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 mm2 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.
I found that my circuit technology can stand Y rads (on silicon) without noticeable damage. But how are these rads given? Knowing that the components in the silicon that can be damaged are just on the surface, it does not look the same to have X keV deposited in a very thin surface, than X keV deposited in a very deep volume, even if both have the same mass.
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 ATE Q/A 4025 and the concept of CPE (charged particle equilibrium) but not sure I understand if they apply in 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 mm2 (0.01 cm2) in cross-sectional area and 0.01 cm thick; these dimensions yield a volume of 10-4 cm3. 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% 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 incident on the 0.01 cm2 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. µen/ρ, as in the discussion accessed through the second link above that you cite. For 35 keV photons in silicon the value of µen/ρ is about 0.724 cm2 g-1. If X is energy incident on the 0.01 cm2 area, the dose at the surface would then be
Dsurface = (X keV/ 0.01 cm2)((0.724 cm2 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 µen/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 cm2 g-1, or 7.3 x 10-3 cm.
I hope this answers your question to your satisfaction.
George Chabot, PhD, CHP