Question

Tungsten, molybdenum, and copper are all are used for X-ray production. For each anode material, calculate the electron energy needed for 100% efficiency of photon production. For the highest value, also determine the electron velocity. Does this seem reasonable for a clinical setting?  

Answer 1

The fraction of electron energy transferred into photon energy is given by

Where E is the energy of the electron in MeV

for 100% efficiency, f = 1.0

So,

For Tungsten, Z = 74

So,

For Molybdenum, Z = 42

So,

For Copper, Z = 28

So,

The maximum energy is 102.041 MeV

E = 102.041*10⁶*1.6*10^-19 = 1.633*10^-11 J

Using

This velocity is impractical in any situation. (the actual velocity will be smaller due to relativistic effects. Still it will be much higher than 10⁶ )

This is because, a 100% efficiency can never be achieved in X ray production. The typical efficiency is in the range of 1%