HAN 395 exam 2

The kinetic energy of the projectile electron in an x-ray tube

If it hits the target, less than 1% of it makes x-ray

while 99% comes off as heat

• Brems and characteristic x-rays

KE depends on tube voltage

Know the characteristics x-ray energy that would be produced with an x-ray tube (new target)

Subtract the vacancies -->

K =80 keV

L= 10 keV

M = 5 keV

K-L or K-M

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Sutbracts the energy from K shell

so:

K-L or K-M

unit is keV

K shell = 80 kev

M shell = 5 kev

80-5 = 75 kev

Bremsstrahlung x rays characteristic

Bremmsstrahlung --> polyenergetic so 0 to max energy keV

Characteristics are specific energies including

60keV and 68 kev for tungsten

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Bremsstrahlung is when high-speed electrons decelerate as it gets close to a nucleus of the anode target (tunsten)

Factors that control bremsstrahlung x-ray efficiency

(E x Z)/1000 = percent --> EFFICIENCY FORMULA

E = tube voltage in kVp

Z= atomic number

Higher efficiency = Bigger Z and more energetic electrons

directly porportional

double Z --> xray efficiency increases by factor of 2

The impact on xray production as mAs is increased

DIRECTLY PORPORTIONAL

mAs doubles, the radiation is doubled

mA increase --> increase in radiation

x-ray production is linear and directly proportional with current (mA) and time

X-ray tube details: electrons are accelerated , the anode is tungsten, impact of filament current and line focus principle.

Electrons are the one accelerated not Xrays.

Filament increases, heating increases, # of electrons increase --> so mA increases

to have mA increase, is when you increase filament.

Anode is tungsten (Z=74), High Z, making it more xrays

Line focus principle -->large actual focal spot and small effective focal spot

• smaller effecitive focal spot and spread heat around

Methods to reduce heat damage to the target of an xray tube :

rotating anode, larger focal spot, smaller anode angle

1 .rotating anode ( spread heats in larger surfaces)

2. larger focal point (increase area for electron to hit and spread heat faster)

3. Smaller anode angle (reduces the size of effective focal spot to make a larger actual focal spot --> help with heat dissipation)

Tungsten is the element of choice for use as a target in an x-ray tube

- High Z = higher efficiency for X-ray production

(atomic number = 74)

- High melting point = 3400 degrees Celsius

- Good conductor for heat and electricity

- higher energy characteristic xrays

Factors that impact the probability of bremsstrahlung production

Energy increases, Z increases, which increases the chances for bremsstrahlung production (X-ray production)

E x Z/1000 = efficieny formula

Compute output change when kV is adjusted.

It is squared (2^2) = 4 so it increases by factors of 4

kv^2, if you double kv, it increase by a factor of 4

----

KeV squared

mR = 10 x mAs x (kV/100)^2 x (100/SSD)^2

Impact of time and mA on exposure

-Radiation exposure is directly proportional to mAs

-double time, exposure double

- mA doubles, exposure doubles

-linear

- mA x time = mAs

Describe the impact of reducing the anode angle

- It reduces the effective focal spot size, improving the image sharpness

- concentrates heat over a small area, increasing risk of heat damage --> heat loading becomes more concentrated

Know the difference between linear attenuation coefficient and mass attenuation coefficient.

attenuation: reduction in the intensity of radiation (ex: X-rays) as it passes through matter. 

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Difference is how DENSITY is handled.

Mass attenuation is linear attenuation divided by density

- attenuation of per unit mass

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Linear attenuation:

More aluminum layers = the beam is more penetrating so it has a BIGGER/GREATER HVL.

- gives you the attenuation per unit path length

Attenuation

Reduction in intensity of the X-ray beam as it passes through materials.

Includes:

- absorption (X-ray absorbed by material)

- scattering (photons deflected out of the beam)

HVL for beam intensity

A beam passes through, the photons would be absorbed and scattered --> reducing the beam intensity

Cut it by 50% --> so in 1/2

Used with materials like aluminum

HVL - Half value layer

The amount of thickness needed to reduce the X-ray beam's intensity by 50% (1/2).

Unit is mm

Know the details of photoelectric interaction :

tightly bound electron, photon is absorbed, photoelectron escapes, characteristic x-rays and Auger electrons are released

1. tightly bound electrons

2. photoelectron escapes

3. Photons are absorbed

4, Auger electrons and characteristic X-rays are released

Know the details of the photoelectric interaction explanation:

- X-rays vanish from the universe --> only photoelectrons left

- photoelectron escapes with the energy of incident photon energy minus the binding energy

-k shell vacancy --> characteristic xrays and auger electrons

DEPENDS ON Z^3 = higher the atomic number, increase in photoelectric absorption

Know the details of Compton Interactions :

loosely bound electrons, independent of Z, proportional to electron density, scattered photon can be in all directions.

-loosely bound electrons (outer)

- Independent Z as its Z^0 (billiard ball effect)

- proportional to electron density (same for everything 1/2 except for hydrogen)

- A scattered photon can be in all directions, but with reduced energy --> especially for diagnostic

Know the details of Compton Interactions (Z^0)

-Independent Z is Z^0

- Independent means the amount of scattering does NOT change when the atomic number changes

Z^0 = 1

Know details of electron density (electrons per gram of material):

nearly the same for all elements except for hydrogen

Z/A for everything --> Z/A = (1/2)

Z= atomic number

An atomic mass number

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- same for all elements except hydrogen (1/1) = since it has Z= 1, A= 1 so the density is higher than more electrons.

Avagadros number --> 6.022 x 10^23

Hydrogen explanation

- Has 1 electron and low atomic mass (1)

- Higher number of electrons per gram

-different electron density compared to others.

Know the details of pair production :

proportional to Z, increases as log E, requires 1.022 MeV, kinetic energy is shared, can happen in the field of a nucleus of electron

High Z more likely to undergo pair production

Increase log E = energy increase, probability of pair production increase slowly

Z log e = based on atomic # and log E --> for pair production

Requires more than 1.022MEV to create pair production of electron and positron (eventually result in annihilation)

Kinetic energy is shared between the positron and electron

Triplet production caused with nucleus of electron (original electron with energy + new electron + positron)

Annihilation (review)

- 2 photons are created going in opposite directions due to conservation of momentum.

- The photons are gamma rays.

- each one is 0.511 MeV or 511 keV

The electron and positron vanish from the universe, causing annihilation.

Compute the energy of a backscatter photon

Ebs = E0/(1+(2E0/0.511))

Ebs = backscatter photo

E0 = initial

0.511 MeV or 511 keV = resting mass energy

Compute the probability of photoelectric interaction when Z is changed

Z^3 --> (Z2/Z1)^3

If Z increases by 2, --> it goes up by a factor of 8

Only use when Z only change and energy stay constant

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E^3 --> (E1/E2)^3

If E increases by 2, --> goes down by a factor of 8

Only use when Z stays constant and only energy changes

Given the HVL, how much will be transmitted through a block of some thickness.

Each HVL is 1/2

1 HVL = 1/2

2 HVL = 1/4

3 HVL = 1/8

4 HVL = 1/16

Ex: HVL = 2 mm

block of thickness = 8mm

How much do you need to get through this block?

8/2 = 4 --> 4 HVL is 1/16

For a given mAs determine the exposure time needed for a specified mA.

Q: 20 mAs to 200 mA what is the time?

20 = 200mA (x)

20/200= x

20/200 = 1/10 s

Time is 1/10 s

Know the purpose of the glass envelope of an X-ray tube

Keep out air molecules --> maintain vacuum so electrons won't hit air molecules causing loss of energy

keep electricity from leaking out

Know what X-ray tube current is

X-ray tube current is measured in mA.

- How many electrons are transversing the tube going from filament to the target.

"banging into the target"

1 mA

6.25 x 10^15 electrons per second

Describe why heel effect occurs

- Reduction of x-ray beam intensity towards the anode side of the X-ray field

- absorption of some xrays to reduce intensity

- xray beam intensity is stronger in cathode side (where electrons come from) compared to anode side (opp)

Why is the line-focus principle used

To make a smaller effective focal size and spread heat around

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- Large Actual focal size = to spread heat

- Small effective focal spot =improve image sharpness and resolution, better spatial resolution

- angled anode gives you both = the line-focus principle

know the impact and source of ripple --> impact

less ripple: higher quality and quantity ==> more xrays and more penetrating radiations since kv doesnt go to 0.

• beam = stronger and consistent

know the impact and source of ripple --> source

comes from waveforms

- single phase (100% ripple)

- three phases (about 15% ripples)

- highest frequency (about a couple of ripples = 4%?)

Know the purpose/impact of filtration

- To reduce low energy photons

- Reduce patient dose

- The beam becomes more penetrative

-beam quality increase

- Quantity decreases --> Intensity decreases

- Quality increases

Compute HVL from a series of transmission measurements

HVL unit = mm

0 HVL is 100

1 HVL is 50

2 HVL is 25

3 HVL IS 12.5

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If you want to find the HVL of 52 mR at 3 (on the graph)

- you have to go to the higher value so 3.1 since it didn't exactly cut it in half.

Compute HVL from a series of transmission measurements --> problem

Ex

0 mm --> 120

2 mm --> 82

4 mm --> 61

6 mm --> 34

Tranmission =

120/120 = 1 | 82/120= 0.683 | 61/120 = 0.508 | 34/120 = 0.283 |

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Step 2 : HVL = x1 + ((I1/2 - I1)/(I2 - I1) x (x2 - x1) = formula

At 2 mm intensity 82 (x1)

AT 4 mm: intensity 61 (x2)

So 60 falls between 4mm and 2mm.

60 = I1/2

82 = I1

61 = I2

HVL = 2+ ((60-82)/(61-82)) x ( 4 x 2 )

Know the impact of increased SID on exposure

SID: Source-to-image distance

Exposure => 1/distance^2

(Inverse square law)

- SID increases = Exposure decrease (lower intensity) --> inversely prop

-decrease in patient dose

- Double distance, exposure goes down by factor of 4

Know the impact of increased SID on exposure - EXAMPLE

SID: Source-to-image distance

If the exposure at 100 cm SID is 80 mR, what will the exposure be at 200 cm SID?

(100/200)^2 = (1/2)^2 = 1/4

1/4 x 80 = 20 mR

Compute the change in exposure from mAs change.

- Exposure is directly proportional to mAs.

- mAs doubles, exposure doubles

- Increase mAs by 50%, exposure increases by 50%

Compute the change in exposure from mAs change. -- Problem (with 0.5) => unlikely question on the exam dont study

EX:

20 mAs, exposure 100mR

40 mAs, exposure 200mR

To find mAs2, you multiple 20 x 0.5 (since it increases by 50%)

Original value + (% incrase x original value)

= 20 + (0.5 x 20) = 30 mAs2

E2/E1 = mAs2/mAs1

= 100 x 30/20

= 100 x 1.5 = 150 mR

Compute the change in exposure from mAs change. -- Problem #2 (this is most likely the question)

If 25 mAs produces 100 mR, what exposure results from 50 mAs?

50/25 = 2 x 100 = 200 mR

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Exposure is 180 mR at 90 mAs. What mAs would produce 60 mR?

60/180 = 1/3 or 0.333 x 90 = 30 mAs

2nd value on top of first value

Impact of Compton scatter on photon energy.

- Photon energy diminishes so it goes down

- Photons collides with loosely bound electrons = absorbs part of the energy

- has less energy after scattering

-less penetrating

- decrease in image quality --> use grid to fix that since it catches scattered photons

- Under 100 keV, energy loss isnt huge

Compton effect's impact on image quality.

- degrade images because random events hit the image receptors

This can be fixed with a grid.

Compute the probability of photoelectric interaction if Z is changed.

- Double Z --> probability is increased by factor of 8

Formula: (Z2/Z1)^3

- Double E --> Probability is decreases by factors of 8

Formula: (E1/E2)^3

Compute the probability of photoelectric interaction if Z is changed. -- Example

Ex: If the atomic number (Z) of a material increases from 10 to 20, by what factor does the probability of photoelectric interaction increase?

USE Z^3

(20/10)^3 = (2)^3 = 8

When will absorption generally be highest (E >/= BE)

Iodine --> 30-32keV = absorbs photons very efficiently

- less photoelectric absorption if its higher than 33.2 keV

- k shell binding energy edge

- Want to be slightly above k shell binding energy

(E >/= BE)

photon energy (E) is greater than or equal to the binding energy(BE) of an electron

When will absorption generally be highest (E >/= BE) ----> Iodine K-shell BE

33.2 kEv

When will absorption generally be highest (E >/= BE) -----> Minimum energy needed to eject K-shell electrons

Photons energy less than BE = CANNOT eject the electron

- No k shell absorption (photon energy too low)

When will absorption generally be highest (E >/= BE) -----> maximum

MAXIMUM ABSORPTION

Photon energy equal or bit higher than BE = has enough energy to cause photoelectric effect

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Ex: 25, 33, 35 or 60 kEv. which one is best absorbed for iodine 33kev or 34 (depends on what prof puts).

- pick something slightly above the kEv so 35 kEv, will be absorbed.

k-edge for absorption

Higher probability for photoelectric interaction right after BE is K-EDGE

When will absorption generally be highest (E >/= BE) -----> Photon energy is much higher than BE

Probability of photoelectric absorption decreases:

- High energy photons most likely to scatter or pass through (compton)

Factors that control tube power (kW) rating

- max kv x max mA

- the physical size controls it

- focal spot size

- the anode angle

-how fast you spin it

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- Make anode bigger (absorbs and dissipates more heat)

- Use a bigger actual focal spot ( electron beams hit a larger aream)

- Rotating anode (spreads heat in a larger area)

Know what causes greater radiation intensity on the cathode side of a tube.

HEEL effect

- xrays travel cathode side, where there is less anode materials causing less attenuation = so higher intensity

What is HEEL effect

The variation of xray beam intensity across the geometry of the anode angle.

Reduction of xray beam intensity towards the anode side as it travels from cathode to anode

Causing less attenuation --> higher intensity

What is the purpose of a small anode angle

- Produce a smaller effective spot = loads better image sharpness

- actual focal spot remains large enough to spread heat

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- SPREAD HEAT

-SMALL EFFECTIVE FOCAL SPOT

----

- increases image detail

- maintains heat capacity

- better resolution

Anode angle vs focal spot size (actual and effective)

anode angle is the angle of the anode is tilted

actual focal = physical area of the anode where electrons strike and xrays form

effective focal = projected focal spot size from the anode angle --> affects image resolution

How are bremsstrahlung x-rays produced?

Produced by the sudden deceleration of high speed electrons by the electric field --> releasing xray photons

-Slowing down electrons

- "braking radiation"

Know the impact of mAs, kV, and distance on quantity and quality. --> mAs

- Increases mAs = Increase quantity (more photons)

- No change in quality

Know the impact of mAs, kV, and distance on quantity and quality. --> kV

increase kV = increase quantity (higher energy/ bremsstrahlung photons produced)

Increase kV = increase quality (penetration increase)

Know the impact of mAs, kV, and distance on quantity and quality. --> Distance

Increase distance = decrease quantity due to 1/distance^2

No change in quality

Know the impact of mAs, kV, and distance on quantity and quality. --> Filtration

Quantity decreases

Quality increases

Routine mamography focal spot size 

0.3

magnification mammography focal spot size

0.1

large radiography focal spot size 

1.2

small radiography focal spot size

0.3 or 0.6 (typically 0.6)

measured large focal spot size of 1.2 can be as big as _______

and ______

• 1.8 (50% bigger in one direction)

• 2.4 (100% bigger in the other)

Find dose equivalent

Dose equivalent = dose x radiation weighting factor

DE= D x Wr

Wr= either 20 (alpha) or 1 (other)

calculate effective dose 

effective dose = dose equivalent x tissue weighting factor

E= DE xWt

you will be given Wt

Know the typical focal spot sizes

- mA blooming makes focal spot look bigger

- Shouldnt be smaller than suggested size due to risk of burning or damage to anode

-actual focal spot is usually bigger than effective

- large focal size: 50% wider for width and 100% larger in length

Actual vs effective focal spot

Actual focal spot = where electrons hit the anode

Effective focal spot = the size projected towards the patient due to anode angle

Know the following regions of the mass attenuation curve: K-edge and the Photoelectric effect

- occurs at low energy (10-100keV)

- k-edge is the drop sharply to a certain point

- falls down, discontinuity is k-edge)

Know the following regions of the mass attenuation curve: Compton Effect

- Medium energy (100keV to 1000 keV)

- stays constant but slightly decrease with increase energy

Know the following regions of the mass attenuation curve: pair production

- High energy photos interact with nuclear field

- Makes 2 photons : electron and positron

- The photon energy needs to be greater than 1.022 MeV

- it is the region where attenuation increase.

Know the unit for exposure

2.58 x 10^-4 Coulombs per kg (c/kg) of air = 1 Roentgen (R)

How beam quality is defined in diagnostic x-ray and therapy. -- XRAY

Diagnostic Xray measured by HVL in Aluminum (mm)

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not important

increase kVp or filtration = HVL increase and beam quality increase

How beam quality is defined in diagnostic x-ray and therapy. -- THERAPY

Beam quality is percentage depth dose (%DD) --> usually 70%

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not important

Higher %DD = higher beam quality

lower %DD = lower beam quality

beam quality in therapy means penetrating ability

Know how heat is removed from the target for a radiographic tube.

REMOVED THROUGH RADIATION

Glows red, yellow or white and gets radiated out in the insulating the world

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Oil is insulator that helps cool the tube and absorbs the heat

Rotating Anode spins causes electron beam to strike larger surface area = spreading heat

Thermal energy radiates off anode --> tungsten used to absorb heat

Apply %DD to compute dose at depth

depth dose= entrance dose x (%DD/100)

entrance Dose = 200 rads

%DD = 70%

200 x ( 70/100)

200 x (0.7)

= 140 rads

Impact of a filter on quality (penetrating; high-energy photons)

INCREASE QUALITY

-Increase quality since there are only high-energy photons

- increase beam quality/penetration due to only high energy

- reduce patient dose

-Increase HVL

Impact of a filter on quantity (intensity; # of photons)

DECREASE QUANTITY

- decrease in beam quality

- decrease in quantity (intensity), but quality increases

-exposure factor (mAs and time) increases

- reduces patient dose

Two methods of x-rays are production.

Bremmstrahlung :(high energy electrons deflect of decelerated near nucleus field.

characteristic radiation/xray:

K-L or K-M = electron ejects inner shell electron

Know the impact of kV and mA on focal spot size --- mA

Increase kV, smaller focal spot size

Increase mA, result in ma blooming so it gets bigger

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mA increases = more electrons hit the target --> generations more heat = increase focal spot

Focal spot size increases = reduces heat concentration

Know the impact of kV and mA on focal spot size --- kV

-Increase kV ==> increase the energy of electrons

- more penetration due to higher energy electrons

- more heat generation due to higher kV

Does not impact the focal spot size

Know radiation units = exposure

Roentgen (R) or coulombs/kg (c/kg) = 2.58x 10^-4

exposure= ionization of air

Know radiation units = absorbed dose (dose)

rads or gray(gy)

1 gray(gy) = 100 rad

100 rads = 1 gray

energy absorbed by tissue

Know radiation units = dose equivalent

rem or sievert (sv)

1 sievert(sv) = 100 rem

absorbed dose adjusted for radiation type (biological effect)

Know radiation units = effective dose

sievert (sv)

dose equivalent adjusted for tissue sensitivity

How many rads in a Gy

1 Gy = 100 rads

Multiply by 100 if you want rads

divide by 100 if you want Gray (Gy)

Know the radiation weighting factors for different radiations

X-rays, Gamma rays, beta particles = 1

Alpha particles = 20

Neutrons = 5 to 20 (depends on energy but averages to 10)

Protons = 2

Discuss the source and ways that the Z^3 dependence is used in medical imaging. ---> source

- the photoelectric effect --> Z^3 dependence

results in contrast greatly enhanced

Discuss the source and ways that the Z^3 dependence is used in medical imaging. ---> medical imaging

Z^3 --> (Z2/Z1)^3

- Z^3 dependence causes bone to absorb more X-rays than soft tissues

- higher the atomic number more X-ray absorption and image contrast

- In tissue contrast allows image contrast due to Z (bone v muscle)

- kVp balances contrast and dose based on Z^3 = decrease dose and reduce photoelectric interaction and contrast

Be able to draw spectra expected from an x-ray tube (filtered and unfiltered) ---> FILTERED

LOOK AT THE DRAWINGS

- filter removes lower energy --> leaving high energy = more penetrating

- curve shifts to the right = higher photon energy (beam hardening) = more pentrating

-decrease image contrast (due to losing low-energy photons)

- intensity decrease due to fewer photons

Be able to draw spectra expected from an x-ray tube (filtered and unfiltered) ---> UNFILTERED

LOOK AT THE DRAWINGS

- contains low and high-energy photons

- increase patient dose

- higher image contrast

-soft beam= less pentrating

Describe the balance required between image resolution and x-ray tube heating.

Image resolution:

- the smaller EFFECTIVE focal spot size (produce sharper images) --> less blur

causing heat to be concentrated in one area

so you do:

- line focus principle

-rotating anode

Describe the conflict between dose and contrast --> DOSE

LOWEST DOSE = HIGH ENERGY (PENETRATES BETTER)

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-Dose decreases (higher kVp) --> increases penetration

-contrast decrease = less dependent on Z, less photoelectric effect

Instead, its Compton scatter increases (degrading image)

Describe the conflict between dose and contrast --> CONTRAST

BETTER CONTRAST WHEN ITS LOW ENERGY --> PHOTOELECTRIC DOMINATES (Z^3)

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SMART TECHS USE SOMETHING IN BETWEN ABOUT 70%

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- Contrast increases (lower kVp) --> increases photoelectric effect

Describe the balance required between image resolution and X-ray tube heating --> balance

- Small focal spot, with low mA --> high resolution

-Large focal spot (to spread heat) --> high output

- Rotating anode (proper cooling) --> prevents overheating