1

Physics Applied to Radiology

Chapter 5 Chapter 5

2

Electromagnetic Energy Spectrum continuous range of energy

spectrum indicates that the distribution of energies exist in an uninterrupted band rather than at specified levels

released by accelerating charged particles moves through space or matter as oscillating

magnetic & electric fields needs no carrier medium but can have one can penetrate or interact with matter

3

Electromagnetic Spectrum

transverse energy waves traveling as magnetic & electric fields to each other maxima and minima of wave occur simultaneously unlike other waves, needs no carrier

4

Electromagnetic Spectrum Chart

ultraviolet

10

10

10

10

10

10

10

10

5

0

-5

-10

-15

-20

energy (eV) frequency (Hz)

1025

1020

1015

1010

105

100

10-5

wavelength (m)

10

10

10

10

10

10

10

-15

-10

-5

0

5

10

15

xray

infrared

microwave

TV/FM radio

radio

long waves

gamma

light

5

EMS RelationshipsWhich of the following has the highest energy?

Radio waves or visible light

frequency = 3.2x1019Hz or 4.9x1014Hz

wavelength = 8.5x10-6m or 4.2x10-12m

6

EMS Relationships (cont.)

Which of the following has the longest wavelength?

microwave or ultraviolet waves

energy = 2.3x10-5eV or 57keV

frequency = 3.1x108Hz or 8.9MHz

7

EMS Relationships (cont.)Which of the following has the lowest frequency?

Red light or yellow light

energy = 980 eV or 6.25x10-2keV

wavelength = 8325 mm or 4.78x10-3m

8

General Characteristics of EMS no mass or physical form travel at speed of light (c) in a vacuum (or air)

c = 3 x 108 m/s travel in a linear path (until interaction occurs)

dual nature: wave vs. particle unaffected by

electric or magnetic fields gravity

9

Characteristics (cont.) obeys the wave equation

c = obeys the inverse square law

I1d12 = I2d2

2

10

EM Interactions with Matter sections may overlap general interactions with matter include

scatter (w or w/o partial absorption) absorption (full attenuation)

11

EM Interactions (cont.) probability if matter size the wavelength

examples: radiowaves vs. TV antena microwaves vs. food light vs. rods & cones in eye x rays vs. atom

ionization occurs only EM energy > 33 to 35 eV high ultraviolet, x-ray, gamma

12

Dual Nature of EM Radiation continuously changing force fields

energy travels as sine WAVE macroscopic level

photon or quantum small bundle of energy acting as a PARTICLE microscopic level

13

PARTICLE vs. WAVE (in general) Wave

extended in space always in motion repeating

Particle (mass) localized in space moving or stationary

14

Wave Characteristics cycle:

one complete wave form or repetition

crest

trough

15

Wave (cont.) amplitude

max. displacement from equilibrium

+

-

0

16

Wave (cont.) wavelength

distance traveled by wave

= d/cycle Unit meter

m

17

Wave (cont.) frequency f or

number of cycles per unit time

Unit hertz Hz #/t Example below: 2 cycles/s = 2 Hz

time = 1 s1 2

18

Wave (cont.) For the wave depicted below, determine the frequency and wavelength.

t = 25 ms d = 58 nm cycles = 4.5

f = #/t = 4.5 cycles/25 ms = 4.5/25 x10-3 = 180 Hz

= d/cycle = 58 nm/ 4.5 cyc. = 58 x 10-9/4.5 = 1.3 x 10-8 m

time = 25 ms

d = 58 nm

19

Wave (cont.) velocity v (general) c (EM radiation)

speed each cycle travels Unit m/s

total distance wave moves in time period v of EM radiation always = c

20

Mathematical Relationships for EM Waves wave equation

general: v = forv EM radiation: c = for c

constant velocity at c

v = c = 3x108m/s

of EM are inversely proportional

for vice versa

21

Inversely Proportional as one goes up other goes down

v = fsame 100 = 1 100100 = 2 50100 = 4 25100 = 5 20100 = 10 10

22

Example An x-ray photon has a wavelength of 2.1nm.

What is its frequency?f= 2.1x10-9m c = 3x108m/s

c = f

f= c / = [3x108m/s] / [2.1x10-9m]

= 1.428571428571 x 1017 /s

= 1.4 x 1017 Hz

23

Example #2 A radio station broadcasts at 104.5 MHz.

What Is the wavelength of the broadcast?

= ?? 104.5 x 106 /s = [c = 3 x 108 m/s]

c = f= c / f= [3 x 108 m/s] / [104.5 x 106 /s]

= 0.028708134 x 102 m

= 2.871 m

24

Example #3 What it the frequency of microwave

radiation that has a wavelength of 10-4 m?

f= ?? 1 x 10-4m = [c = 3 x 108 m/s]

c = f= c /

= [3 x 108 m/s] / [1 x 10-4m ]

= 3 x 1012 Hz

25

Particle Nature (Quantum Physics) Photon (quantum)

view as if a single unit of EM radiation indivisible

Views EM radiation as a particle "bundle of energy" acts like a particle (but is not particle)

relates E to (direct relationship) "count" # of photons per unit time f = E

26

Mathematics E f E = h f

h = Planck’s constant

= 4.15 x 10-15 eVs units

usual energy units = J EM energy units = variation of J

[eVs][/s] = eV

x rays & gamma rays usually in keV or MeV

27

Example What is the energy (keV) of an x-ray photon

with a frequency of 1.6 x 1019 Hz?

E = ?? 1.6x1019Hz = f [h = 4.15 x 10-15 eVs]

E = h f

= [4.15 x 10-15 eVs] [1.6 x 1019 Hz]

= 6.64 x 104 eV

= [6.64 x 104 eV] / [103 ev/keV ]

= 6.64 x101keV = 66 keV

28

Example #2 What is the energy in MeV of an x-ray

photon with a frequency of 2.85 x 1021 Hz?

E = ?? 2.85x1021Hz = f [h = 4.15 x 10-15 eVs]

E = h f= [4.15 x 10-15 eVs] [2.85 x 1021 Hz]= 11.8275 x 106 eV

= [11.8275 x 106 eV] / [106 ev/MeV]= 11.8 MeV

29

Wave & Particle Theories Combined

= inverse relationship =

E & = direct relationship = E

E & should have ??? .

inverse relationship = E

30

Combination of Wave & Practical Theories

combine formulas: c = E = h solve wave wave equation for frequency: = c / insert solution in quantum formula:

[4.15 x 10-15 eVs] [3 x 108m/s]

m

[12.4 x 10-7 eVm]

m

hcm

EeV =

31

Shortcut FormulaeEeV = hc/ = [12.4x10-7eVm] / m

by incorporating changes in prefixes you can

arrive at the following shortcut formulae:

[12.4 keV A]

A

EkeV = A = 10-10m

[1.24 keVnm]

nm

EkeV = nm = 10-9m

32

ExampleWhat is the of an 85 keV x-ray photon?

?? 85 keV = energy need h & c

EeV= hc m

m = [4.15 x 10-15eVs] [3 x 108m/s] EeV

= [12.4x10-7eVm] 85 x 103 eV

= 0.1458823529412 x 10-10m = .15 x 10-10m or .15A

33

Shortcut method?? 85 keV = energy

shortcut h & c

EkeV = 12.4 / A

A = 12.4 / EkeV

= 12.4 / 85

= 0.1458823529412

= .15 A

EkeV = 1.24 / nm

nm = 1.24 / EkeV

=1.24/85

= 0.01458823529412

= .015 nm

34

Example #2What is the energy of a .062nm x-ray photon?

keV = ?? .062 nm = nm shortcut h & c for nm

EkeV = 1.24 / nm

= 1.24 / .062 nm

= 20 keV

35

Matter and Energy Relativity Formula

2cmE Enables calculation of matter equivalence

for any photon Must convert E in keV to E in J

1 J = 6.24x1018eV

36

Relativity problem example: What is the matter equivalence of a 86keV x-ray photon?

? = mass E = 86keV [c = 3x108 m/s]

kg

kg

J

eV

c

EmcmE

sm

sm

eVJ

32

32

16

15

28

1024.613

22

104.1

104250.1

109

108205.1

103

1086

2

2

18

37

Relativity problem example: How many electron-volts are contained in .25kg of matter?

? = E m = .25 kg [c = 3x108 m/s]

eV

J

eVJ

J

kg

cmE

s

mkg

sm

35

1816

1616

28

2

104.1

1

1024.61025.2

103.21025.2

)103(25.

2

2