A. What is an anode?  

B.  What is the anode's primary function in an x-ray tube?

C. What material is typically used as the target in an Anode.

D. Give two reasons why this is an ideal target material. 

A. An anode is a positive electrode.

B. The anode serves as target for the high-velocity electrons produced by the cathode.  The interactions between the bombarding electrons and the target produce x-rays.

C. Tungsten, W (Z = 74) 

D.

1. Tungten has high atomic number (74):  Probability of Bremsstrahlung increases with Z², and efficiency of x-ray production increases with Z.

2. Tungsten has high melting poing (3,370ºC): High MP necessary to withstand intense heat produced by electron bombardment.

1) The tungsten target (W) is attached to a thick copper anode (rod) that conducts heat from the target to the outside of the tube, where it is cooled by oil, water or air.

2) The anode can be made to rotate, in order to prevent the temperature from becoming too high at any one spot.

A. As focal spot increases, image sharpness ______.

B. What apparant focal spot sizes are used in diagnostic and therapeutic x-ray machines, respectively?

C. Describe the principle of line focus.  Include the equation that relates apparant focal size to actual focal size.

A. decreases

B. Diagnostic: 0.1 - 0.2 mm
Therapeutic: 5 - 7 mm 

C. Line focus is a means of reducing the apparant size of the focal spot by placing the anode target at a steep angle with respect to the electron beam.

a = A sinθ,

where a = apparant focal size, A = actual focal size on target, θ = angle with respect to electron beam.

A. What is a cathode?

B. What is the primary function of the cathode in an x-ray tube?

C. What are the main components of the cathode?

D. What is the name of the process by which electrons are produced in the cathode?

E. The size of the focal spot depends on ______.

A. A cathode is a negative electrode.

B. The cathode serves to produce the electrons that are accelerated towards the anode target by the potential difference created between the electrodes.

C. The cathode consists of a tungsten filament that is heated to produce electrons.  The filament current is provided by a circuit.  The electron beam is focused towards a well-defined focal spot by a negatively-charged cathode cup.

D. When the tungsten filament is heated, it emits electrons.  This is known as thermionic emission.

E. filament size

The x-ray circuit can be divided into two parts:
A.
B.

C. Which ultimately dictates tube current? 

A. A high-voltage circuit that provides accelerating potential for the electrons.

B. A low-voltage circuit to supply heating current to the filament.

C. Because the voltage between the electrodes is sufficient to accelerate all electrons emitted towards the target, the filament current ultimately controls tube current (and thus x-ray intensity).

1. Bremmstrahlung

2. Characteristic x-rays

A. Describe the process of Bremsstrahlung.

B. Is the Bremsstrahlung energy spectrum discrete or continuous?

C. How does the probability of Bremsstrahlung correlate with Z?

D. Describe the relationship between electron energy and spatial distribution of x-rays around the target.

A. Bremsstahlung (braking) radiation is produced when a "collision" occurs between an incident electron and an atomic nucleus in the target.  The electron is deflected, and the resulting change in energy is paired with the emission of an x-ray.

B. Continuous.  A bremsstrahlung photon can have any energy up to the energy of the incident electron.

C. Probability of bremsstrahlung is proportional to Z².

D. At low electron energies (<100 keV), spatial distribution is fairly uniform.  As energy increases, x-ray direction becomes increasingly forward. 

A. Compare the spatial distribution of x-rays produced in a low-voltage diagnostic x-ray tube to those produced in a megavoltage linear accelerator.

B. In each situation, where is the x-ray beam obtained?

A. In low-voltage x-ray tubes, the spatial distribution is relatively uniform.  In a megavoltage tube, the distribution is forward-peaked.

B. In a low-voltage tube, it is advantageous to obtain the beam at 90º with respect to the electron beam.  

Linear accelerators use transmission-type targets, where the beam is collected on the other side of the target.

A. Define efficiency, as it relates to x-ray production.

B. What formula describes the efficiency of x-ray production?

A. Efficiency is the ratio of output energy emitted as x-rays to the input energy deposited on the target by electrons.

B. Efficiency = 9 x 10^-10 ZV

Describe the energies of the incident and the ejected electrons after they collide, with relation to the incident electron's initial energy and the binding energy of the orbital electron.

Incident electron has initial energy = E₀.

Orbital electron has a binding energy of E_B.

In the collision, ΔE is transferred from the bombarding electron to the orbital electron.

The primary electron is thus left with E₀ - ΔE, while the ejected electron has energy ΔE - E_B.

A. The x-ray energy spectrum produced by an x-ray machine is ______.

B. The intensity of the beam with respect to energy is given by what equation?

C. What is the average photon energy with respect to E_max?

A. Heterogeneous

B. Kramer's Equation:

I_E = KZ(E_m - E),

where I_E is the intensity of photons with energy E, K is the Kramer constant, and E_m is the maximum photon energy.

C. E_avg ~ 1/3 E_max

A. What is the purpose of added filtration?

B. Increased filtration is said to ______ the x-ray beam.

A. Added filtration tends to absorb low-energy photons from the beam, enriching it with higher-energy photons.

B. harden

A. What are the three fundamental settings for an x-ray generator?

B. Give the formula that relates output to these settings.

C. How can output of an x-ray machine be measured?

A.

(1) Kilovolt-peak, kVp: the maximum (peak) voltage across the x-ray tube.

(2) Tube current, i: the current through the tube, measured in milliamps (mA)

(3) Time: the length of time that x-rays are being emitted.  

B. Output --> (kVp)² x mA*s

C. Output can be expressed in terms of exposure: a measure of ionization per unit mass of air.

A. What is the heel effect?

B. How can it be compensated for?

A. The heel effect, due to the anode being placed at an angle, causes the beam to decrease in intensity from the cathode end of the apparant focal spot to the anode end.  X-rays are produced at various depths within the target, and must pass through greater target thickness, on average, to reach the anode end.

B. The heel effect can be minimized with the use of a compensating filter to improve beam uniformity.

20 kVp --> 120 kVp

*Low voltage (~20 kVp) is typically used in mammography.

(1) Increase filtration

(2) Increase kVp