6.5 Medical Physics

Medical Physics

Gamma Camera

After photons pass through collimator they go into photomultiplier tube which absorbs the photon and emits an electron which can be detected by a computer to produce and image.

Photomultiplier tube

A photon enter a vacuum tube and strikes a metal plate of $0V$, it then emits some electrons (dependant on the intensity of the photon). The electron is attracted to another plate of $100V$, it gets absorbed and causes 2 electrons to be emitted, this continues to increase the number of electrons which makes the current more detectable.

Causes of anomalies:

Photoelectric Effect

A photon collides with a metal plate causing an electron to be emitted. $E_{photon} = hf$ $E_{electron} = \frac{1}{2}mv^2$ Electron must overcome work function $\phi$ to reach the surface of the metal and be emitted.

Can be modelled with an “electron well” where the depth is $\phi$ and the electron needs enough energy to leave the well.

The wavelength of the light is directly propertional to the amount of kinetic energy, $hf-\phi = E_{k,max}$

Energy levels of different elements

Ground state $n=0$

______________ $n=4$ ______________ $n=3$

______________ $n=2$

______________ $n=1$

______________ $n=0$

Compton Scattering

An energetic photon of short wavelength $\lambda$ can interact with an electron and cause it to “scatter” with some angle $\theta$. A photon with a longer wavelength can also be emitted (Longer because some energy was lost on the scattered electron. It will have angle $-\theta$)

Pair Production

Extremely high energy photon, when interacting with an atom (Through gravity) can be split into a pair of particles - positron electron pair($e^-\text{ and }e^+$).

$m_e=9.11\times 10^{-31}$ $E_e=m_ec^2=9.11\times10^{-31};\times;3\times10^8$ $E_e=0.512\text{Mev/c}^2$

Therefore the minimum energy required for pair production = $2\times0.512\text{Mev/c}^2=1.024\text{Mev/c}^2$

It would practically need slightly more to give the particles some energy.

$f_{min} = \frac{h}{E_p} = 2.47\times10^{20}$ $\lambda_{max}=\frac{c}{f_{min}} = 1.21\times10^{-12}$



A filter to only allow light travelling from the direction of the xray source, this prevents scattered photons from being detected and massively reduces noise in the image.

Positron Emission Tomography

Radioactive isotope is injected with glucose into patient. $\beta^+$ is emitted which annihilates an electron and produces two gamma rays which can be detected and the location of the emission can be calculated.

Average half life of 20 minutes to a few hours to test. (2 hours)