X-rays: structure, uses and attenuation

​​In a nutshell

X-rays are high energy photons which are used to diagnose medical conditions without subjecting patients to invasive techniques. X-rays are produced by bombarding a tungsten plate with high energy electrons. CAT scans are x-ray machines which take a two dimensional slice through a patient. X-rays are attenuated in three different ways, the photoelectric effect, Compton scattering and pair production.

Equations

Constants

Variable definitions

X-rays

X-rays are high energy photons in the penultimate group of the electromagnetic spectrum. X-rays are used as a non-invasive diagnostic technique.

X-rays are non-invasive as no instruments or equipment enter the patients body and it is a relatively quick and painless procedure. 

X-ray tube

X-rays are produced by bombarding tungsten with high energy electrons. 

The x-ray tube looks like this:

1 High-voltage power supply 2 Evacuated glass tube 3 Heater acting as cathode 4 Beam of electrons 5 Tungsten acting as anode 6 Motor 7 Lead shield 8 X-rays

There are two main electrical circuits occurring in the x-ray tube. There is the high potential difference circuit which acts as the accelerating potential for the electrons. This is connected to the heater and the tungsten plate. The heater is the cathode, which is the negative terminal, and the tungsten plate is the anode, which is the positive terminal.

The heater is connected to a high current circuit. The high current heats up the filament which thermionically emits electrons. The filament is normally encased in a focussing parabola, which focusses the electrons in one direction.

As the electrons are produced, they accelerate rapidly towards the tungsten anode before smashing into the tungsten with a lot of energy. When the electrons hit the tungsten plate, they decelerate and some of their kinetic energy is converted into x-ray photons. 

The amount of energy the photons have can be calculated using the equation for a charge accelerated in a potential difference:

​$$W=VQ$$​​

The charge of an electron is given by $$e$$ and the potential difference is given by the tube voltage, which is the accelerating potential between the anode and cathode: 

$$E=eV$$​​

Example

If the tube voltage is $$40\, kV$$ then the energy of the emitted x-ray photons will be $$40 \, keV$$.

Note: The tube is encased in a lead shield which only allows the x-rays to leave through a directed gap. This is to prevent unwanted x-rays leaving, to protect the patients and radiographers as x-rays are ionising radiation. It also allows the radiographers to be very accurate with the imaging. The x-ray tube is also evacuated (so there is no air) which minimises the energy loss and prevents the filament from burning up.  

The efficiency of producing x-rays is incredibly small. Only around $$1\%$$ of the electrons energy goes into producing x-rays, the rest is transferred to the thermal energy of the tungsten plate. 

This means the tungsten gets incredibly hot very quickly, so it is normally mounted to a heatsink such as copper and attached to a motor. The heatsink absorbs a lot of the thermal energy and dissipates it quickly, whilst the motor ensures that the same part of the tungsten isn't being constantly bombarded. 

X-ray production

X-rays are produced when the electrons smash into the tungsten converting some of their kinetic energy into high energy x-ray photons. X-rays can also be produced with how the electrons interact with the electrons within the tungsten atoms.

When the high energy electrons hit the tungsten, the high energy electrons can knock other electrons out from the inner electron shells of the tungsten atoms. This causes an electron from the outer shell to drop down and fill the gap, releasing a high energy x-ray photon. 

​​Example

An x-ray machine has a tube voltage of $$44\,kV$$. Calculate the wavelength of the x-ray photons being produced.

Firstly, write down the known values: 

$$V=44\,kV=44\times 10^3 \, V \newline \\[0.1in]Q_e=1.6\times 10^{-19} \, C \newline \\[0.1in]h=6.63\times 10^{-34} \,J\,s \newline \\[0.1in]c=3\times 10^8 \,m\,s^{-1}$$

​Next, write down the equations needed and rearrange if necessary:

$$E=QV \newline \\[0.1in]E=\frac {hc}\lambda \rightarrow \lambda = \frac {hc}E$$

Calculate the energy of the photon:

$$E=1.6\times10^{-19}\times 44\times 10^3 \newline \\[0.1in]E=7.04\times 10^{-15} \, J$$

Calculate the wavelength of the photon:

​$$\lambda = \frac {6.63\times 10^{-34}\times 3\times 10^8}{7.04\times 10^{-15}} \newline \\[0.1in] \lambda = 2.82528...\times 10^{-11} \, m$$​

Make sure to include units and round to the lowest number of significant figures of the values given by the question:

$$wavelength \ of \ photon, \lambda =2.8\times 10^{-11} \, m \newline \\[0.1in]\lambda =28\, pm$$​​​​​

The wavelength of the x-ray photon is $$\underline{28 \, pm}.$$​

​

Intensity of the x-ray beam

Intensity is defined as the power per unit area perpendicular to a surface. Therefore the two ways of increasing the intensity of the beam, is to increase the tube voltage and the filament current.

By increasing the tube voltage, the electrons will have a greater kinetic energy. The higher energy electrons can knock out electrons from shells deeper within the tungsten atoms. 

By increasing the current supplied to the filament there will be more heat,  and therefore more electrons will be liberated from the filament per second. So, more x-ray photons will also be produced per second as there are more electrons to convert their energy into photons. 

X-ray attenuation

X-rays interact differently with different mediums. So, when an x-ray passes through a patient, the different parts of the patient will absorb more or less x-rays depending on their composition.

The intensity of the x-ray beam decreases exponentially with the distance from the surface. It also depends on the materials attenuation coefficient $$\mu$$.

$$I=I_0e^{-\mu x}$$​​

​​Example

An x-ray passes through a medium with an attenuation coefficient of $$0.2\,cm^{-1}$$. Calculate the intensity of the x-ray after $$7\, cm$$ as a percentage of the original beam.

Firstly, write down the known values: 

$$\mu =0.2\,cm^{-1} \newline \\[0.1in]x=7\,cm$$

​Next, write down the equations needed and rearrange if necessary:

$$I=I_0e^{-\mu x}\rightarrow \frac I{I_0}=e^{- \mu x}$$

Then, substitute the values into the equation:

$$\frac I{I_0}=e^{-0.2\times 7} \newline \\[0.1in]\frac I{I_0}=0.24659...\newline \\[0.1in]\frac I{I_0}=24.659 \%$$

Make sure to include units and round to the lowest number of significant figures of the values given by the question:

$$percentage \ of \ original \ intentsity, \ I=25\%$$​​​​​

The x-ray beam will have $$\underline{25 \%}$$ of its original intensity.

The x-rays are attenuated by absorption and scattering. The three ways in which x-rays are scattered are:

1 Photoelectric effect 2 Simple scattering 3 Compton scattering 4 Pair production 5 Attenuation coefficient 6 ​$$\log E_\gamma$$(Photon energy / eV)​

The attenuation of different materials depends on the atomic number of what it is made from. Tissues containing a variety of atomic numbers, will attenuate the x-rays by different amounts, which will ultimately contrast the image. 

If the x-ray is being used to observe tissue with similar attenuation coefficient, then an artificial medium can be introduced which will contrast the image.

Example

A doctor wants to x-ray a patient's stomach but the stomach and tissue all have similar x-ray attenuation coefficients. A barium meal can be consumed, as barium has a relatively high atomic number. The stomach and digestive tract will now absorb a lot of the x-rays by showing the path the barium meal took through the patient. 

Cat scans

Computerised axial tomography (CAT or CT) scans take a two dimensional "slice" through the body. The x-ray beam is directed through the patient into x-ray detectors on the opposite side.

The x-ray detectors then measure the intensity of the received x-rays and a computer builds a composite image. The x-ray beam and detectors can rotate around the patient to build up a two dimensional image.

The patient can also be moved in and out of the scanner which allows for a three dimensional image to be constructed. 

CAT scans normally produce a much more detailed image than a typical x-ray scan. It is particularly useful for soft tissue x-rays. 

The disadvantage of CAT scans are the much larger exposure to x-rays, along with the cost of running and maintaining a CAT scanner.