Factors controlling the X-ray beam intensity

Question 1

Intensity:

Answer 1

Total amount of energy contained in the x-ray beam

Question 2

Intensity: equation

Answer 2

Quantity X Quality

Question 3

o Quantity:

Answer 3

Number of photons in the beam. Primarily related to tube current and exposure
time and less so by tube voltage (kVp)

Question 4

Quantity
Expressed as:

Answer 4

tube current (mA) X time (sec) → mAs

Question 5

oQuality:

Answer 5

Average energy the x-ray beam.

Question 6

Quality
▪ Controlled primarily by the

Answer 6

tube voltage (kVp)

Question 7

X-ray of beam is —.

Answer 7

heterogeneous

Question 8

Factors
controlling the X-ray beam
intensity (6)

Answer 8

1. Tube voltage
2. Exposure time
   3.Tube current
3. Filtration
   5.Collimation
4. Source-receptor distance

Question 9

As tube voltage increases:
(3)

Answer 9

o Number of photons generated increases
(increased quantity)
o Mean energy of photons increases
(increased quality)
o Maximum energy of photons increases

Question 10

As exposure time increases:
(3)

Answer 10

o Number of photons generated increases
(increased quantity)
o Mean energy of photons is unchanged
(quality unchanged)
o Maximum energy of photons is unchanged
(quality unchanged)

Question 11

As mA increases:
(3)

Answer 11

o Number of photons generated increases
(increased quantity)
o Mean energy of photons is unchanged (quality
unchanged)
o Maximum energy of photons is unchanged
(quality unchanged)

Question 12

Filtration
Selectively removes

Answer 12

long wavelength (low energy) x-rays.

Question 13

Total Filtration =

Answer 13

inherent filtration + added filtration

Question 14

Inherent Filtration
(4)

Answer 14

o Glass envelope
o Immersion oil
o Metal housing
o Tube window

Question 15

Added Filtration
(1)

Answer 15

o Aluminum disks

Question 16

As filtration increases:
(3)

Answer 16

o Number of photons decreases (reduced quantity)
o Mean energy of photons increases (increased
quality)
o Maximum energy of photons is unchanged

Question 17

Collimation
Restricts the

Answer 17

size and shape of the beam.

Question 18

Collimation
(3)

Answer 18

o Number of photons decreases
o Mean energy of photons is unchanged
o Maximum energy of photons is unchanged

Question 19

Inverse Square Law

Answer 19

Intensity of the beam varies inversely to the square of the source-to-receptor distance.

Question 20

Inverse Square Law
One method of calculating the new beam intensity when changing the source-toreceptor distance:
If distance is doubled (8” to 16”) =
If distance is tripled (4” to 12”) =
If distance is halved (16” to 8”) =

Answer 20

new intensity will be ¼ (inverse of 22)
new intensity will be 1/9 (inverse of 32)
new intensity will be 4x (inverse of 1/22)

Question 21

As source-to-receptor distance increases:
(3)

Answer 21

o Number of photons decreases (decreased quantity)
o Mean energy of photons is unchanged
(unchanged quality)
o Maximum energy of photons is unchanged

Question 22

Increase in: Tube Voltage (kVp)
Quantity:
Quality:

Answer 22

increase
increase

Question 23

Increase in: Tube current (mA)
Quantity:
Quality:

Answer 23

increase
no change

Question 24

Increase in: Exposure time (s)
Quantity:
Quality:

Answer 24

increase
no change

Question 25

Increase in: mAs
Quantity:
Quality:

Answer 25

increase
no change

Question 26

Increase in: Distance
Quantity:
Quality:

Answer 26

decrease
no change

Question 27

Increase in: Filtration
Quantity:
Quality:

Answer 27

decrease
increase

Question 28

DENSITY – altered by factors
affecting — of the beam

Answer 28

Quantity

Question 29

CONTRAST – altered by factors
affecting — of the beam

Answer 29

Quality

Question 30

Amount of blackness of an image → Related to

Answer 30

how many x-rays reach the receptor

Question 31

Density
o Primarily controlled by —, and less so by —

Answer 31

mAs
kVp

Question 32

Primarily controlled by mAs, and less so by kVp
Why?

Answer 32

Because increasing mAs and kVp will increase the
quantity. Increasing quantity more x-rays will reach the
receptor → the darker the image will be (higher density)

Question 33

mAs rule:

Answer 33

mA and exposure time are inversely proportional.

Question 34

If mA is increased, exposure time must be — to
maintain the same density of the image.

Answer 34

decreased

Question 35

oIf mA is decreased, exposure time must be — to
maintain the same density of the image.

Answer 35

increased

Question 36

Example: The mA has been increased from 5 to 10. In order to
keep the same density in this image, the exposure time should
be reduced from 6 to _____

Answer 36

?

Question 37

Contrast

Answer 37

o The difference in densities between light
and dark regions of a radiograph

Question 38

Contrast is
o Primarily controlled by the

Answer 38

voltage

Question 39

High Contrast
(Short Gray Scale)
(2)

Answer 39

▪ Low kVp (long wavelengths, less penetrating)
▪ Density differences between adjacent areas
are greater; fewer shades of gray.

Question 40

Low Contrast
(Long Gray Scale)
(2)

Answer 40

▪ High kVp (short wavelengths, more
penetrating)
▪ Density differences between adjacent areas
are more subtle; more shades of gray

Question 41

Effect of kVp
on contrast
Low kVp:
High kVp:

Answer 41

High contrast (low gray scale)
Low contrast (large gray scale)