# Answer to Question #9530 Submitted to "Ask the Experts"

*Category: Instrumentation and Measurements*

The following question was answered by an expert in the appropriate field:

I need to radiate a lead plate with a noncollimated electron beam (150 k V 10 mA) for dose measurements in a sterilization process. Could the x rays emitted pass the lead shield of the instrument? How is it possible to assess the minimum thickness of the lead shield needed in this case?

Since the bremsstrahlung x rays produced as electrons are slowed down in the lead and are considerably more penetrating than are the electrons, a significant number of x rays may emerge from the lead and produce a response on the dose-measuring instrument. The magnitude of such an effect will depend strongly on the thickness of lead used and its distance from the instrument. From the information you give, I cannot judge what the geometrical setup is that you are using, but we can illustrate the possible effects of some parameters through a simple example in which we will make some assumptions about lead thickness and distance from the instrument.

The range of 150 keV electrons in lead is 0.005203 cm. If we assume the lead is at least this thick, we can calculate the number of electrons per second stopping in the lead, N_{s}, given the 10 mA current:

N_{s} = (10^{-2} A)(1 C s^{-1}/A)(1 electron/1.6 x 10^{-19} C) = 6.25 x 10^{16 }electrons s^{-1}.

The radiation yield of a 150 keV electron stopping in lead is 0.01664, meaning that this fraction of the electron energy appears as bremsstrahlung radiation. Thus, each electron produces an average expected bremsstrahlung energy yield of (150 keV)(0.01664) = 2.5 keV. If we assume an effective bremsstrahlung photon energy of about 50 keV (1/3 of the maximum electron energy) we can estimate the number of bremsstrahlung photons emitted per second, B_{s}, as

B_{s} = (6.25 x 10^{16 }electrons s^{-1})(2.5 keV/electrons/50 keV/photon) = 3.12 x 10^{15} photons s^{-1}.

If, for estimation purposes, we assume that the electron-beam diameter is small compared to the distance of interest from the lead of thickness, T, to the instrument and that the bremsstrahlung emission is isotropic, we can estimate the primary photon fluence rate, f, at distance R from the point where the bremsstrahlung is produced as

F = (3.12 x 10^{15} photons s^{-1}/4πR^{2}) (e^{-µT}), (1)

where µ is the effective linear attenuation coefficient of the bremsstrahlung photons in lead. If we assume an effective energy of 50 keV we find the value of µ is 21.5 cm^{-1}. If the target spot on the lead is 1 cm from the instrument and the thickness of lead is taken as a nominal 0.25 mm (0.025 cm), the primary photon fluence rate at the instrument from equation 1 is

F = (3.12 x 10^{15} photons s^{-1}/4π(1 cm)^{2})(e^{-(21.5)(0.025)}) = 1.45 x 10^{14} photons cm^{-2} s^{-1}.

The fluence rate to soft tissue equivalent dose rate conversion factor for the assumed 50 keV photons is about 2.7 x 10^{6} photons cm^{-2 }s^{-1} per μSv s^{-1}. Applying this factor, we obtain an expected equivalent dose rate, H, at the instrument location of

H = 1.45 x 10^{14} photons cm^{-2} s^{-1}/2.7 x 10^{6} photons cm^{-2 }s^{-1} per µSv^{-1} = 5.37 x 10^{7 }µSv s^{-1},

clearly a high value, although it may be small compared to the unattenuated dose rate from the electron beam itself. Naturally, the thickness of lead and the distance from the lead to the instrument are important factors that affect any calculated value. If we want to know what thickness of lead, x, must be to yield some desired exposure rate, we can estimate using the same approach. For example, for the 1 cm distance used above, if we had wished to reduce the exposure rate to 100 R h-1, we would write

280 µSv s^{-1 }= (F/2.7 x 10^{6} photons cm^{-2 }s^{-1} per µSv s^{-1}).

This yields F = 7.5 x 10^{8} photons cm^{-2 }s^{-1}.

We then write

7.5 x 10^{8}photons cm^{-2 }s^{-1} = (3.12 x 10^{15} photons s^{-1}/4π(1 cm)^{2})(e^{-(21.5)(x)})

and rearranging this we obtain

e^{-(21.5)(x)} = 3.021 x 10^{-6}.

This is solved by taking the natural logarithm of each side of the equation and solving for x. We obtain x = 0.59 cm, which implies a thickness of lead of almost 0.6 cm would be required to reduce the primary bremsstrahlung exposure rate at the instrument to 280 µSv s^{-1}.

The values used here and the assumptions made are intended simply to aid in providing an example of how some estimations may be made to establish the significance of bremsstrahlung production and penetration. You can use your own values of lead thickness and distance to make better estimates, but keep in mind that the results are not exact since they do not take account of the actual bremsstrahlung energy distribution, nonisotropic emission, photon scatter, and actual electron beam geometry.

I hope this is useful.

George Chabot, PhD, CHP