the more work it can perform. • Every kind of energy falls into one of three general categories: kinetic energy, potential energy, and rest energy. Energy, Work, Power • The units of energy are the same as those of work, namely, the joule and the foot-pound. KINETIC ENERGY • The energy a body has by virtue of its motion is called kinetic energy. • lf the body's mass is m and its velocity is v, its kinetic energy is Kinetic energy = KE = 1 /2. mv2 POTENTIAL ENERGY • The energy a body has by virtue of its position is called potential energy. A nail held near a magnet has magnetic potential energy because the nail can do work as it moves toward the magnet. Energy, Work, Power • The gravitational potential energy of a body of mass m that is at a height h above some reference level is Gravitational potential energy = PE = m g h where g is the acceleration of gravity. In terms of the weight w of the body, PE = wh REST ENERGY • Matter can be converted into energy, and energy into matter. • The rest energy of a body is the energy it has by virtue of its mass alone. Thus mass can be regarded as a form of energy. Energy, Work, Power • The rest energy of a body is in addition to any KE or PE it might also have. • If the mass of a body is mo when it is at rest, its rest energy is Rest energy = Eo = moc2 • In this formula c is the velocity of light, the value of which is c = 3.00 x 108 m/s = 9.83 x 108 ft/s = 186,000 mi/s • The rest mass mo, is specified here because the mass of a moving body increases with its velocity; the increase is only significant at extremely high velocities, however. Energy, Work, Power CONSERVATION OF ENERGY • According to the law of conservation of energy, energy cannot be created or destroyed, although it can be transformed from one kind into another. • The total amount of energy in the universe is constant. A falling stone provides a simple example: more and more of its initial potential energy turns into kinetic energy as its velocity increases, until really all of its PE has become KE when it strikes the ground. The KE of the stone is then transferred to the ground by the impact. Energy, Work, Power • In general, Work done on an object = change in the object's KE + change in the object's PE + work done by the object Solved Problems • 1. A force of 60 lb is used to push a 150-lb crate a distance of 10 ft across a level warehouse floor. (a) How much work is done? (b) What is the change in the crate's potential energy? • Solution: (a) The weight of the crate docs not matter here since its height does not change. The work done is W= Fs = (60 lb)(10 ft) = 600 ft.lb (b) The crate's height does not change, so its potential energy remains the same. Solved Problems • 2. (a) How much work is done in raising a 2000-lb elevator cab to a height of 80 ft? (b) How much potential energy does the cab have in its new position? • Solution: (a) The force needed is equal to the weight of the cab. Hence W= Fs = wh = (2000 lb)(80 ft) = 1.6 x 105 ft.lb (b) PE = wh = 1.6 x 105 ft.lb Solved Problems • 3. A 150-1b man runs up a staircase 10 ft high in 5 s. Find his minimum power output in hp. • Solution: The minimum downward force the man's legs must exert is equal to his weight of 150 lb. Hence Since 1 hp = 550 ft.lb/s, Solved Problems • 4. A motorboat requires 80 hp to move at the constant velocity of 10 mi/h. How much resistive force does the water exert on the boat at this velocity? • Solution: Since v = s / t, P =W /t = Fs /t = Fv. Here Solved Problems • 5. Find the kinetic energy of a 1000-kg car whose velocity is 20 m/s. • Solution: KE = ½.mv2 = ½(1000 kg)(20 m/s)2 = 2 x 105 J • 6. What velocity does a 1-kg object have when its kinetic energy is 1 J? • Solution: Since KE = ½ mv2, Solved Problems • 7. How much mass is converted into energy per day in a nuclear power plant operated at a level of 100 megawatts (100 x 106 W)? • Solution: There are 60 x 60 x 24 = 86,400 s/day, so the energy liberated per day is Eo = Pt = (108 w)(8.64 x 104 s) = 8.64 x 1012 J Since Eo = moc2, Fluids DENSITY • The density (d) of a substance is its mass per unit volume. • The SI unit of density is the kilogram per cubic meter (kg/m3). • The density of aluminum, for instance, is 2700 km/m3. Another common unit of density is the gram per cubic centimeter (g/cm3). Since 1 k = 1000 g and 1 m3 = (100 cm)3 = 106 cm3, 1 g/cm3 = 103 kg/m3 • Hence the density of aluminum can also be given as 2.7 g/cm3. Fluids SPECIFIC GRAVITY • The specific gravity (or relative density) of a substance is its density relative to that of pure water, which is d(water) = 1000 kg/m3 = 1.00 g/cm3 = 1.94 slugs/ft3 • The weight density of water is dg(water) = 62 lb/ft3 • Since the density of water is 1 g/cm3, the specific gravity of a substance is the same as the numerical value of its density when given in g/cm3 Thus the specific gravity of aluminum is 2.7. Fluids PRESSSURE • When a force act: perpendicular to a surface, the pressure exerted is the ratio between the magnitude of the force and the area of the surface: P = F / A Pressure = Force / Area • Pressures are properly expressed in Rascals (1 Pa = 1 N/m2 ) or in lb/ft2, but other units are often used: 1 lb/in2 = 144 lb/ft2 1 atmosphere (atm) = average pressure exerted by the earth's atmosphere at sea level = 1.013 x 105 Pa = 14.7 lb/in2 . Fluids • 1 bar = 105 Pa (slightly less than 1 atm) • 1 millibar (mb) = 100 Pa (widely used in meteorology) • 1 torr = 133 Pa (used in medicine for blood pressure) • The conversion factor is: 1 mm Hg = 133 pascals = 0.02 pounds per square inch. Blood pressure ¾ Because pressure is commonly measured by its ability to displace a column of liquid in a manometer, pressures are often expressed as a depth of a particular fluid (e.g., inches of water). The most common choices are mercury (Hg) and water; water is nontoxic and readily available, while mercury's high density allows for a shorter column (and so a smaller manometer) to measure a given pressure Blood pressure ¾ Blood pressure (BP) is the pressure exerted by circulating blood on the walls of blood vessels, and is one of the principal vital signs. During each heartbeat, BP varies between a maximum (systolic) and a minimum (diastolic) pressure. ¾ The doctor measures the maximum pressure (systolic) and the lowest pressure (diastolic) made by the beating of the heart. ¾ The systolic pressure is the maximum pressure in an artery at the moment when the heart is beating and pumping blood through the body. ¾ The diastolic pressure is the lowest pressure in an artery in the moments between beats when the heart is resting. Blood pressure ¾ A mercury sphygmomanometer is operated by inflating a rubber cuff placed around a patient's arm until blood flow stops. The cuff pressure is measured via the mercury column. The inflating bulb is used to inflate the cuff. It contains two one- way valves. Valve A allows air to enter the back of the bulb. When the bulb is squeezed this valve closes and the air is propelled through valve B to the cuff. Valve B stops the air going back into the bulb. Blood pressure ¾ After the cuff has been inflated and the blood pressure taken, the cufy may be deflated by opening valve C. The reservoir contains the supply of mercury which rises up the measurement tube. Normally the apparatus is contained within a box. When opened the graduated tube becomes vertical, and the mercury reservoir is at the bottom. As the pressure within the cuff increases the mercury is displaced from the reservoir into the graduated tube. The two leather discs (D and E) allow air to pass in and out of the column, but prevent mercury escaping from the sphygmomanometer. Blood pressure ¾ Normal values: In a study of 100 subjects with no known history of hypertension, an average blood pressure of 112/64 mmHg was found, which is in the normal range Fluids GAUGE PRESSURE • Pressure gauges measure the difference between an unknown pressure and atmospheric pressure. What they measure is known as gauge pressure, and the true pressure is known as absolute pressured: P = Pgauge + Patm Absolute pressure = gauge pressure + atmospheric pressure • A tire whose gauge pressure is 30 lb/in2 contains air at an absolute pressure of 45 lb/in2, since sea-level atmospheric pressure is about 15 lb/in2. Fluids ARCHIMEDES PRINCIPLE • An object immersed in a fluid is acted upon by an upward force that arises because pressures in a fluid increase with depth. Hence the upward force on the bottom of the object is more than the downward force on its top. The difference between the two, called the buoyant force, is equal to the weight of a body of the fluid whose volume is the same as that of the object. • This is Archimedes' principle: The buoyant force on a submerged object is equal to the weight of fluid the object displaces. Fluids • lf the buoyant force is less than the weight of the object itself, the object sinks', if the buoyant force equals the weight of the object, the object floats in equilibrium at any depth in the fluid; if the buoyant force is more than the weight of the object, the object floats with part of its volume above the surface. Solved Problems • 1. The specific gravity of gold is 19. (a) What is the mass of a cubic centimeter of gold? (b) What is the weight of a cubic inch of gold? • Solution: (a) Since the density of water is 1 g/cm3, the density of gold is 19 g/cm3 and 1 cm3 has a mass of 19 g. • (b) Since the weight density of water is 62 lb/ft3 the weight density of gold is dg = 19 x 62 lb/ft3 = 1200 lb/ft3. Because 1 ft3 = 12 in. x 12 in. x 12 in. = 1728 in3, a cubic inch of gold weighs Solved Problems • 2. An oak beam 10 am by 20 cm by 4 m has a mass of 58 kg. (a) Find the density and specific gravity of oak. (b) Does oak boat in water? • Solution: (a) The volume of the beam is V = 0.1 m x 0.2 m x 4 m = 0.08 m3 and so its density is d = m / V = 58 kg / 0.08 m3 = 725 kg/m3 since the density of water is 1000 kg/m3, the specific gravity of oak is sp gr = doak /dwater = 725 /1000 = 0.725 (b) Any material whose specific gravity is less than 1 floats in water, so oak does. Solved Problems • 3. How much does the air in a room 12 ft square and 10 ft high weigh? The weight density of air is 0.08 lb/ft3 at sea level. • Solution: The volume of the room is V = (12 ft)(12 ft)(10 ft) = 1440 ft3. Hence the weight of the air is w = (dg)V = (0.08 lb/ft3)(1440 ft3) = 115 lb Solved Problems • 4. A 130-lb woman balances on the heel of her right shoe, which is 1 in. in radius. How much pressure' does she exert on the ground? How does this compare with atmospheric pressure? • Answer: The area of the heel is A = πr2 = 3.14 in2, so the pressure is P = F/A = 130 lb / 3.14 in2 = 41.4 lb/in2 Since patm. = 14.7 lb/in2, this pressure is 2.8 times greater. Solved Problems • 5. An airplane whose mass is 20,000 kg and whose wing area is 60 m2 is in level flight. What is the average difference in pressure between the upper and lower surfaces of its wings? Express the answer in rascals and in atmospheres. • Answer: The upward force on an airplane in level fight is equal to its weight, which here is F = w = mg = (20,000 kg)(9.8 m/s2) = 1.96 x 105 N The pressure difference p is therefore P = F /A = 1.96 x 105 N / 60 m2 = 3267 Pa Since 1 atm = 1.013 x 105 Pa, Solved Problems • 6. The interior of a submarine located at a depth of 50 m in seawater is maintained at sea-level atmospheric pressure. Find the force acting on a window 20 cm square. The density of seawater is 1.03 x 103 kg/m3. • Answer: The pressure outside the submarine is p = patm + dgh and the pressure inside it is patm. Hence the net pressure p' acting on the window is p'= dgh = (1.03 x 103 kg/m3)(9.8 m/s2)(50 m) = 5.05 x 105 Pa Since the area of the window is A = (0.2 m)(0.2 m) = 0.04 m2, the force acting on it is F = p'.A = (5.05 x 105 Pa)(4 x 10-2 m2) = 2.02 x 104 N Solved Problems • 7. An iron anchor weighs 200 lb in air. How much force is required to support the anchor when it immersed in seawater? The weight density of iron is 480 lb/ft3 and that of seawater is 64 lb/ft3 . • Answer: Since dg = w / V, the volume of the anchor is V = w / dg = 200 lb / 480 lb/ft3 = 0.417 ft3 The weight of seawater displaced by the anchor is w = (dg)V = (64 lb/ft3)(0.417 ft3) = 27 lb Thus the buoyant force on the anchor is 27 lb and the net force needed to support it is 200 lb - 27 lb = 173 lb • 8. A l00-gallon steel tank weighs 50 lb when empty. Will it float in seawater when filled with gasoline? The weight density of gasoline is 42 lb/ft3 , that of seawater is 64 lb/ft3, and 1 gallon = 0.134 ft3 . • Answer: The volume of the tank is V = (100 gal)(0.134 ft3/gal) = 13.4 ft3. The total weight of the tank when filled with gasoline is w = 50 lb + (dg)gasoline V = 50 lb + (42 lb/ft3)(13.4 ft3) = 50 lb + 563 lb = 613 lb The maximum buoyant force on the tank is exerted when the tank is completely submerged. Thus Fmax = (dg)water V = (64 lb/ft3)(13.4 ft3) = 858 lb Since the weight of the filled tank is less than 858 lb, it will float. Heat INTERNAL ENERGY • Every body of matter, whether solid, liquid, or gas, consists of atoms or molecules which are in rapid motion. The kinetic energies of these particles constitute the internal energy of the body of matter. • The temperature of the body is a measure of the average kinetic energy of its particles. • Heat may be thought of as internal energy in transit. When heat is added to a body, its internal energy increases and its temperature rises; when heat is removed from a body, its internal energy decreases and its temperature falls. Heat TEMPERATURE • Temperature is familiar as the property of a body of matter responsible for sensations of hot or cold when it is touched. • Temperature provides an indicator of the direction of internal energy flow: when two objects are in contact, internal energy goes from the one at the higher temperature to the one at the lower temperature, regardless of the total amounts of internal energy in each one. • Thus if hot coffee is poured into a cold cup, the coffee becomes cooler and the cup becomes warmer. Heat • A thermometer is a device for measuring temperature. • Matter usually expands when heated and contracts when cooled, the relative amount of change being different for different substances. • This behavior is the basis of most thermometers, which make use of the different rates of expansion of mercury and glass, or of two metal strips joined together, to indicate temperature. TEMPERATURE SCALES • The Celsius (or centrigrade) temperature scale assigns 00 to the freezing point of water and 1000 to its boiling point. Heat • On the Fahrenheit scale these points are, respectively, 320 and 2120. • A Fahrenheit degree is therefore 5/9 as large as a Celsius degree. • The following formulas give the procedure for converting a temperature expressed in one scale to the corresponding value in the other. 9 o T f = Tc + 32 5 5 Tc = (T f − 32 o ) 9 Heat HEAT • Heat is a form of energy which, when added to a body of matter, increases its internal energy content and thereby causes its temperature to rise. • The customary symbol for heat is Q. • Because heat is a form of energy, the proper SI unit of heat is the joule. • However, the kilocalorie is still sometimes used with SI units: 1 kilocalorie (kcal) is the amount of heat needed to raise the temperature of 1 kg of water by 1 oC. Heat • The calorie itself is the amount of heat needed to raise the temperature of 1 g of water by 1 oC; hence 1 kcal = 1000 calories. • (The calorie used by dieticians to measure the energy content of foods is the same as the kilocalorie.) • The British unit of heat is the British thermal unit (Btu): 1 Btu is the amount of heat needed to raise the temperature of 1 lb of water by 1 oF. • To convert heat figures from one system to the other we note that 1 J = 2.39 x 10-4 kcal = 9.48 x 10-4 Btu 1 kcal = 3.97 Btu = 4185 J = 3077 ft.lb 1 Btu = 0.252 kcal = 778 ft.lb = 1054 J Heat • Although weight rather than mass is specified in the British system when dealing with heat, in practice this makes no difference in the various calculations. Whenever m appears in the equations of heat, it is understood to refer to mass in kilograms when metric units are used and to weight in pounds when British units are used. SPECIFIC HEAT CAPACITY • Different substances respond differently to the addition or removal of heat. • For instance, 1 kg of water increases in temperature by 1 oC when 1 kcal of heat is added, but 1 kg of aluminum increases in temperature by 4.5 oC when this is done. Heat • The specific heat capacity of a substance is the amount of heat needed to change the temperature of a unit quantity of it by 1o. • The symbol of specific heat capacity is c; its SI unit is the joule/(kg. oC) [although the kcal/(kg. oC) is still sometimes used], and its British unit is the Btu/(lb . oF). • Among common materials, water has the highest specific heat capacity, namely, c water = 4185 J /(kg.o C ) = 1.00 kcal /(kg.o C ) = 1.00 Btu /(lb.o F ) • Ice and steam have lower specific heat capacities than water: cice = 2090 J /(kg.o C ) = 0.50 kcal /(kg.o C ) = 0.50 Btu /(lb.o F ) Heat c steam = 2010 J /(kg.o C ) = 0.48 kcal /(kg.o C ) = 0.48 Btu /(lb.o F ) • Metals usually have low specific heat capacities; thus lead and iron have c = 130 and 460 J /(kg. oC) , respectively. • When an amount of heat Q is transferred to or from a mass m of a substance whose specific heat capacity is c, the resulting temperature change ∆T is related to Q, m, and c by the formula Q = mc ∆T Heat transferred = (mass).(specific heat capacity) .(temperature change) Heat CHANGE OF STATE • When heat is continuously added to a solid, it grows hotter and hotter and really begins to melt. • While it is melting, the material remains at the same temperature and the absorbed heat goes into changing its state from solid to liquid. • After all the solid is melted, the temperature of the resulting liquid then increases as more heat is supplied until it begins to boil. • Now the material again stays at a constant temperature until all of it has become a gas, after which the gas temperature rises. Heat • The amount of heat that must be added to a unit quantity (1 kg or 1 lb) of a substance at its melting point to change it from a solid to a liquid is called its heat of fusion (Lf). • The same amount of heat must be removed from a unit quantity of the substance when it is a liquid at its melting point to change it to a solid. • The amount of heat that must be added to a unit quantity of a substance at its boiling point to change it from a liquid to a gas is called its heat of vaporization (Lv). • The same amount of heat must be removed from a unit quantity of the substance when it is a gas at its boiling point to change it to a liquid. Heat • The heat of fusion of water is Lf = 335 kJ/kg = 80 kcal/kg = 144 Btu/lb, and its heat of vaporization is Lv = 2260 kJ/kg = 540 kcal/kg = 972 Btu/lb. PRESSURE AND BOILING POINT • The boiling point of a liquid depends on the pressure applied to it: the higher the pressure, the higher the boiling point. • Thus water under a pressure of 2 atm boils at 121 oC instead of at 100 oC as it does at sea-level atmospheric pressure. • At high altitudes, where the atmospheric pressure is less than at sea level, water boils at a lower temperature than 100 oC. • At an elevation of 2000 m, for instance, atmospheric pressure is about three-quarters of its sea-level value, and water boils at 93 oC there. Solved Problems • 1. A person is dissatisfied with the rate at which eggs cook in a pan of boiling water. Would they cook faster if he (a) turns up the gas flame; (b) uses a pressure cooker? • Answer: (a) No. The maximum temperature that water can have while in the liquid state is its boiling point. Increasing the rate at which heat is supplied to a pan of water increases the rate at which steam is produced, but does not raise the temperature of the water beyond 100 oC (212 oF). (b) Yes. In a pressure cooker, the pressure is greater than normal atmospheric pressure, which elevates the boiling point and so causes the eggs to cook faster. Solved Problems • 2. What is the Celsius equivalent of 80 oF? • Answer: o 5 5 o o o Tc = (T f − 32 ) = (80 − 32 ) = 26.7 C 9 9 • 3. What is the Fahrenheit equivalent of 80 oC? • Answer: 9 9 o o T f = T c + 32 = (80 o ) + 32 o = 176 F 5 5 • 4. Oxygen freezes at -362 oF. What is the Celsius equivalent of this temperature? • Answer: o 5 5 o o o Tc = (T f − 32 ) = ( − 362 − 32 ) = − 219 C 9 9 Solved Problems • 5. Nitrogen freezes at -210 oC. What is the Fahrenheit equivalent of this temperature? • Answer: 9 9 o o T f = T c + 32 = ( − 210 o ) + 32 o = − 346 F 5 5 • 6.Two hundred Btu of heat is removed from a 50lb block of ice initially at 25 oF. What is its final temperature [cice = 0.5 Btu/(lb. oF)]? • Answer: Q = mc ∆T 200 Btu Q ∆T = = = 8 oF o mc (50 lb)[0.5 Btu /(lb. F )] Final temperature is therefore 25 oF - 8 oF = 17 oF. Solved Problems • 7. Ten kcal of heat is added to a 1-kg sample of wood and its temperature is found to rise from 20 oC to 44 oC. What is the specific heat capacity of the wood? • Answer: Q = mc ∆T 10 kcal Q o = = 0 . 42 kcal /( kg C) c= o m ∆T (1 kg )(24 C ) • 8. Three lb of water at 100 oF is added to 5 lb of water at 40 oF. What is the anal temperature of the mixture? • Answer: lf T is the final temperature, then the 5 lb of water initially at 40 oF undergoes a temperature change of ∆T1 = T - 40 oF and the 3 lb of water initially at 100 oF undergoes a temperature change of ∆T2=100 oF - T. We proceed as follows: • 9. In order to raise the temperature of 5 kg of water from 20 oC to 30 oC a 2-kg iron bar is heated and then dropped into the water. What should the temperature of the bar be [ciron = 0.11 kcal/(kg. oC)]? • Answer: Let the temperature of the iron bar be T. Then the change in the water's temperature is ∆Tw = 30 oC - 20 oC = 10 oC and the change in the bar's temperature is ∆Tiron = T - 30 oC. We proceed in the usual way: Solved Problems • 10. How much heat must be added to 200 lb of lead at 70 oF to cause it to melt? The specific heat capacity of lead is 0.03 Btu/(lb . oF), it melts at 626 oF, and its heat of fusion is 10.6 Btu/lb. • Answer: Solved Problems • 11. Five kg of water at 40 oC is poured on a large block of ice at 0 oC. How much ice melts? • Answer: Solved Problems • 12. A 30-g ice cube at 0 oC is dropped into 200 g of water at 30 oC. What is the final temperature? • Answer: If T is the final temperature, then ∆Tice = T - 0 oC and ∆Twater = 30 oC - T. Therefore Thermodynamics HEAT ENGINES • To convert internal energy into mechanical energy is much more difficult than the reverse, and perfect efficiency is impossible. • A heat engine is a device or system that can perform this conversion; the human body and the earth's atmosphere are heat engines, as well as gasoline and diesel motors, aircraft jet engines, and steam turbines. Thermodynamics All heat engines operate by absorbing heat from a reservoir of some kind at a high temperature, performing work, and then giving off heat to a reservoir of some kind at a lower temperature (see Fig.). Thermodynamics • According to the principle of conservation of energy, the work done in a complete cycle that returns the engine to its original state is equal to the difference between the heat absorbed and the heat given off; this statement constitutes the first law of thermodynamics. SECOND LAW OF THERMODYNAMICS • Internal energy resides in the kinetic energies of randomly moving atoms and molecules, whereas the output of a heat engine appears ill the ordered motions of a piston or a wheel. • Since all physical systems in the universe tend to go in the opposite direction, from order to disorder, no heat engine can completely convert heat into mechanical energy or, in general, into work. Thermodynamics • This fundamental principle leads to the second law of thermodynamics: It is impossible to construct a continuously operating engine that takes heat from a source and performs an exadly equivalent amount of work. • Because some of the heat input to a heat nine must be wasted, and because heat bows from a hot reservoir to a cold one, every heat engine must have a low-temperature reservoir for exhaust heat to go to as well as a high-temperature reservoir from which the input heat is to come. Thermodynamics ENGINE EFFICIENCY • The efficiency of an ideal heat engine (often called a Carnot engine) in which there are no losses due to such practical difficulties as friction depends only on the temperatures at which heat is absorbed and exhausted. • If the heat Q1 is absorbed at the absolute temperature T1 and the heat Q2 is given off at the absolute temperature T2, Q1/Q2 = T1/T2 in such an engine. Its efficiency is therefore w Q1 − Q2 = = Efficiency (ideal ) = Q1 Q1 heat input Q T = 1− 2 = 1− 1 Q1 T2 work output • The smaller the ratio between T2 and T1, the more efficient the engine. Because no reservoir can exist at a temperature of 0 K or 0 oR, which is absolute zero, no heat engine can be 100 percent efficient. REFRIGERATION A refrigerator is a heat engine that operates backward to extract heat from a low-temperature reservoir and transfer it to a high-temperature reservoir (see Fig.). Thermodynamics • Because the natural tendency of heat is to now from a hot region to a cold one, energy must be provided to a refrigerator to reverse the now, and this energy adds to the heat exhausted by the refrigerator. HEAT TRANSFER: CONDUCTION • The three mechanisms by which heat can be transferred from one place to another are conduction, convection, and radiation. • In conduction, heat is carried by means of collisions between rapidly moving molecules at the hot end of a body of matter and the slower molecules at the cold end. Thermodynamics • Some of the kinetic energy of the fast molecules passes to the slow molecules, and the result of successive collisions is a flow of heat through the body of matter. • Solids, liquids, and gases all conduct heat. • Conduction is poorest in gases because their molecules are relatively far apart and so interact less frequently than in the case of solids and liquids. • Metals are the best conductors of heat because some of their electrons are able to move about relatively freely and can travel past many atoms between collisions. Thermodynamics CONVECTION • In convection, a volume of hot fluid (gas or liquid) moves from one region to another carrying internal energy with it. • When a pan of water is heated on a stove, for instance, the hot water at the bottom expands slightly so that its density decreases, and the buoyancy of this water causes it to rise to the surface while colder, denser water descends to take its place at the bottom. Thermodynamics RADIATION • In radiation, energy is carried by the electromagnetic waves emitted by every object. • Electromagnetic waves, of which light, radio waves, and x-rays are examples, travel at the velocity of light (3 x 108 m/s = 186,000 mi/s) and require no material medium for their passage. • The higher the temperature of an object, the greater the rate at which it radiates energy; the rate is proportional to T4, where T is the object absolute temperature. • With increasing temperature, the predominant wavelength of the radiation emitted by a body decreases. Thermodynamics • Thus a hot body that glows red is cooler than one that glows bluish-white since red light has a longer wavelength than blue light. • A body at room temperature emits radiation that is chiefly in the infrared part of the spectrum, to which the eye is not sensitive. • More details about Radiation and its applications in Medicine will discuss later Solved Problems • 1. lf all objects radiate electromagnetic energy, why do the objects around us in everyday life not grow colder and colder? • Answer: Every object also absorbs electromagnetic energy from its surroundings, and if both object and surroundings are at the same temperature, energy is emitted and absorbed at the same rate. When an object is at a higher temperature than its surroundings and heat is not supplied to it, it radiates more energy than it absorbs and cools down to the temperature of its surroundings. • 2. A 1-MW (106 W) generating plant has an overall efficiency of 40%. How much fuel oil whose heat of combustion is 11,000 kcal/kg does the plant burn each day? • Answer: Solved Problems • 3. (a) Find the maximum possible efficiency of an engine that absorbs heat at a temperature of 327 oC and exhausts heat at a temperature of 127 oC. (b) What is the maximum amount of work (in joules) the engine can perform per kcal of heat input? • Answer: Solved Problems • 4. An engine is being planned which is to have an efficiency of 25 percent and which will absorb heat at a temperature of 267 oC. Find the maximum temperature at which it can exhaust heat. • Answer: Principles of Thermography, ¾ Infrared thermography, thermal imaging, thermographic imaging, or thermal video, is a type of infrared imaging science. ¾ Thermographic cameras detect radiation in the infrared range of the electromagnetic spectrum (roughly 900–14,000 nanometers or 0.9–14 µm) and produce images of that radiation, called thermograms. Thermogram of a small dog taken in mid-infrared Principles of Thermography, ¾ Since infrared radiation is emitted by all objects near room temperature, according to the black body radiation law, thermography makes it possible to "see" one's environment with or without visible illumination. ¾ The amount of radiation emitted by an object increases with temperature, therefore thermography allows one to see variations in temperature (hence the name). Thermogram of two ostriches Principles of Thermography, ¾ The use of thermal imaging has increased dramatically with governments and airports staff using the technology to detect suspected swine flu cases during the 2009 pandemic. ¾ Other uses include, firefighters use it to see through smoke, find persons, and localize the base of a fire. ¾ Thermal imaging cameras are also installed in some luxury cars to aid the driver, the first being the 2000 Cadillac DeVille. Thermogram of lion Principles of Thermography, ¾ It is important to note that thermal imaging displays the amount of infrared energy emitted, transmitted, and reflected by an object. Because of this, it is quite difficult to get an accurate temperature of an object using this method. ¾ Thus, Incident Energy = Emitted Transmitted Energy + Reflected Energy Energy + where Incident Energy is the energy profile when viewed through a thermal imaging device, Emitted Energy is generally what is intended to be measured, Transmitted Energy is the energy that passes through the subject from a remote thermal source, and Reflected Energy is the amount of energy that reflects off the surface of the object from a remote thermal source. Principles of Thermography, ¾ Thermal imaging camera & screen, photographed in an airport terminal in Greece. Thermal imaging can detect elevated body temperature, one of the signs of the virus H1N1 (Swine influenza). Principles of Thermography, ¾ Advantages of thermography: 1. It shows a visual picture so temperatures over a large area can be compared 2. It is capable of catching moving targets in real time 3. It is able to find deteriorating, i.e., higher temperature components prior to their failure 4. It can be used to measure or observe in areas inaccessible or hazardous for other methods 5. It is a non-destructive test method 6. It can be used to find defects in shafts, pipes, and other metal or plastic parts. 7. It can be used to see better in dark areas Solved Problems • 5. In the operation of a thermograph, the radiation from each small area of a person's skin is measured and shown by different shades of gray or by different colors in a thermogram. Because the skin over a tumor is warmer than elsewhere, thermograph are widely used in screening for breast cancer. What is the percentage difference between the radiation rates from skin at 34 oC and 35 oC? • Answer: Optics FOCAL LENGTH • Figure 1 shows how a converging lens brings a parallel beam of light to a real focal point F, and Fig. 2 shows how a diverging lens spreads out a parallel beam of light so that the refracted rays appear to come from a virtual focal point F. • A positive focal length corresponds to a converging lens, and a negative focal length to a diverging lens. Fig. 1 Fig. 2 RAY TRACING • The position and size of the image of an object formed by a lens can be found by constructing a scale drawing. • What is done is to trace two different light rays from each point of interest in the object to where they (or their extensions in the case of a virtual image) intersect after being refracted by the lens. • Three rays that are especially useful for this purpose are shown in Fig. 3 although any two of these are sufficient. Fig. 3 Optics • They are: 1. A ray that leaves the object parallel to the axis of the lens. After refraction, this ray passes through the far focal point of a converging lens or seems to come from the near focal point of a diverging lens. 2. A ray that passes through the near focal point of a converging lens or is directed toward the far focal point of a diverging lens. After refraction, this ray travels parallel to the axis of the lens. 3. A ray that leaves the object and proceeds toward the center of the lens. This ray is not deviated by refraction. Optics LENS EQUATION • The object distance p, image distance q, and focal length f of a lens (Fig. 4) are related by the lens equation: • The equation holds for both converging and diverging lenses. The lens equation is readily solved for p, q, Or f: Fig. 4 Optics • • A positive value of p or q denotes a real object or image, and a negative value denotes a virtual object or image. A real image of a real object is always on the opposite side of the lens from the object, and a virtual image is on the same side; thus if a real object is on the left of a lens, a positive image distance q signifies a real image to the right of the lens whereas a negative image distance q denotes a virtual image to the left of the lens. Optics MAGNIFICATION • The linear magnification m produced by a lens is given by the formula: • A positive magnification signifies an erect image, a negative one signifies an inverted image. Optics • Table below is a summary of the sign conventions used in connection with lenses. Optics LENS SYSTEMS • When a system of lenses is used to produce an image of an object, for instance in a telescope or microscope, the procedure for finding the position and nature of the final image it to let the image formed by each lens in turn be the object for the next lens in the system. Thus to and the image produced by a system of two lenses, the first step is to determine the image formed by the lens nearest the object. • This image then serves as the object for the second lens, with the usual sign convention: if the image is on the front side of the second lens, the object distance is considered positive, whereas if the image is on the back side, the object distance is considered negative. Optics • • The total magnification produced by a system of lenses is equal to the product of the magnifications of the individual lenses. Thus if the magnification of the objective lens of a microscope or telescope is m1 and that of the eyepiece is m2 , the total magnification is m = m 1m2 . Solved Problems • 1. What is the nature of the image formed by a diverging lens of a real object? • Answer: It is virtual, erect, and smaller than the object, as in Fig. 3(b). • 2. Describe the image formed by a converging lens of an object located at the focal point of the lens. • Answer: Here p = f, and so As below shows, the refracted rays are parallel to each other and so no image is formed. Solved Problems • • 3. The focal length f of a combination of two thin lenses in contact whose individual lengths are f1 and f2 is given by Use this formula to find the focal length of a combination of a converging lens of f = + 10 cm and a diverging lens of f = -20 cm that are in contact. Answer: The above formula can be rewritten in the more convenient form Solved Problems Since f1 = 10 cm and f2 = -20 cm, The combination acts as a converging lens of focal length +20 cm. Basic of radiation physics ¾Atoms, Isotopes, ¾Binding energy, ¾Radiation types and production, ¾Radiation decay and half life ¾Biological effects of radiation , ¾Radiation units, ¾Limits of radiation doses, ¾Principles of radiation protection Atoms ﺸﺤﻨﺔ ﺍﻹﻟﻜﺘﺭﻭﻥ 1.6X10-19C Atom Z = Proton No. A = Proton + Neutron ﻜﺘﻠﺔ ﺍﻹﻟﻜﺘﺭﻭﻥ 9.1X10-31 kg Atoms Natural atoms Atom Atom + Atom = 92 Compound Example: H2O = 2 Hydrogen atoms + 1 Oxygen atom Atoms Nucleus Radius 10-13 cm Proton mass = 1.67 x 10-24 gm Electron mass = 9.11 x 10-28 gm 1/1840 amu Electron mass less than Proton mass with 1840 Atoms Neutron live = 12 min & decay to proton and electron • C = 3 x 108 m/sec • Unit of energy is joule, but in atomic field = eV Einstein prove that energy and matter are equivalent. That energy can change to matter & via versa. Eo = moxC2 • 1 eV = 1.6 x 10-19 joule • eV = amount of energy lost or gain by electron (or proton) when pass potential volt equivalent to one volt. Atoms • 1 eV = 1.6 x 10-19 joule • 1 KeV = 103 eV = 1.6 x 10-16 joule • 1 MeV = 106 eV = 1.6 x 10-13 joule Isotopes Nuclides with the same atomic number Isotopes Many application of using radioactive isotopes اﻟﻄﺐ اﻟﺰراﻋﺔ اﻟﺼﻨﺎﻋﺔ اﻟﺒﺤﺚ اﻟﻌﻠﻤﻲ Binding energy ¾The energy required to remove an electron completely from the atom is called its binding energy. ¾Binding energy of nucleus is different ¾Protons of positive charges Î + + + + + + Î Î but strong attractive force + + Nuclear force ¾All nuclear force are equivalent so we can consider proton and neutron are one object called nucleon. ¾All nucleons attract each other if the distance between them less than 10-13 cm ¾Nucleus can decay to nucleons by energy called binding energy + + + Binding energy ¾ B of nucleus = (NMn + ZMp – M ) C2 M = mass of nucleus Mn = neutron mass Mn = proton mass N = neutron no. ¾The energy required to separate an atom into its constituent parts is the atomic binding energy. ¾It is electron binding energy + the nuclear binding energy. Radiation types and production ¾What it is the meaning of radiation? Radiation is energy that travels through space or matter. e- Radiation types and production ¾ Two types of radiation: ionizing and nonionizing radiation. ¾Visible light, radio waves, and x-rays are different types of electromagnetic (EM) radiation. ¾EM radiation has no mass, is unaffected by either electrical or magnetic fields, and has a constant speed in a given medium. ¾EM radiation travels in straight lines. Radiation types and production ¾There are two equally correct ways of describing EM radiation-as waves and as particle-like units of energy called photons or quanta. ¾In some situations EM radiation behaves like waves and in other situations like particles. ¾All waves (mechanical or electromagnetic) are characterized by their amplitude (maximal height), wavelength (A), frequency (v), and period. ¾The amplitude is the intensity of the wave. Radiation types and production ¾ The wavelength is the distance between any two identical points on adjacent cycles. The time required to complete one cycle of a wave (i.e., one A) is the period. The number of periods that occur per second is the frequency (1/period). ¾Phase is the temporal shift of one wave with respect to the other. Electromagnetic spectrum Electric and magnetic field components of electromagnetic radiation Ionizing Vs. Nonionizing Radiation ¾EM radiation of higher frequency than the near-ultraviolet region of the spectrum carries sufficient energy per photon to remove bound electrons from atomic shells, thus producing ionized atoms and molecules. Radiation in this portion of the spectrum (e.g., ultraviolet radiation, x-rays, and gamma rays) is called ionizing radiation. ¾EM radiation with energy below the farultraviolet region (e.g., visible light, infrared, radio and TV broadcasts) is called nonionizing radiation. Most important type of radiation Alpha Particles gamma radiation neutrons outside nucleus (x-ray) Radiation types •Neutron no. = Proton no. => stable nucleus + + + + + + + + + •Neutron no. ≠ Proton no. => unstable nucleus γ,α, β α Decay modes Ra226 = Rn222 + α ﻨﻭﺍﺓ ﺍﻟﻬﻴﻠﻴﻭﻡ A Z XN → YN − 2 + 2 He 2 A− 4 Z −2 4 • • • • α = 2n + 2P Positive charge Short range at air Stopped by paper Beta decay types Positron decay X N → Z −A1YN + 1 + β + + ν Electron decay A Z p → n + β+ + ν − + ν% X Y → N N − 1 + β Z Z +1 A A n → p + β − + ν% Electron Capture 0e + 1P Æ 1n ﺗﺤﻮل أﺣﺪ ﺑﺮوﺗﻮﻧﺎت اﻟﻨﻮاة اﻟﻲ ﻧﻴﺘﺮون .و ﻳﺘﻢ ذﻟﻚ ﺑﺎن ﺗﺄﺳﺮ اﻟﻨﻮاﻩ اﻟﻜﺘﺮوﻧﺎ ﻣﻦ اﻻﻟﻜﺘﺮوﻧﺎت اﻟﻤﺪارﻳﺔ اﻟﻘﺮﻳﺒﺔ ﻣﻦ اﻟﻨﻮاة و ﻳﺘﺤﺪ هﺬا اﻻﻟﻜﺘﺮون اﻟﻤﺄﺳﻮر ﻣﻊ اﺣﺪ اﻟﺒﺮوﺗﻮﻧﺎت ﻓﻴﺘﻜﻮن اﻟﻨﻴﻮﺗﺮون .ﻓﻲ ﺣﺎﻟﺔ اﻷﺳﺮ اﻻﻟﻜﺘﺮوﻧﻲ ﻻ ﺗﺼﺪر اﻟﻨﻮاﻩ اﻳﺎ ﻣﻦ ﺟﺴﻴﻤﺎت ﺑﻴﺘﺎ. Energy calculation for beta decay ﺍﻟﺘﻔﻜﻙ ﺍﻻﻟﻜﺘﺭﻭﻨﻲ ﺍﻟﺘﻔﻜﻙ ﺍﻟﺒﻭﺯﻴﺘﺭﻭﻨﻲ E = {ZM – (Z+1M + me)}C2 E = {ZM – (Z-1M + me)}C2 E = {ZM –Z-1M }C2 • me ﻜﺘﻠﺔ ﺍﻻﻟﻜﺘﺭﻭﻥ ﺍﻷﺴﺭ ﺍﻻﻟﻜﺘﺭﻭﻨﻲ ﻧﻔﺎذﻳﺔ اﻷﺷﻌﺔ Decay concept Radioactivity The quantity of radioactive material, expressed as the number of radioactive atoms undergoing nuclear transformation per unit time (t), is called activity (A). Activity ¾ Described mathematically, activity is equal to the change (dN) in the total number of radioactive atoms (N) in a given period of time (dt), or ¾ The minus sign indicates that the number of radioactive atoms decreases with time. ¾ Activity is traditionally expressed in units of curies (Ci). ¾ The curie is defined as 3.70 X 1010 disintegrations per second (dps). Activity ¾ A curie is a large amount of radioactivity. ¾ In nuclear medicine, activities from 0.1 to 30 mCi of a variety of radionuclides are typically used for imaging studies, and up to 300 mCi of iodine 131 are used for therapy. ¾ Although the curie is still the most common unit of radioactivity in the United States, the majority of the world's scientific literature uses the Systeme International (SI) units. Activity ¾ The SI unit for radioactivity is the becquerel (Bq), named for Henti Becquerel, who discovered radioactivity in 1896. ¾ The becquerel is defined as 1 dps. ¾ 1 millicurie (mCi) is equal to 37 megabecquerels ¾ (1 mCi = 37 MBq). Activity Radioactive decay ¾ Radioactive decay is a random process. ¾ The number of atoms decaying per unit time (dN/dt) is proportional to the number of unstable atoms (N) that are present at any given time: ¾ A proportionality can be transformed into an equality by introducing a constant. This constant is called the decay constant (A). 100 Physical Half-Life 50 25 12.5 0 T1 2 2T 1 2 3T 1 2 6.25 4T 1 2 3.125 5T 1 t 2 Half life (t1/2) High activity Very short half life 238U 14C 18F 131I 92Ir : T = 4,5 109 years : 5730 years : 1,8 hours : 8 days : 74 days Low activity Very long half life Ex: U-238 (T = 4,5 109 years) But Considered as very dangerous !! Half life (t1/2) ¾ The half-life is defined as the time required for the number of radioactive atoms in a sample to decrease by one half. ¾ The number of radioactive atoms remaining in a sample and the number of elapsed half-lives are related by the following equation: ¾ where N is number of radioactive atoms remaining, No is the initial number of radioactive atoms, and n is the number of half-lives that have elapsed. ¾ t 1/2 = ln 2 / λ Fundamental Decay Equation Example: Radioactivity • Calculate activity for sample with t ½ = 30 days. 3 day left from reference activity date (100 mCi). A = 100.e A = 100.e −0.693 X 3 30 0.693.t − T1 2 = 93.3mCi First exercice : determination of the activity at time t • A sample contains I-131 (T1/2 = 8 days). Measure at time to, A (activity) is equal to 1 MBq. After 12 days, how much activity is left in the sample ? • Result A = Ao e-λt – λ = 0,693/T1/2 – A = 1 MBq e -0,693/8 t → t = 12 days – A = 1 MBq e-1,0395 = 0,35 MBq after 12 days Second exercice • We have a sample 131I (T1/2 = 8 days) with a number of nuclei of N0 = 6,0x1020 at initial time (to) obtained by sampling a mass of 0,1 g. • Calculate the activity of this sample at initial time • How many I atoms are still there after 1 T1/2? 2 T1/2? • Calculate the time after which 80% of the nuclei have disintegrated. Results • Activity of the sample at initial time Activity: A0 = λ N0 (where λ = 0,693 / T½) λ = 0,693 / (8 * 24 * 3600) = 10-6 sec-1. So A0 = 10-6 * 6,0x1020 = 6,0 1014 Bq. • How many I atoms are still there after 1 T½? after 2 T½? After 1 T½ , N = ½N0 = 3,0 10 20 nuclei. After 2 T ½ , 0,5 N = 1,5 10 20 nuclei. • Calculate the time after which 80% of the nuclei have disintegrated N = 0,2 No N /N0 = 0,2 = e -λ t so ln 0,2 = -λt = - 10-6 t t = ln 0,2 / (- 10-6 ) = 1,61 10 6 s = 447,2 h = 19 days Third exercice • The radioactive constant of Tc-99m is λ = 0,1155 h-1, calculate: – T1/2 (Tc-99m) in hours; – Mass of Tc-99m (µg) which corresponds to an activity of A = 3,7x1010 Bq (remember NA = 6,022x1023) – Time (hours) after which the activity of Tc-99m is equal to 1/16 of the initial activity. • Results – T1/2 = 0,693/λ → T1/2 = 6 hours – A = λ N (λ = 0,1155/3600 sec-1 = 3,21x10-5 sec-1) → 3,7x1010 = 3,21x10-5 x N → N = 1,15x1015 atoms → mass of 1 atom = 99/NA = 16,44x10-23 g → 16,44x10-23 x 1,15x1015 = 0,19µg – A = Ao/16 = Ao e - λ t → 1/16 = e-0,1155t → ln 1/16 = -0,1155t → t = 24h Suggested reading • Friedlander G, Kennedy JW; Miller JM. Nuclear and radiochemistry, 3rd ed. New York: Wiley, 1981. • Patton JA. Introduction to nuclear physics. Radiographies 1998;18:995-1007. • Sorensen JA, Phelps ME. Physics in nuclear medicine, 2nd ed. New York: Grune & Stratton, 1987. Biological effects of radiation nucleus Nucleus membrane cytoplasm Cell membrane Contain chromosomes (46 chromosome) Biological effects of radiation indirect Radiation interactions that produce biologic changes are classified as Direct Direct action if a biologic macromolecule such as DNA, RNA, or protein becomes ionized or excited by an ionizing particle or photon passing through or near it. Biological effects of radiation indirect Radiation interactions that produce biologic changes are classified as Direct Indirect effects are the result of radiation interactions within the medium (e.g., cytoplasm) which create reactive chemical species that in turn interact with the target molecule. Biological effects of radiation 70% to 85% composed of water majority of radiation-induced damage through indirect action on water molecules. Interaction with a water molecule results in an ion pair (H2O+, H2O-). Produced by the ionization of H2O. Produced via capture of a free electron by a water molecule. Biological effects of radiation These ions dissociates to form another ion and a free radical: z Free radicals are atomic or molecular species that have unpaired orbital electrons. Do not produce significant biologic damage because of their extremely short lifetimes (≈10-10 sec) and their tendency to recombine to form water. Free radicals can combine with other free radicals to form other molecules such as hydrogen peroxide (e.g., OH· + OH· = H2O2), which are highly toxic to the cell. Damage cell can occur during following stage: 1. - Physical Stage: Finish with short time (≈10-16 sec) Ionization occur according to : H2O Æ H2O+ + ewhere H2O+ positive water ion, e- negative electron 2. - - 3. - Physcio-chemical stage: Finish with short time (≈10-6 sec) Positive & negative ions interact with water molecules New compounds product See previous slit Chemical Stage: Finish with short seconds H2O2, OH or H interact with other organic part of the cell (e.g. chromosomes) Damage cell can occur during following stage: 4. a. b. c. Biological Stage: Range between few minutes or ten years Some effects appear such as: Cell death Increase cell division or stop it. Change at cells lead to hereditary effects. Biological effects of radiation Biological effects of radiation 3 Gy of x-ray lead to Erythema Radiation effect on DNA May occur as single-strand breaks, double-strand breaks, base loss, or Base changes. 3 principles govern the professionnal use of radioactivity • Justification of their use Every human activity involving an exposition should be justified by the advantages that can give ( Advantages > inconvenients) • Optimisation of the radiation protection The exposition of the individuals and of the population should be maintained as low as reasonably achievable (ALARA) • Limitation of the individuals doses The limits are chosen to be sufficiently low to avoid any serious effects or the probability of aleatory effects To avoid routine regarding an invisible risk……… How to protect against external irradiation ? Source (radiopharmaceutical) Total Dose = Dose rate x lenght of exposition D=Dxt Increase distance Reduce exposition time Use shielding devices External exposition can be reduced by respecting these 3 golden rules during waste management Increase distance as much as possible… • Excellent method, simple, easy, not costly Mathematic Law: inverse of the square distance D2 = D1 x (d1)2 (d2)2 Do D1 D2 1 mSv/h (1 m) 0,25 mSv/h (2 m) The dose rate decreases rapidly with the increase of the distance ⇒ Just place yourself as far as possible… (place the sources as far as possible) Dose limitation ⇒ required by the legislation Organ or tissue Worker Public (mSv)/year (mSv)/year Whole body 20 1 Skin 500 50 Hands 500 50 Other organs 500 50 Lens 150 15 These limits of dose are security levels….not danger level ! Introduction to Nuclear Medicine Different kind of radioisotopes produced by nuclear reactors or cyclotrons Shielding protection are used intensively in nuclear medicine …. Cyclotron-Produced Radionuclides ¾ Cyclotrons and other charged-particle accelerators produce radionuclides by bombarding stable nuclei with high-energy charged particles. ¾ Protons, deuterons (2H nuclei), tritons (3H nuclei), and alpha particles (4He nuclei) are commonly used to produce radionuclides used in medicine. ¾ Heavy charged particles must be accelerated to high kinetic energies. Molybdenum 99/Technetium-99m Radionuclide Generator ¾ Technetium-99m pertechnetate (99mTcO4-) is produced in a sterile, pyrogen-free form with high specific activity and a pH (~5.5) that is ideally suited for radiopharmaceutical preparations. ¾ The Mo-99 (produced by nuclear fission of U-235 to yield a high-specific-activity, carrier-free parent) is loaded, in the form of ammonium molybdenate (NH4+)(MoO4-), onto a porous column containing 5 to 10 g of an inorganic alumina (Al2O3) resin. ¾ The ammonium molybdenate becomes attached to the surface of the alumina molecules (a process called adsorption). Molybdenum 99/Technetium-99m Radionuclide Generator NaCl 0,9% RADIOPHARMACEUTICALS ¾ The vast majority of radiopharmaceuticals in nuclear medicine today use T c-99m as the radionuclide. ¾ Most Tc-99m radiopharmaceuticals are easily prepared by aseptically injecting a known quantity of Tc-99m pertechnetate into a sterile vial containing the pharmaceutical. ¾ Radiopharmaceuticals can be called "kits’’ ¾ Although most T c-99m radio pharmaceuticals can be prepared rapidly and easily at room temperature, several products (e.g., Tc-99m macroaggregated albumin [MAA]), require multiple steps such as boiling the Tc-99m reagent complex for several minutes. RADIOPHARMACEUTICALS Gamma Camera Planar nuclear imaging: the anger scintillation camera ¾ Developed by Hal O. Anger at the Donner Laboratory in Berkeley, California, in the 1950s. ¾ Most of the advantages of the scintillation camera over the rectilinear scanner stem from its ability simultaneously to collect data over a large area of the patient, rather than one small area at a time. ¾ A scintillation camera, contains a disk-shaped or sodium iodide NaI(TI) crystal, typically 0.95 cm (3/8 inch) thick, optically coupled to a large number of 5.1- to 7.6cm diameter photomultiplier tubes (PMTs). Gamma Camera ¾ In most cameras, a preamplifier is connected to the output of each PMT. Between the patient and the crystal is a collimator, usually made of lead, that only allows x- or gamma rays approaching from certain directions to reach the crystal. Gamma Camera ¾ The lead walls, called septa, between the holes in the collimator absorb most photons approaching the collimator from directions that are not aligned with the holes. Most photons approaching the collimator from a nearly perpendicular direction pass through the holes; many of these are absorbed in the sodium iodide crystal, causing the emission of visible and ultraviolet light. ¾ The light photons are converted into electrical signals and amplified by the PMTs. These signals are further amplified by the preamplifiers (preamps). The amplitude of the pulse produced by each PMT is proportional to the amount of light it received from an x- or gammaray interaction in the crystal. PMT ¾ Photomultiplier Tubes (PMTs) perform two functionsconversion of ultraviolet and visible light photons into an electrical signal and signal amplification, on the order of millions to billions. PMT ¾ PMT consists of an evacuated glass tube containing a photocathode, typically 10 to 12 electrodes called dynodes, and an anode. ¾ The photocathode is a very thin electrode, located just inside the glass entrance window of the PMT, which emits electrons when struck by visible light. (Approximately one electron is emitted from the photocathode for every five light photons incident upon it.) Introduction to Diagnostic Radiology • • • • X-ray Fluoroscopy MRI Ultrasound ﺃﺸــﻌﺔ ـ Essential physics of diagnostic radiology X اآﺘﺸﻔﺖ ﻣﻦ ﻗﺒﻞ: .اﻟﻌﺎﻟﻢ وﻟﻴﻢ روﻧﺘﺠﻦ ﻋﺎم 1895 وﻓﻲ ﻋﺎم 1895ﻧﺸﺮت أول ورﻗﺔ ﻋﻠﻤﻴﺔ ﻋﻦ ﺗﻄﺒﻴﻘﺎت اﻷﺷﻌﺔ اﻟﺴﻴﻨﻴﺔ Wilhelm Conrad Rontgen 1845 - 1923 NP 1901 X-ray production, x-ray tubes, and generators • X-rays are produced when highly energetic electrons interact with matter and convert their kinetic energy into electromagnetic radiation. • A device that accomplishes such a task consists of an electron source, an evacuated path for electron acceleration, a target electrode, and an external energy source to accelerate the electrons. • x-ray tube insert • tube housing X-ray • collimators device • generator PRODUCTION OF X-RAYS PRODUCTION OF X-RAYS A large voltage is applied between two electrodes PRODUCTION OF X-RAYS The anode is positively charged The cathode is negatively charged PRODUCTION OF X-RAYS An x-ray photon with energy equal to the kinetic energy lost by the electron is produced (conservation of energy). This radiation is termed bremsstrahlung, a German word meaning "braking radiation." PRODUCTION OF X-RAYS At relatively "large" distances from the nucleus, the coulombic attraction force is weak; these encounters produce low x-ray energies (Fig. electron no. 3). Factors that affect x-ray production • Major factors that affect x-ray production efficiency include the atomic number of the target material and the kinetic energy of the incident electrons (which is determined by the accelerating potential difference). • The approximate ratio of collisional energy loss is expressed as follows: kinetic energy of the incident electrons in keV Atomic number X-Ray Tubes cathode, anode, rotor/stator, glass (or metal) envelope, and tube housing. • Major components are the Fluoroscopy • • Fluoroscopy is an imaging procedure that allows real-time x-ray viewing of the patient with high temporal resolution. Before the 1950s, fluoroscopy was performed in a darkened room with the radiologist viewing the faint scintillations from a thick fluorescent screen. Fluoroscopic Imaging Components • • The x-ray tube, filters, and collimation are similar technologies to those used in radiography and are not discussed in detail here. The principal component of the imaging chain that distinguishes fluoroscopy from radiography is the image intensifier. 10 min ≈ 18000 images The Image Intensifier (II) • There are four principal components of an II: (a) a vacuum bottle to keep the air out, (b) an input layer that converts the x-ray signal to electrons, (c) electronic lenses that focus the electrons, and (d) an output phosphor that converts the accelerated electrons into visible light. Magnetization Properties ¾ Magnetism is a fundamental property of matter; it is generated by moving charges, usually electrons. ¾ Atoms and molecules have electron orbitals that can be paired (an even number of electrons cancels the magnetic field) or unpaired (the magnetic field is present). Magnetization Properties ¾ The magnetic field strength, B, (also called the magnetic flux density) can be conceptualized as the number of magnetic lines of force per unit area. ¾ The SI unit for B is the tesla (T), and as a benchmark, the earth's magnetic field is about 1/20,000 T.. Magnetization Properties ¾ Magnetic fields can be induced by a moving charge in a wire. ¾ The direction of the magnetic field depends on the sign and the direction of the charge in the wire, as described by the "right hand rule": The fingers point in the direction of the magnetic field when the thumb points in the direction of a moving positive charge. Magnetization Properties Magnetic Characteristics of the Nucleus ¾ The nucleus is comprised of protons and neutrons with characteristics listed in table below. ¾ The magnetic moment, represented as a vector indicating magnitude and direction, describes the magnetic field characteristics of the nucleus. Magnetic Characteristics of the Elements Under the influence of a strong external magnetic field, Bo, however, the spins are distributed into two energy states: alignment with (parallel to) the applied field at a low-energy level, and alignment against (antiparallel to) the field at a slightly higher energy level (see Fig. B). Magnatisium in medicine ¾ The first effects of magnetism were observed when the smelted iron was brought close to the iron oxide in the chemical form of FeO.Fe2O3 (Fe3O4), a natural iron ore which came to be known as lodestone or magnetite. ¾ The origin of the term ‘‘magnetite’’ is unclear, but two explanations appear most frequently in the literature. In one of these, magnetite was named after the Greek shepherd Magnes, who discovered it when the nails on the soles of his shoes adhered to the ore. In the other explanation, magnetite was named after the ancient county of Magnesia in Asia Minor, where it was found in abundance. Magnatisium in medicine ¾ First Medical Uses of Magnets: Thales of Miletus, the first Greek speculative scientist and astronomer was also the first to make a connection between man and magnet. ¾ He believed that the soul somehow produced motion and concluded that, as a magnet also produces motion in that it moves iron, it must also possess a soul. It is likely that this belief led to the many claims throughout history of the miraculous healing properties of the lodestone. Hand-held electromagnets used for the removal of ¾ magnetic objects from the eye. Left: The original Hirschberg magnet. Right: A further development of Dr. Hubbell. The needle-like tip is placed, preferably through the entry wound, as close as possible to the foreign iron or steel particle. The magnet is then turned on and the foreign body pulled out. Magnatisium in medicine ¾ Removal of an open safety pin from a patient’s stomach. ¾ (Photograph courtesy of F.E. Luborsky; Luborsky et al., 1964). Treatment of Nervous Diseases and Mesmerism ¾ The first person to mention the topical application of a magnet in nervous diseases was Aetius of Amida (550–600), who recommended this approach primarily for the treatment of hysteria, and also for gout, spasm, and other painful diseases. ¾ Some five centuries later, abbess Hildegard of Bingen (1098–1179) – who was use of plants and minerals (stones) for medical purposes and devoted a whole chapter to the lodestone. ¾ Her method of using the lodestone was somewhat new, in that the magnet had to be held in the patient’s mouth to remedy fits of anger or rage, to make fasting bearable, and to keep lies and maliciousness at bay (Riethe, 1961). Magnetic Resonance Imaging (MRI) ¾ Nuclear magnetic resonance (NMR) is the spectroscopic study of the magnetic properties of the nucleus of the atom. Magnetic field with neutron & protons nuclear spin and charge distribution ¾ Resonance is an energy coupling that causes the individual nuclei, when placed in a strong external magnetic field, to selectively absorb, and later release, energy unique to those nuclei and their surrounding environment. Magnetic Resonance Imaging (MRI) ¾ NMR start since 1940s as an analytic tool in chemistry and biochemistry research. ¾ NMR is not an imaging technique but rather a method to provide spectroscopic data concerning a sample placed in the device. ¾ In the early 1970s, NMR can use to generate images that display magnetic properties of the proton, reflecting clinically relevant information. NMR mid 1980 MRI As clinical imaging applications increased MRI ¾ The protons in a material, with the use of an external uniform magnetic field and RF energy of specific frequency, are excited and subsequently produce signals with amplitudes dependent on relaxation characteristics and spin density, as previously discussed. ¾ Conventional MRI involves RF excitations combined with magnetic field gradients to localize the signal from individual volume elements (voxels) in the patient. Magnetic Field Gradients ¾ Magnetic fields are produced in a coil wire energized with a direct electric current of specific polarity and amplitude. ¾ Magnetic field gradients are obtained by superimposing the magnetic fields of one or more coils with a precisely defined geometry. ¾With appropriate design, the gradient coils create a magnetic field that linearly varies in strength versus distance over a predefined field of view (FOY). Magnetic Field Gradients ¾ Inside the magnet bore, three sets of gradients reside along the coordinate axes-x, y, and z-and produce a magnetic field variation determined by the magnitude of the applied current in each coil set. Slice select Gradients ¾ The RF antennas that produce the RF pulses do not have the ability to direct the RF energy. Thus, the slice select gradient (SSG) determines the slice of tissue to be imaged in the body. ¾ For axial MR images, this gradient is applied along the long (cranial-caudal) axis of the body. A narrow band RF pulse is applied to the whole volume, but only those spins along the gradient that have frequency equal to the frequency of the RF will absorb energy due to the resonance phenomenon. MRI ¾ The control interfaces, RF source, detector, and amplifier, analog to digital converter (digitizer), pulse programmer, computer system, gradient power supplies, and image display are crucial components of the MR system. Suggested Reading ¾ NessAiver M. All you really need to know about MR! physics. Baltimore, MD: Simply Physics, 1997. ¾ Price RR. The MPM/RSNA physics tutorial for residents: MR imaging safety considerations. RadioGraphies 1999;19:1641-1651. ¾ Saloner D. The MPM/RSNA physics tutorial for residents. An introduction to MR angiography. RadioGraphies 1995; 15:453-465. ¾ Shellock F, Kanal E. Magnetic resonanceimaging bioefficts, safety, and patient management, 2nd ed. New York: Lippincott-Raven, 1996. ¾ Smith HJ, Ranallo FN. A non-mathematical approach to basic MR!. Madison, WI: Medical Physics, 1989. Home Work (solved problems) ¾ 1. In what ways are electric and magnetic melds similar? In what ways are they different? • Answer: Similarities. Both fields originate in electric charges, and both melds can exert forces on electric charges. Differences: All electric charges give rise to electric fields, but only a charge in motion relative to an observer gives rise to a magnetic field. Electric fields exert forces on all charges, but magnetic fields exert forces only on moving charges. Home Work (solved problems) ¾ 2. A positive charge is moving vertically upward when it enters a magnetic field directed to the north. In what direction is the force on the charge? ¾ Answer: To apply the right-hand rule here, the fingers of the right hand are pointed north and the thumb of the hand is pointed upward. The palm of the hand faces west, which is therefore the direction of the force on the charge. Home Work (solved problems) • 3. The ends of a bar magnet are traditionally called its “poles,'' with the end that tends to point north called the “north pole'' and the end that tends to point south called the south Pole. It is observed that like poles of nearby magnets repel each other and that unlike poles attract. Explain this behavior in terms of the interaction of current loops. • Answer: Bar magnets with like poles facing each other are equivalent to parallel current loops whose currents are in opposite directions (see next Fig-a). Such loops repel. Bar magnets with opposite poles facing each other are equivalent to parallel current loops whose currents are in the same direction (see next Fig. b). Home Work (solved problems) • 4. A cable 5 m above the ground carries a current of 100 A from east to west. Find the direction and magnitude of the magnetic field on the ground directly beneath the cable. (Neglect the earth's magnetic fielded). • Answer: From the right-hand rule, the direction of the field is south. The magnitude of the field is Ultrasound ¾ Ultrasound is the term that describes sound waves of frequencies exceeding the range of human hearing and their propagation in a medium. ¾ Medical diagnostic ultrasound is a modality that uses ultrasound energy and the acoustic properties of the body to produce an image from stationary and moving tissues. ¾ Generation of the sound pulses and detection of the echoes is accomplished with a transducer, which also directs the ultrasound pulse along a linear path through the patient. Characteristics of sound Propagation of Sound Sound is mechanical energy that propagates through a continuous, elastic medium by the compression and rarefaction of "particles" that compose it. Wavelength, Frequency, and Speed ¾Ultrasound represents the frequency range above 20 kHz. ¾Medical ultrasound uses frequencies in the range of 2 to 10 MHz, with specialized ultrasound applications up to 50 MHz. ¾The speed of sound is the distance traveled by the wave per unit time and is equal to the wavelength divided by the period. ¾ The relationship between speed, wavelength, and frequency for sound waves is Wavelength, Frequency, and Speed ¾ The speed of sound is dependent on the propagation medium and varies widely in different materials. ¾ The wave speed is determined by the ratio of the bulk modulus (B) (a measure of the stiffness of a medium and its resistance to being compressed), and the density (ρ) of the medium: ¾ S1 units are kg/(m-sec2), kg/m3, and m/sec for B, ρ, and c, respectively. ¾ A highly compressible medium, such as air, has a low speed of sound, while a less compressible medium, such as bone, has a higher speed of sound. Wavelength, Frequency, and Speed ¾ The difference in the speed of sound at tissue boundaries is a fundamental cause of contrast in an ultrasound image. ¾ Medical ultrasound machines assume a speed of sound of 1,540 m/sec. The speed of sound in soft tissue can be expressed in other units such as 154,000 cm/sec and 1.54 mm/µsec, and these values are often helpful in simplifying calculations. ¾ The ultrasound frequency is unaffected by changes in sound speed as the acoustic beam propagates through various media. Ultrasound wave Wavelength, Frequency, and Speed ¾ Example: A 5-MHz beam travels from soft tissue into fat. Calculate the wavelength in each medium, and determine the percent wavelength change. Answer: In soft tissue, λ = c\f = (1.540 m/sec) /(5x106/sec) = 3.08 x 10-6 = 0.31 mm In fat, = (1.450 m/sec) /(5x106/sec) = 2.9 x 10-6 = 0.29 mm Interaction of Ultrasound with matter ¾ Refraction describes the change in direction of the transmitted ultrasound energy with nonperpendicular incidence. ¾ Scattering occurs by reflection or refraction, usually by small particles within the tissue medium, causes the beam to diffuse in many directions, and gives rise to the characteristic texture and gray scale in the acoustic image. ¾ Absorption is the process whereby acoustic energy is converted to heat energy. In this situation, sound energy is lost and cannot be recovered. Transducer ¾ Ultrasound is produced and detected with a transducer, composed of one or more ceramic elements with electromechanical properties. ¾ The ceramic element converts electrical energy into mechanical energy to produce ultrasound and mechanical energy into electrical energy for ultrasound detection. ¾ Major components include Transducer ¾ A piezoelectric material (often a crystal or ceramic). It converts electrical energy into mechanical (sound) energy. ¾ Resonance transducers for pulse echo ultrasound imaging are manufactured to operate in a "resonance" mode, whereby a voltage (commonly 150 V) of very short duration (a voltage spike of ~1 µsec) is applied, causing the piezoelectric material vibrate at a natural resonance frequency. Image data acquisition Introduction to Radiotherapy What is Radiotherapy? • Radiation therapy is one of several treatments used to treat cancer by itself or in combination with other forms of treatment, most often surgery or chemotherapy. • Radiation therapy is also called radiotherapy. • You've probably seen an X-ray of your teeth or some other part of your body. At high doses many times greater than those used for X-ray exams - radiation can kill cancer cells and shrink tumors. • More than half of all cancer patients receive some radiation therapy as part of their treatment. What is Radiotherapy? • Radiation is given either externally, through external beam radiation, or increasingly through internal radiation, also called brachytherapy. Radiotherapy Externally therapy Brachtherapy Capsulated isotopes Cobalt –60 Unit LINEAR ACCELERATORS • Medical linear accelerators (linacs) are accelerators which accelerate electrons to kinetic energies from 4 MeV to 25 MeV using non-conservative microwave RF fields in the frequency range from 103 MHz to 104 MHz, with the vast majority running at 2856 MHz linear accelerator ….. • In a linear accelerator the electrons are accelerated following straight trajectories in special evacuated structures called accelerating waveguides. Electrons follow a linear path through the same, relatively low, potential difference several times; hence, linacs also fall into the class of cyclic accelerators just like the other cyclic machines that provide curved paths for the accelerated particles (e.g., betatron) • The high power RF fields, used for electron acceleration in the accelerating waveguides, are produced through the process of decelerating electrons in retarding potentials in special evacuated devices called magnetrons and klystrons linear accelerator ….. • Various types of linacs are available for clinical use. Some provide x-rays only in the low megavoltage range (4 MV or 6 MV) others provide both x-rays and electrons at various megavoltage energies. A typical modern high energy linac will provide two photon energies (6 MV and 18 MV) and several electron energies (e.g., 6, 9, 12, 16, 22 MeV) Components of modern linacs • • • • (1) gantry; (2) gantry stand or support; (3) modulator cabinet; (4) patient support assembly, i.e., treatment couch; • (5) control console. linear accelerator Percentage depth dose (PDD) Tissue Air Ratio (TAR) PDD & TAR TRS - 398 Solved Problems • 1. A 4-MV linear accelerator is calibrated to give 1 rad (10-2 Gy) per MU in phantom at a reference depth of maximum dose of 1 cm, 100-cm SSD, and 10 x 10 cm field size. Determine the MU values to deliver 200 rads to a patient at 100-cm SSD, 10-cm depth, and 15 x 15 cm field size, given Sc=1.020, Sp = 1.010, %DD = 65.1. • Answer: Solved Problems • 2. A tumor dose of 200 rads is to be delivered at the isocenter which is located at a depth of 8 cm, given 4-MV x-ray beam, field size at the isocenter = 6 x 6 cm, Sc= 0.970, Sp = 0.990, machine calibrated at SCD = 100 cm, TMR(0.787)أ.Since the calibration point is at the SAD, SAD factor = 1. • Answer: