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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
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(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: