# 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.

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

## Solutions

### Collimator

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)