# Answer to Question #7613 Submitted to "Ask the Experts"

*Category: Radiation Basics*

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

I have an x-ray source (CT scanner) that deposits X keV of energy per mm^{2 }in one slice of silicon, 100 microns deep; i.e., the x-ray source sends the x rays perpendicular to this surface, transversing (if no interaction) the 100 microns of silicon.

How do I go from X keVs on the described volume to rads on silicon? Can I just go from keV to Joules and volume to mass and be done? Or do I need to take into account the shape of the volume?

I found http://hps.org/publicinformation/ate/q4025.html and the concept of CPE (charged particle equilibrium) http://hps.org/publicinformation/ate/faqs/gammaandexposure.html but not sure I understand if they apply on this case.

The
degree of attenuation of the x rays in silicon (or any other material) depends
on the energies of the x rays and the thickness of the material. As an example,
if we assume an effective x-ray energy of 35 keV (which might be realistic for
a 100 kVp x-ray generator), the linear attenuation coefficient in silicon for
this energy is about 1.45 cm^{-1}. The fractional penetration of a
parallel beam of 35 keV photons though a thickness T of silicon is given by

F = e^{-1.45T},

and for T = 0.01 cm (100 microns),

F = e^{-(1.45)(0.01)
}= 0.986.

This
implies that the photon fluence rate and dose rate at the surface of the
silicon will differ from that at the back side of the silicon by only about 1.4
percent, not a very great difference. This estimation assumes that the silicon
thickness is uniform over the area being irradiated

The
dose estimation is straightforward, as you have inferred it might be, if X
actually represents energy deposited. You state that X keV of energy is
deposited in the slice of silicon that measures 1 mm^{2} (0.01 cm^{2})
in cross sectional area and 0.01 cm thick; these dimensions yield a volume of
10^{-4} cm^{3}. For a typical silicon mass density of 2.42 g cm^{-3},
the mass of this volume would be 2.42 x 10^{-4} g. By definition, dose
is the energy deposited per unit mass, thus the dose, in rads, is obtained:

D
= ((X keV)/2.42 x 10^{-4} g )(1.602 x 10^{-9} ergs/keV)(1 rad/
100 ergs g^{-1}) =

6.62 x 10^{-8} X rads.

This
is the dose averaged over the specified volume of silicon. For the assumed
effective energy of 35 keV, the dose near the surface of x-ray incidence would
be only about 1.4 percent higher than that near the back side of the silicon.

Please
note that the above calculation assumes what you stated—namely, that the energy
__deposited__ in the silicon volume was X keV. The result would naturally be
different if X were to represent the x-ray energy i__ncident __on the 0.01
cm^{2} silicon surface. Then the dose at the incident surface, under
conditions of secondary charged particle equilibrium (SCPE), would be given by
the product of the photon energy fluence and the effective mass energy
absorption coefficient. µ_{en}/ρ,
as in the discussion accessed through the second link above that you cite. For 35 keV
photons in silicon the value of µ_{en}/ρ
is about 0.724 cm^{2} g^{-1}. If X is energy incident on the
0.01 cm^{2} area, the dose at the surface would then be

D^{surface} = (X
keV/ 0.01 cm^{2})((0.724 cm^{2} g^{-1})(1.602x10^{-9}
ergs/keV)(1 rad/100 ergs g^{-1}) = 1.16x10^{-9} X rads.

This dose is much lower than
the dose calculated first by assuming X was actual energy deposited in the
silicon volume; this is to be expected since, as we showed earlier, most of the
x rays pass though the 100 microns of silicon without interacting.

The concept of SCPE becomes
important if you were to perform a calculation similar to the second one done
above, since the use of µ_{en}/ρ is legitimate only if SCPE applies. If
you are performing measurements it is rather simple to ensure that SCPE exists
at the surface of the silicon by covering the surface with a thin covering of
silicon or other material close in atomic number to silicon. The thickness of
this silicon covering should be at least as thick as the range of the most
energetic electrons that could be produced by the photon interactions. For 35
keV photons in silicon this range is 0.00302 cm^{2} g^{-1}, or
7.3 x 10^{-3} cm.

I
hope this answers your question to your satisfaction.