# Calculation of energy deposit of photon beam

Let's say I shoot a 1 MeV photon beam onto a volume. I want to know the ratio of deposited dose per path length for different materials (water and air) at a certain position (e.g. x=0) in the volume.

My approach so far: Use Beer-Lambert law $$I = I_0 \exp( - (\frac{\mu}{\rho})_\text{material} \cdot \rho_\text{material} \cdot x)$$ to calculate the intensity that is released from the photon beam by the derivative.

Therefore

$$\frac{\frac{d}{dx}(I_\text{water})}{\frac{d}{dx}(I_\text{air})}|_{x=0} = \frac{(\frac{\mu}{\rho})_\text{water}\cdot \rho_\text{water}}{(\frac{\mu}{\rho})_\text{material}\cdot\rho_\text{air}} \approx 1000$$

Next, one is able to interpret the released intensity as proportional to the number of electrons released due to Compton effect (Compton is most probable in this energy domain).

Now, if I assume an energy loss of approximately $2\,\text{MeV}\,\text{cm}^2/\text{g}$ for ~1 MeV electrons in water and air, the ratio of $1000$ appears also for the energy loss of one electron. Now, for the dose, this ratio again cancels out due to the normalization to the unit mass. Therefore, electrons deposit approximately the same dose in air and water.

However, as we release $1000$ times more electrons in water, the dose should be roughly a factor of $1000$ higher.

Is this correct? In my simulation I see a ratio of 50-60 maybe, but not 1000. Where is my mistake?

• Is the path-length in water and in air the same? Next, what means 1 MeV electrons? Whose energy is 1MeV? – Sofia Feb 10 '15 at 21:34
• yes, path length in water and air the same. For the energy of the electrons, I assumed O(1 MeV) due to Compton scattering. I am only interested in order of magnitude calculation. – mad Feb 11 '15 at 9:04