PRIMARY EXPOSURE FACTORS (mA, exposure time (s), kVp, SID)
A technologist makes an exposure using 300 mA, 0.1 second exposure time (s) and 100 kVp. calculate the mAs.
300 mA x 0.1 s = 30 mAs
Why is a shorter exposure time a good thing?
A shorter exposure time (s) is VERY helpful for minimizing motion artifact. See image for example
An exposure is made at 60" and results in a receptor exposure of 200 microgram. If the exposure is repeated at 40 inches, how will the receptor exposure change? A. Increased receptor exposure B. Decreased receptor exposure C. No change in receptor exposure
A. Increased receptor exposure
What happens to intensity when you increase mA?
Increasing mA will increase intensity
What happens to patient dose when you increase mA?
Increasing mA will increase patient dose
What happens to receptor exposure when you increase mA?
Increasing mA will increase receptor exposure
What is exposure time (s)?
Its the amount of time that electrons are flowing through the x-ray tube and x-rays are being created
At an SID of 40" the beam intensity is measured at 400 microgram. If the exposure is repeated using the same technical factors, what is the new beam intensity at 72"?
I₂ = I₁ x (D₁² ÷ D₂²) I₂ = 400 x (40² ÷ 72²) I₂ = 400 x ( 1600 ÷ 5184 ) I₂ = 400 x 0.3086 I₂ = ~123.46 I₂ = 123 microgray
What does kVp stand for?
KiloVoltage Peak = kVp
Tube current is kind of like a river, if the river has a SMALL current then there is a?
LOWER amount of water flowing down the river and vise versa
A LONG exposure time results in?
MORE total electrons and MORE total x-rays
What is the primary factor for controlling the amperage applied to the filament?
Setting the mA at the control panel is the primary factor controlling the amperage applied to the filament.
When you INCREASE mA, this increases the filament amperage which increases....
THERMIONIC EMMISIONS which increases the number of electrons created which are then available to collide with the anode target and produce x-rays.
x-rays are produced when electrons travel from the cathode (-) filaments and interact with atoms in the anode (+) target. This MOVEMENT of electrons is what we can call the?
TUBE CURRENT (mA)
What is the 15% rule?
The 15% rule states that changing the kilovoltage peak by 15% will essentially double exposure and vise versa
During an x-ray exposure the cathode (-) is negatively charged and the anode (+) is positively charged. This difference in charge is called?
The Tube potential or potential difference
What is the TUBE POTENTIAL?
The difference in charge between the cathode and the anode measured in kilovolts. That's why this technical factor is called the kilovoltage peak because the kVp describes the MAXIMUM voltage difference between the cathode and the anode during an x-ray exposure.
Why is the tube potential so important?
The voltage difference that makes up the tube potential is important because this is what FORCES the electrons to MOVE across the x-ray tube.
What happens when you DECREASE mA?
When you decrease mA: -Tube current decreases -Rate of electron flow decreases -The number or x-ray photons produced decreases
If you INCREASE mA, what happens?
When you increase mA: -Tube current increases -Rate of electron flow increases -The number or x-ray photons produced increases
The kVp is what causes the electrons to ______________________ and forces them through the x-ray tube
accelerate
The process of creating the x-ray tube current starts at the?
cathode filament
So tube current is simply the MOVEMENT or flow of?
electrons in the x-ray tube
Which of the following prime exposure factors controls the ENERGY of the x-ray beam? -distance -exposure time -mA -kVp
kVp
Tube current is measured in units of ?
milliamperes (mA)
We measure exposure time in units of?
seconds (s)
mA and exposure time (s) which together make up the mAs have a reciprocal relationship. This means that?
There are many combinations of mA and exposure time (s) that can be combined to produce the same outcome. For example: 50mA x 0.20sec = 10 mAs 100mA x 0.10sec = 10 mAs 200mA x 0.05sec = 10 mAs All of these combinations have different mA and exposure times but the total mAs is the same.
When is it appropriate to INCREASE exposure time?
For example when imaging a lateral T-spin or a sternum, you would need the blur in order to blur out ribs so you can see the anatomy of interest. See image for example
A technologist makes an exposure using 300 mA, 0.1 second exposure time, and 100 kVp. If the resulting image shows patient motion, what new technique will reduce motion artifact while maintaining the same total exposure? A. 600 mA and 0.1s B. 300 mA and 0.05s C. 150 mA and 0.2s D. 600 mA and 0.05s
D. 600 mA and 0.05s
