# How to calculate the distance that 1-10 KeV x-rays have to travel in air to lose 90% of its energy?

I'm trying to calculate the distance of air that can strip a beam of 1-10 KeV x-rays from 90% of its original energy.

I came across this graph which shows the mass attenuation coefficient of air, but I'm note sure how to use it to get the distance that I need.

For simplicity, let's just limit the energy range to 1 KeV and 10 KeV. So how long does a 1 KeV or 10 KeV x-rays beam have to travel in air to lose 90% of its energy?

## 1 Answer

The mass attenuation coefficient is simply the ordinary attenuation coefficient divided by the mass density of the medium.

$$\mu_{m}=\mu/\rho=$$

All you need to do is substitute in the mass density of air, which is 1.225 kg/m$$^3$$. Using the value from your graph at 1 keV of 3.5e3 cm^2/g, I get an attenuation coefficient of about 4.4 cm^-1. Plug that number into the Beer-Lambert formula below and I got that it takes 0.5cm to attenuate the 1 keV xrays by 90%. You can do the same thing for the energy by using $$\mu_{en}$$.

$$I(x)=e^{-\mu x}$$