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1

BASIC SKILLS FOR RADIOTHERAPY PHYSICS Elaine Ryan

Chapter 1

Mathematical s kills r elevant to r adiotherapy

Fractions

A fraction is one number divided by another. The top number, a , is the numerator and the bottom number, b , is the denominator:

x

ab

=

Addition

To add two fractions together:

ab

cd

a db d

c db d

a d c db d

+ = ××

+ ××

= ×( ) + ×( )×

Example

Question: add together two - thirds and four - fi fths:

Answer:

two-thirds four-fifths+ = +

=××

+××

=×( ) + ×( )

×

=

23

45

2 53 5

4 33 5

2 5 4 33 5

22215

1715

=

Subtraction

To take one fraction away from another:

ab

cd

a db d

c bb d

a d c bb d

− = ××

− ××

= ×( ) − ×( )×

Aims and o bjectives

The aim of this chapter is to provide a review of the mathematics and basic physics that will help with the understanding of radiotherapy physics. This section aims to serve as a quick reference guide to help when confronted with problems further in the book.

After completing this chapter the reader should: • have a clear understanding of the basic math-

ematical skills required to understand the physics of radiotherapy

• be able to complete simple mathematical prob-lems applied to radiotherapy physics

• understand the theories and concepts of basic physics

• be able to perform simple calculations using basic physics formulae.

COPYRIG

HTED M

ATERIAL

2 Practical Radiotherapy

Multiplication

To multiply two fractions together:

ab

cd

a cb d

× = ××

Ratios

A ratio is another way of expressing a fraction, but is more useful when trying to compare two quantities. A ratio is expressed as m : n (the ratio of m to n )

To divide a quantity A into a ratio m : n to fi nd p : q :

pA

m nm=

+( )×

q

Am n

n=+( )

×

Example

Question: take three - tenths away from four - ninths:

Answer:

four-ninths three-tenths− = −

= ××

− ××

= ×( ) − ×

49

310

4 109 10

3 99 10

4 10 3 999 10

1390

( )

×

=

Example

Question: multiply together two - thirds and three - fi fths

Answer:

23

35

2 33 5

615

25

× =××

= =

Division

To divide one fraction by another, you have to invert one fraction and then multiply them together:

ab

cd

ab

dc

a db c

÷ = × = ××

Example

Question: divide two - fi fths by three - eighths Answer:

25

38

25

83

2 85 3

1615

11

15÷ = × = ×

×= =

Example

Question: a treatment regimen with a total dose of 20 Gy must be given in two phases. The ratio of the dose in each phase is 3 : 1. What dose must be given for each phase? Answer:

Phase 1 dose Gy=+

× =206 2

6 15

Phase 2 dose Gy=+

× =206 2

2 5

The ratio of the phase 1 to phase 2 dose is 15 : 5.

To fi nd a ratio in the form p : 1 or 1 : q :

q

nm

=

pmn

=

Basic Skills for Radiotherapy Physics 3

Proportionality

If two quantities are linearly proportional, then they have a constant scaling or multiplication factor, k . When a a b then a = kb where k is a constant of proportionality.

Standard f orm

A very large or very small number needs to be expressed in standard form, or scientifi c nota-tion. A number in standard form has a number between 1 and 10 multiplied by 10 raised to a power. The power is easy to work out by count-ing the number of times that you shift the sig-nifi cant fi gure.

Example

Question: if an anterior fi eld of length 4 cm appears as 3 cm on a simulator fi lm fi nd the fi lm magnifi cation. Answer: the measurements are in the ratio 4 : 3, so the magnifi cation is:

143

1 0 75: : .=

Example

Question: a treatment machine delivers a dose of 0.75 Gy in 15 seconds. How long will it take to deliver a dose of 2 Gy? Answer:

k = = ( )15 0 75 20. s Gy

t k= × =2 40s

Percentages

Percentages are often used to indicate the amount by which a value has changed or by how much a value should be changed. Per cent means ‘ per 100 ’ or 1/100. Therefore 50% = 50/100 or one half.

To change a fraction ( a / b ) to a percentage ( x ):

xab

= ×( )1001

%

Example

Question: a CT plan shows a 95% coverage of the prostate on 21 slices out of 30. What percentage of the prostate volume is covered by the 95% isodose?

Answer: x = ×( ) =2130

1001

70% %

To work out x % of y :

xx y

% = ×100 1

Example

Question: the spinal cord is covered by the 90% isodose in a plan. If the patient is given 1.75 Gy for the fi rst fraction, what dose did the spinal cord receive?

Answer: z = ×( ) =90100

1 751

1 575.

. Gy

To fi nd z , an increase of y by x %:

zx

y y= ×( ) +100

Example

Question: a fi eld size of 30 cm needs to be increased by 12% in order to allow for organ motion. What is the correct enlarged fi eld size?

Answer: z = ×( ) + =12100

30 30 33 6. cm


Dividing numbers in standard form:

a bab

x y x y×( ) ÷ ×( ) = ( ) × −( )10 10 10

y 10 x x

10 10 1 1 100 10 2 2

1000 10 3 3 10 000 10 4 4

100 000 10 5 5

This is a ‘ log ’ table to the ‘ base ’ 10. Therefore:

• log 10 10 = 1 • log 10 100 = 2 • log 10 1000 = 3.

The value for y gets very large, but when it is expressed as a logarithm it stays small.

Exponentials

A special logarithmic case is ‘ e ’ , where a constant change in x results in the same fractional change in y . The example, y = e x can also be expressed as log e y = x or as ln y = x . A log to base e is a ‘ natural logarithm ’ (ln) and ‘ e ’ has the value 2.718:

e x x y

e 1 1 2.718 e 2 2 7.389 e 3 3 20.086 e 4 4 54.598 e 5 5 148.413

If e x is plotted against x , an exponential curve is obtained. This type of curve describes many naturally occurring events such as radioactive decay. Any number would produce a similarly shaped curve, but, with ‘ e ’ , the gradient at any point is proportional to the value of x , i.e. if the value of x is halved the rate of decrease of y is also halved.

Similar t riangles

If one triangle is an enlargement of another they are similar triangles. The triangles must have the

Examples

Question: fi nd 0.0895 in standard form. Answer: move the decimal place twice to the right so the power = − 2: 8.95 × 10 − 2 Question: fi nd 800 000 in standard form. Answer: move the decimal place fi ve places to the left so the power = 5: 8.0 × 10 5 Question: fi nd 47 000 000 000 in standard form. Answer: move the decimal point a total of 10 places to the left so the power = 10: 4.7 × 10 10

Multiplying numbers in standard form:

a b a bx y x y×( ) × ×( ) = ×( ) × +( )10 10 10

Example

Question: fi nd 6.4 × 10 8 multiplied by 7.2 × 10 4 . Answer: (6.4 × 10 8 ) × (7.2 × 10 4 ) = (6.4 × 7.2) × 10 (8+4) = 46.08 × 10 12 = 4.608 × 10 13

Example

Question: Find 6.4 × 10 8 divided by 8.0 × 10 − 4 . Answer: (6.4 × 10 8 ) ÷ (8.0 × 10 − 4 ) = (6.4/8) × 10 (8 − − 4 ) = 0.8 × 10 12 = 8.0 × 10 11

Logarithms

A log is just a way of expressing a number as the ‘ power ’ of another number (the ‘ base ’ ).

For the general equation:

y b y xxb= =log

If b is given the value 10, then y = 10 x . Now, if x is given the values 1, 2, 3, etc.:


same corresponding angles. Figure 1.1 shows two similar triangles.

In radiotherapy, similar triangles are used in three ways: fi eld size; inverse square law; and magnifi cation.

Field s ize To fi nd the fi eld size (FS) variation with the source to skin distance (SSD).

In Figure 1.2 the ratio SSD 1 : SSD 2 is equal to the ratio FS 1 : FS 2 , therefore:

SSDSSD

FSFS

2

1

2

1

⎛⎝⎜

⎞⎠⎟

= ⎛⎝⎜

⎞⎠⎟

Provided that any three of these variables are known, it is possible to calculate the fourth, unknown value.

The i nverse s quare l aw This relationship considers similar triangles in three dimensions. In Figure 1.3 the top of the object is a point source of X - rays. All photons travel in straight lines. Photons in the shaded area ‘ A ’ have come from the point source, are at a distance ‘ d ’ from it and are travelling directly away from it. Photons in the shaded area ‘ B ’ are further away from the point source at a distance ‘ 2 d ’ , but the number of photons in area B is the same as in area A. Therefore, the number of photons per unit area, or the intensity, is less in B than in A. The decrease in intensity can be derived using similar triangles.

In Figure 1.3 , if the distance from the source is doubled, then all sides of the triangles will double. Therefore:

Figure 1.1 Two similar triangles.

cb

a 2a

2c2b

Figure 1.2 Similar triangles to illustrate how fi eld size changes with source to skin distance (SSD).

FS1

FS2

SSD2

SSD1

Example

Question: at an SSD of 100 cm, the fi eld length is 12 cm. What is the fi eld length if the SSD is increased to 150 cm? Answer:

SSDSSD

FSFS

2

1

2

1( ) = ( )

FS FSSSDSSD

cm2 12

1

12150100

18= ( ) = × ( ) =

Figure 1.3 A diagram of the inverse square law.

d

ba

2a

B

A

Point source

2b

2d


Area 1 = × =a b ab

Area 2 2 2 4= × =a b ab.

Therefore, the area is four times greater but the number of photons is the same and:

II= 0

4

where I is the intensity and I 0 is the initial inten-sity of the beam.

I Ia b

a b= × ×

× ×( )04

If the distance is increased from d to 10 d :

Area 1 = × =a b ab

Area 2 10 10 100= × =a b ab

I I

a ba b

= × ×× ×( )0

100

When the distance from the source increases, I always decreases by:

I Idd

= × ⎛⎝

⎞⎠0

1

2

2

or it can be said that I is inversely proportional to d 2 .

Id

∝ ( )1 2

For the inverse square law to work, all photons must be travelling from the same point and in a straight line. Therefore, there must be a point source and no scatter (change of direction).

Magnifi cation Figure 1.4 shows an object being imaged onto a fi lm using a point source. The magnifi cation ( M ) is defi ned as

M

IO

=( )( )

size of imagesize of object

Where FFD is the focus to fi lm distance and FOD is the focus to object distance, similar tri-angles can be used to work out dimensions, if three quantities are known.

M

IO

= = FFDFOD

Pythagoras ’ s t heorem

For a right - angled triangle the longest side is called the hypotenuse. The square of the hypo-tenuse is equal to the sum of the squares of the other two sides. For the triangle shown in Figure 1.5 :

a b c2 2 2+ =

Figure 1.4 An object being imaged using a point source.

FOD

FFD

Image

Object

Example

For a triangle with sides a = 3.5 cm and b = 4.2 cm, what is the length of the hypotenuse?

c = ( ) + ( ) =3 5 4 2 5 52 2. . . cm


Trigonometry

A right - angled triangle has a hypotenuse, an adjacent side and an opposite side.

If one of the angles apart from the right angle is chosen as the marked angle ( θ ), then the sides are as follows: the hypotenuse is opposite the right angle. It is always the longest side. The adjacent side joins θ to the right angle. The oppo-site side is opposite θ .

The trigonometric ratios for calculating unknown sides or angles are as follows:

Sinθ = oppositehypotenuse

Cosθ = adjacent

hypotenuse

Tan

oppositeadjacent

θ =

Figure 1.5 Pythagoras ’ s theorem.

Example 1

Question: in Figure 1.6 a – c fi nd the length of the side marked x Answer:

(a) S opphyp

inθ =

therefore opp = Sin θ × hyp = Sin 37 94.2 = 56.7

(b) Cosadjhyp

θ =

therefore hypadj

Cos Cos= = =

θ58 2

2463 7

..

(c) Tanoppadj

θ =

therefore opp = adj × Tan θ = 30.7 × Tan52 = 39.3

Example 2

Question: in Figure 1.6 d,e fi nd the unknown angle θ . Answer:

(d) Sinθ = opphyp

therefore

(e) Tanoppadj

θ =

therefore

θ = ⎛⎝⎜

⎞⎠⎟

= ( ) = °− −Sin Sin1 1 54 267 8

53opphyp

.

.

θ = ⎛⎝⎜

⎞⎠⎟

= ( ) = °− −Tanoppadj

Tan1 1 38 442 9

42..

Figure 1.6 Examples of trigonometry.

94.2

(a) (b) (c) (d) (e)

58.2

30.7

42.9

38.454.2

67.8 θ

θ

XX

X

37° 24°

52°


Basic p hysics r elevant to r adiotherapy

Units of m easurement

It is often necessary to present the results of measurements or calculations in a numerical fashion. The number used to do this requires two parts: a pure number and the unit in which the quantity has been measured or calculated, e.g. 100 Gy where the unit is the ‘ gray ’ and the pure number is 100. There are, however, some basic units on which all measurements are based and a number of derived units that are based on combinations of the basic units.

Basic u nits All measurements used in science are based on three basic units of measurement (Table 1.1 ). These are the basic units of measurement because they are independent of each other and cannot be converted from one to another.

Derived u nits Any physical quantities other than mass, length or time are measured in derived units, some of which are listed in Table 2.2 , Chapter 2 .

Prefi xes As discussed earlier, it is easier to present very large or small numbers in standard form. A way of simplifying this further is by the use of univer-sal prefi xes. Table 1.2 shows common prefi xes.

For example, 0.000007 m = 7 × 10 − 6 = 7 μ m and 3000 m = 3 × 10 3 = 3 km.

Suffi xes Suffi xes are used to identify a specifi c value of a given quantity. For example, the activity of a radionuclide may be denoted using the symbol A . The activity of the radionuclide will decay over a period of time and it may be necessary to specify the activity at a given time. Time may be denoted using the symbol t and activity at time t is identifi ed as A t .

A special case is the suffi x ‘ 0 ’ which is nor-mally used to indicate the initial value of a quan-tity. The initial activity of the radionuclide would therefore be A 0 .

Classical m echanics ( T able 1.3 )

Classical, or newtonian, mechanics is the study of the connection of matter, forces and motion.

Table 1.1 Basic units of measurement

Unit Symbol SI unit

Mass m kilogram (kg)

Length l metre (m)

Time t second (s)

Table 1.2 Common prefi xes

pico p 10 − 12 kilo k 10 3

nano n 10 − 9 mega M 10 6

micro μ 10 − 6 giga G 10 9

milli m 10 − 3 tera T 10 12

Table 1.3 A summary of units used in classical mechanics

Unit Symbol Equation SI Unit

Velocity v v = l / t m/s

Acceleration a a = v / t m/s 2

Weight w w = mg newton (N)

Force F F = ma newton (N)

Work and energy

E E = Fl joule (J)

Power P P = E / t watt (W)

Velocity The average velocity ( v ) of a body in motion is the total distance ( l ) travelled by the body divided by the time ( t ) taken for this travel in a given direction. It is given as:


v m

sl mt s

( ) =( )( )

Although the SI units for distance and time are metres and seconds, respectively, velocity can be expressed in units such as miles per hour or kilometres per hour.

Acceleration Linear acceleration ( a ) is the rate of change in velocity ( v ) or the ratio of a change in velocity to a corresponding change in time ( t ). This is when a body in motion is speeding up or slowing down. It is given as:

a ms

v ms

t s2( ) =( )( )

Acceleration is expressed in units of metres per second squared (m/s 2 ).

Force Force ( F ) is the mass ( m ) of a body multiplied by the acceleration ( a ) acting on it, which is expressed as:

F m aN kg m s( ) = ( ) × ( )2 .

This is Newton ’ s second law of motion. Weight is the name given to the force exerted

on a body by the gravitational acceleration of the earth, g . Therefore, weight is given as:

W m gN kg m s( ) = ( ) × ( )2

and is just a special case of Newton ’ s second law of motion. This gravitational acceleration is measured as about 10 m/s 2 . Force is expressed in newtons (N), where 1 N = 1 kg m/s 2 .

It is important to distinguish between mass and weight. If something is weighed as 0.5 kg, this is actually its mass. The weight of this body is 0.5 kg × 10 m/s 2 = 5 N.

Work, e nergy and p ower Work is defi ned as force (F) multiplied by the distance (d) moved, i.e. if a force is applied to a body then work is performed:

W F dJ N( ) = ( ) × ( )m .

The unit of work is the joule (J) and 1 J of work is performed when a force of 1 N moves through a distance of 1 m: 1 newton metre or joule = 1 kg m 2 /s 2 .

Energy is defi ned as the ability to do work. Two special forms of energy must be considered: potential and kinetic.

Potential energy (PE) is the energy a body pos-sesses by virtue of its condition or state. Potential energy is therefore ‘ stored ’ energy such as that in a battery or coiled spring. A body ’ s potential energy is equal to the amount of work that has been performed to put it in its particular posi-tion, i.e. the force applied multiplied by the dis-tance moved. For a body of mass m above the ground by a distance h and under the effect of gravity, g , its potential energy = mg h.

Kinetic energy (KE) is the energy that a body possesses by virtue of its motion. A body ’ s kinetic energy is that energy that would have to be applied to the body to bring it completely to rest. A body of mass m moving at a velocity v will have kinetic energy

= 12

2mv

Power is defi ned as the rate of doing work or the rate at which energy is used. The unit of power is the watt (W) and 1 watt (W) = 1 J s − 1 .

Waves

Electromagnetic waves all travel in a vacuum with a velocity equal to the speed of light ( c ) where c = 3 × 10 8 m/s. All waves have a wave-length, λ , the distance travelled in one cycle, and a frequency, v , the number of cycles per second, measured in hertz (Hz). Figure 1.7 shows the


wavelength, frequency and amplitude of a wave. The wave in this example has a frequency of 3 Hz.

For all waves the speed (c) is equal to the fre-quency, v , times the wavelength, λ :

c m vm s s( ) = ( ) × ( )λ 1 .

Electricity, m agnetism and e lectromagnetic r adiation

Electricity and magnetism are very important to the understanding of radiation. The movement of charges in an electric fi eld and the combina-tion of electric and magnetic fi elds in the form of electromagnetic waves underpin X - ray produc-tion and interactions.

Electric fi elds

Separate but similar charges repel each other and unlike charges attract each other. As two elec-trons are both negatively charged they repel each other. The force with which they do this is the same as the force between any two charges ( q 1 and q 2 ), and is given as:

F kq qr

N( ) = 1 22

with units of newtons ( N ). The electric fi eld strength at a point in space

is given by:

NC

Fq C

( ) = ( )( )N

An electric fi eld can be created across two electrodes – a cathode, which is positive, and an anode, which is negative – with the application of a potential.

Current

Current is the motion of electrons from one elec-trode to another through a conductor. Current is defi ned as the net charge ( Q ) fl owing through an area A per unit time ( t ), which is expressed as

I AQ Ct s

( ) =( )( )

The unit of current is the ampere ( A ), which is equal to Coulombs per second ( C/s ). This is shortened to ‘ amp ’ . Currents in X - ray tubes are of the magnitude of milliamps (mA).

Voltage

The force that drives current around a circuit is the potential difference. This is measured in volts. An electron moving through a potential difference experiences a net change in energy, measured in electron - volts. This effect is analo-gous to a mass falling through a given height difference in a gravitational fi eld. One electron volt is the energy required by an electron to move through a potential difference of 1 V.

Resistance

The voltage of a circuit divided by its current is known as the resistance of the circuit. This is given as:

RV VI A

Ω( ) =( )( )

This is Ohm ’ s law. The unit of resistance is the ohm ( Ω ), which is equal to 1 volt per ampere ( V/A ).

Figure 1.7 The wavelength, frequency and amplitude of a wave.

Am

plitu

de

–1

+1

1 s

t

λ


Magnetism

A magnetic fi eld produced by a magnet has a north and a south pole. As with electric charges, opposite poles attract and like poles repel; however, unlike electric charges the opposite charges in a magnet cannot be isolated, but always come in pairs. The magnetic fi eld lines produced by a magnet run from north to south.

Electromagnetic i nduction

When a magnet is moved through a coil of wire, a current is induced in the coil. The opposite is also true: a current passing through a coil gener-ates a magnetic fi eld through the centre of the coil. This is known as electromagnetic induction.

Electromagnetic w aves

The phenomenon of electromagnetic induction tells us that electric and magnetic fi elds that vary with time are not independent – one affects the other. When a magnetic fi eld moves with time it induces an electric fi eld perpendicular to it, and the same is true of the inverse situation when a time - varying electric fi eld induces a magnetic fi eld. Electromagnetic waves are so called because they are waves of energy transmitted by oscillat-ing interdependent electric and magnetic fi elds. As an electromagnetic wave is energy radiating away from an oscillating source it is also called electromagnetic radiation.

Radio waves, microwaves, ultraviolet light, infrared light, visible light, X - rays and gamma rays are all types of electromagnetic radiation. Electromagnetic radiation can be considered to behave as both a wave and a particle because it demonstrates both wave - and particle - like properties.

Wave - l ike p roperties Electromagnetic radiation is usually thought of as being a wave. How is it known that it behaves like a wave? If visible light is considered, it can

be demonstrated that visible light can undergo refl ection, refraction, diffraction, interference, etc., all of which are properties of waves. Exper-imental work with mass spectrometers has also demonstrated the diffraction of X - rays.

Particle - l ike p roperties Some phenomena associated with electromag-netic radiation, such as the photoelectric effect and Compton scattering, cannot be explained by the wave theory. To explain these phenomena, electromagnetic radiation must be considered to behave as particles or packets of energy rather than as waves.

Wave p article d uality The wave - and particle - like properties of electro-magnetic radiation can be related. From the wave - like properties, it is known that the wave can be described by the following relationship:

E v= λ

If particle - like properties are considered, it can be shown that the energy carried by the particles or photons is given by:

E hv=

where E is the energy of the photon, expressed in joules, h is known as Planck ’ s constant and is equal to 6.62 × 10 − 34 J s, and v is the frequency expressed in cycles per second.

These two equations can be combined because it is known that v = c / λ and we can therefore substitute for v in the equation E = hv which results in the relationship:

Ehc=λ

It is also known that h = 6.62 × 10 − 34 J s and c = 3 × 10 8 m s − 1 and, if energy is converted to electron volts, where 1 eV = 1.602 × 10 − 19 J, then the relationship simplifi es to:


E = × −1 24 10 6. λ

where E is expressed in electron volts (eV) and λ is expressed in metres (m). From this relation-ship it can be seen that, as the energy increases, the wavelength will decrease.

The e lectromagnetic s pectrum

The electromagnetic spectrum includes radiation from the longest radio waves to the shortest X - rays. All forms of electromagnetic radiation have an associated frequency and energy. It should be noted that there is often a degree of overlap between the radiation types and that there are therefore no distinct boundaries between them. Figure 1.8 shows the electromag-netic spectrum.

All radiation types share some common properties:

• They are composed of transverse electric and magnetic waves.

• They travel at the speed of light in a vacuum. • In free space they travel in straight lines. • In free space they obey the inverse square law.

Atomic s tructure

The atom is the basic component of an element and elements are the individual entities from

which all matter is composed. An atom consists of a central nucleus, which is positively charged. It also has electrons orbiting this nucleus, which are negatively charged.

Nucleus

The nucleus can be broken down further and is composed of two kinds of elementary particles or nucleons: protons and neutrons. Protons (p) are positively charged with a charge of 1.602 × 10 − 19 C whereas neutrons (n) have no charge, thus giving the nucleus its overall positive charge. The electrons (e) orbiting the nucleus are negatively charged and their charge is equal in magnitude to that of the proton, i.e. − 1.602 × 10 − 19 C. The number of electrons orbiting the atom is equal to the number of protons in the nucleus and the overall charge of the atom is therefore balanced.

The mass of a proton is the same as that of a neutron and is 1.67261 × 10 − 27 kg, while the mass of an electron is much less, 9.109 × 10 − 31 kg, which is approximately 1/1840 of the mass of the proton or neutron.

Mass n umbers and a tomic n umbers

For an element X, with mass number A and atomic number Z , its chemical symbol is written as:

ZA X

Figure 1.8 The effect on the tube spectrum when fi ltration has been added to the exit beam.

10–16 10–14 10–12 10–10

102010221024

10–8 108

1081018

10–6 106

1061016

10–4 104

1041014

10–2

Increasing wavelength (λ)

Increasing frequency (ν)Visiblclight

Gamma raysUV

X-raysInfrared

Microwaves

Radio-waves

λ (m)

ν (Hz)

102

1021012

100

1001010


A is equal to the number of nucleons, i.e. the total number of protons and neutrons in the nucleus, whereas Z denotes the total number of protons in the nucleus. As the number of protons and electrons is equal, the atomic number also indicates the number of electrons outside the nucleus. For example, an atom of carbon would be written as:

612C

where the mass number 12 indicates that the total number of protons and neutrons in the nucleus is 12 and the atomic number 6 indicates that there are 6 protons (and 6 electrons). As the mass number indicates that there are a total of 12 nucleons, and from the atomic number it is known that there are 6 protons, it can be calcu-lated that there are also 6 neutrons.

Isotopes

Atoms of the same element can exist with differ-ent mass numbers but the same atomic number, i.e. the number of neutrons present in the nucleus is different but the number of protons remains the same. Such elements are known as isotopes. For example, cobalt has 11 isotopes, of which two are 27

59Co and 2760Co. Of the 11 isotopes, only 27

59Co is stable. 27

60Co is used in cobalt treatment machines and some brachytherapy afterloading machines. The other isotopes are also radioactive.

The important point to remember about iso-topes is that, as the number of protons and there-fore the number of electrons is unchanged, the chemical properties of the element are unchanged. When the imbalance in the number of protons and neutrons becomes too great, the atom and therefore the isotope may become radioactive and the isotope is called a radioisotope.

Electron o rbits

Within the atom, the electrons orbiting the nucleus are arranged in shells, each being identi-

fi ed by a letter K, L, M, etc. starting from the shell closest to the nucleus. Each shell has a limit to the number of electrons that it can contain and the inner shells are always occupied fi rst. The K - shell can accept up to 2 electrons, the L - shell up to 8, the M - shell up to 18, etc. For example, carbon with its atomic number of 6 has 2 elec-trons in the K - shell and the remaining 4 electrons in the L - shell. A diagram of a carbon atom is illustrated in Figure 1.9 .

The electron confi guration of the atom deter-mines its chemical properties. Neon ( Z = 10), for example, is chemically inert because both the K - and L - shells are full. Fluorine ( Z = 9), however, has one electron missing from its outer shell and is chemically reactive. Another example, argon ( Z = 18), suggests the presence of subshells. It is chemically inert but has only eight electrons in its M - shell, suggesting that atoms are also stable when subshells are full.

Atomic e nergy l evels

Electron orbits also have an energy associated with them. This is known as the binding energy and is dependent on the force of attraction between the nucleus and the orbiting electrons. The binding energy associated with the K - shell is greater than for shells further away from the nucleus because the force of attraction on the K - shell electrons will be greater.

It is the binding energy that must be overcome if an electron is to be removed from its shell. The

Figure 1.9 Representation of the particles in a carbon atom.

Nucleus consisitingof six protons and six neutrons

Electrons


potential energy associated with shells further from the nucleus will be greater than for those shells close to the nucleus, which is analogous to the greater potential energy associated with an object the further it is away from the earth.

Heat and t emperature

Heat

The atoms and molecules of any material are in constant motion. The type of motion varies depending on the form of the material. Within a solid, particles vibrate about a fi xed position, whereas in liquids and gases motion is much more random as the particles have much greater freedom of movement. All materials therefore possess kinetic energy due to this motion. Heat is the form of kinetic energy possessed by a mate-rial resulting from the motion of its particles and, as heat is a form of energy, the SI unit for it is the joule.

Temperature

Temperature is a measure of the level of kinetic energy of the atoms and molecules of a material. As the speed of vibrations increases within a body, so the kinetic energy due to these vibra-tions will increase and the material will increase in temperature. The temperature determines the direction of heat fl ow when one object is brought into thermal contact within another. Heat will fl ow from a region of higher to a region of lower temperature.

Temperature can be expressed in several units of measurement:

• Celsius (C): a temperature scale based on certain properties of water. By defi nition, water freezes at 0 ° C and boils at 100 ° C.

• Kelvin (K): water freezes at 273K and boils at 373K. This is known as the absolute or Kelvin temperature scale and is important because at

0K all particles within a material should be a rest. Therefore, at 0K, or absolute zero, a body will have no heat energy. Conversion between the Kelvin and Celsius scales can be made using the following relationship: Temperature (K) = temperature ( ° C) + 273.

• Fahrenheit (F): water freezes at 32 ° F and boils at 212 ° F. Conversion between the Fahrenheit and Celsius scales can be made using the following relationship: Temperature ( ° F) = (temperature ( ° C) + 32) × (8/5).

Heat t ransfer

Heat energy can be transferred from one body to another by the processes of conduction, convec-tion and radiation.

Conduction This process applies principally to solids or objects that are in direct contact with each other. Transfer of heat is by collision of particles in one body with those in the other body. If the tem-perature of one end of a body is increased, this heat will fl ow along the body due to the colli-sions of neighbouring particles. The fl ow of heat in a body will be affected by a number of physi-cal factors: cross - sectional area and length of the body and the difference in temperature between the two ends of the body.

Heat flow cross-sectional area∝ ( )A

Heat flow length∝ ( )1 l

Heat flow temperature difference∝ −( )T T1 2

Heat fl ow is also dependent on the material itself, in particular its thermal conductivity ( k ). The rate of heat fl ow in a body is given by:

kA T Tl2 1−( )


Convection Convection is the main process by which heat is transferred in a fl uid (liquids and gases). Heat energy is moved around by circulation of the heated fl uid. As a fl uid is heated it becomes less dense and will therefore rise. As it rises, it is replaced by cooler fl uid, which is then heated and rises, so the process continues creating convec-tion currents in the fl uid by virtue of this movement.

Radiation

This is the only form of heat transfer that can take place in a vacuum. As no particles are present in a vacuum, transfer due to neither col-lision of particles nor convection currents can take place. As particles in a body vibrate they emit energy in the form of electromagnetic waves. Heat radiation occurs in the electromagnetic spectrum just beyond the red part of the visible spectrum. One of the properties of electromag-netic radiation is that it can travel in a vacuum. Therefore it is possible to transfer heat across a vacuum. It is by this process that heat from the sun is felt, because it has to pass through a vacuum before it reaches the earth.

Examples of these processes of heat transfer can be seen in the cooling of the anode/target of a fi xed anode X - ray tube. By one route heat builds up in the tungsten target and is transferred by conduction to the copper anode. From the copper anode, the heat is transferred to the sur-rounding oil and from there by convection to the surrounding housing. Heat is conducted through

the housing and then transferred by convection to the air. Heat will also be radiated from the tungsten target across the inside of the evacuated glass envelope to the wall of the glass envelope, from where it is transferred by conduction to the oil surrounding the glass envelope.

Self - t est q uestions

1. The spinal cord is covered by the 90% isodose in a plan. If the patient is given 2.00 Gy for their fi rst fraction, what dose did the spinal cord receive?

2. The electromagnetic spectrum includes radia-tion from the longest radio waves to the short-est X - rays. List three common properties.

3. Defi ne the term electromagnetic induction. 4. What is wave particle duality? 5. What is meant by electron - binding energy?

Further r eading

Bomford CK , Kunkler IH , Sherriff S . Walter and Miller ’ s Textbook of Radiotherapy , 6th edn . Edinburgh : Churchill Livingstone , 2003 .

Graham DT , Cloke P . Principles of Radiological Physics , 4th edn . Edinburgh : Churchill Livingstone , 2003 .

Meredith WJ , Massey JB . Fundamental Physics of Radiology , 3rd edn . Chichester : John Wright & Sons , 1977 .

Williams JR , Thwaites DI . Radiotherapy Physics in Practice , 2nd edn . Oxford : Oxford University Press , 2000 .

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