Units used in Radiography

Question 1

➢Kilo Volt peak (KVp

Answer 1

• KVP is the maximum voltage applied across the X-ray tube during an exposure.
• KVP determines the energy of the x-ray photons. penetration

Question 2

Effects of KVP on X-ray images and Patei

Answer 2

1. over or under Penetration: Higher KVP produces X-ray photons with higher energy, allowing them to penetrate denser materials like bone.
2. Contrast: Lower KVP results in higher contrast (greater difference between light and dark areas) in the image, which is often desirable for soft tissue imaging. Higher KVP can reduce contrast, making it more suitable for bone imaging.
3. Scattered radiation: Higher KVP increases the amount of scattered radiation, which can degrade image quality.
4. Patient radiation dose

Question 3

mA and mAs

Answer 3

• Both relates to the quantity of x-ray produced during exposure.
• Constant mAs: If you want to maintain a constant X-ray exposure, you can adjust mA and exposure time inversely. For example, if you double the mA, you can halve the exposure time to keep the mAs constant.
• Image quality: Both mA and mAs affect image quality. Higher mA settings can reduce image noise, while higher mAs settings can increase image density (darkness).

Question 4

mA

Answer 4

The number of electrons flowing per second

Question 5

What is an electron volt

Answer 5

• This is a unit of energy equal to the energy gained or lost by an electron when it moves through an electrical potential difference of one volt.
• If an electron is accelerated from rest across a PD of 1 volt, then the kinetic energy it attains is defined as 1 eV
• If an x-ray tube is subject to a PD of 75kVp, then we know that the energy of
  the electrons striking the anode must be 75keV

Question 6

Heat Units

Answer 6

• When electrons strike the anode they deposit all their energy onto the target where 99% is converted into heat
• HU gives a measure of the heat energy imparted to the anode
• The total energy dissipated by a current I driven by a PD V is given by
  E = VIt joule
  Where V is in volts I in amperes and t in seconds
• mA x t = mAs, hence Heat Unit = kVp x mAs