Methodology for Calculating Radiation Source Terms and Shielding Requirements

I. Bremsstrahlung Source Terms

Electrons stopping in matter will generate bremsstrahlung radiation. Figure 12 in the SAD was used to estimate all bremsstrahlung dose rates at 0^o and 90^o to the direction of the incident electron. Values used at 230 MeV and 300 MeV have been determined by a linear extension of the curves in the figure to the higher energies. The values that were used are as follows:

Table I

Bremsstrahlung dose rates from high Z targets at 0^o and 90^o from Fig. 12 [R-m² /mA-min]

Electron Energy [MeV]	0^o	90^o
6	2 x 10^4	1 x 10^3
70	3 x 10^6	4 x 10^3
230	2 x 10^7	6 x 10^3
300	4 x 10^7	8 x 10^3

Using 20 nA, the design current for the SDL, these values convert to a dose rate at one meter as follows:

0^o at 230 MeV - 2 x 10⁷ R-m² /mA-min x 20 x 10^-6 mA x 60 min/hr = 2.4 x 10⁴ R/hr

Likewise

90^o at 230 MeV = 7.2 R/hr, 0^o at 300 MeV = 4.8 x 10⁴ R/hr, 90^o at 300 MeV = 9.6 R/hr

The same approach is used for all other electron energies or currents used in this document.

The dose rates in Figure 12 are based on a high Z target such as tungsten or lead. Electrons interacting in lower Z materials have lower bremsstrahlung production rates. From Swanson (Ref. 4, fig. 16, p. 51), the following correction factors are developed and applied for electrons interacting in a medium Z material (e.g. copper, steel) and low Z material (e.g. aluminum).

Table II

Bremsstrahlung Dose rates [R/hr @ 1 meter]

Energy [MeV]	Current [nA]	Material & Correction factor	Dose rate @ 0^o	Dose rate @ 90^o
6	10	Steel, copper (0.8)	9.6	0..48
70	5	Steel, copper (0.8)	7.2 x 10^2	0.96
230	20	Al (0.6)	1.44 x 10^4	4.3
230	20	Steel, copper (0.8)	1.92 x 10^4	5.8
230	20	lead	2.4 x 10^4	7.2
300	20	Al (0.6)	2.88 x 10^4	5.8
300	20	Steel, copper (0.8)	3.84 x 10^4	7.7
300	20	lead	4.8 x 10^4	9.6

II. Neutron source terms

Neutron yields from electron interactions in matter can be divided into two energy categories: giant resonance (GRN) and high energy (HEN). GRN range in energy from low energy up to about 15 MeV with an average energy of 1 - 2 MeV, and dominate the total neutron emission rate. HEN range in energy from about 15 MeV up to the maximum energy of the electron. HEN are a small fraction of the total neutron dose equivalent rate in an unshielded configuration, but may dominate after penetration through a thick shield or in a skyshine consideration.

GRN Yields

The total GRN neutron yield can be estimated from Swanson (Ref 4, fig. 34, p. 87). For a thick high Z target the total emission rate is ~ 2 x 10¹² n/s-kw . Production rates (same figure) are about 1/4 that for low Z (e.g. Aluminum) and 1/2 for medium Z (e.g. steel, copper) targets. Using these parameters the GRN production rate from a 20 nA 230 MeV electron beam striking a thick Z target can be estimated as follows:

Power = 230 x 10⁶ V x 20 x 10^-9 A = 4.6 watts (1)

Therefore, the GRN emission rate F in thick high z targets is

F = 2 x 10¹² n/s-kw x 4.6 x 10^-3 kw = 9.2 x 10⁹ n/s (2)

Assuming an isotopic source, the fluence at one meter will be

F = 9.2 x 10⁹ n/s / 4p (100 cm)² = 7.32 x 10⁴ n/cm²- s (3)

Likewise at 300 MeV,

F = 9.5 x 10⁴ n/cm²- s (4)

Applying the correction factors for different atomic numbers, the GRN fluences are

Table III

GRN Production Rates for 20 nA beam loss @ 1 meter [n/cm²- s (x 10⁴)]

Energy [MeV]	Aluminum	Copper, Steel	Lead
230	1.8	3.7	7.3
300	2.4	4.8	9.5

GRN production rates for other energies or currents are calculated in a similar manner.

HEN Yields

As electron energy increases above 100 MeV, neutrons with energies higher than the GRN are increasingly produced. These yields are a small fraction (< 10%) of the GRN yields, but HEN become more important behind shield walls of sufficient thickness to remove the less penetrating GRN. HEN production rates are proportional to beam power. Above 140 MeV photo-pions are produced which provide an additional mechanism for production of HEN.

HEN spectra at 90^o produced by electrons in lead in the 150 to 300 MeV range are available from Swanson (ref. 4, Fig. 36, p. 89). This figure provides the HEN spectrum and production rates for incident electrons with energy at 150, 170, 182, 202, 234, and 266 MeV; and can be used to calculate HEN yields for SDL operating at 230 & 300 MeV. In Table IV below, the yields calculated for 234 MeV have been used for 230 MeV electrons. The 300 MeV yield was calculated by adjusting the 266 MeV yield by the ratio of the energies. In addition, the yields for a thick copper or aluminum target were calculated by increasing the yield determined in lead by a factor of 2 for aluminum and 1.4 for copper (see Swanson, Ref 4, Table XVI, p. 94) The yield at 0^o was increased by 1.5 (see Swanson Ref. 4, fig. 38, p. 92).

Table IV

HEN Yields at 1 m from 20 nA beam loss in thick target Fluence [n/cm²- s (x 10³)]

Energy [MeV]	Al	Steel, Cu	Pb
230 (0^o)	1.5	1.05	0.75
230 (90^o)	1	0.7	0.5
300 (0^o)	1.95	1.35	0.98
300 (90^o)	1.3	0.9	0.65

III. Shielding Calculations

The effectiveness of the shielding was calculated using the source terms calculated in Sections I & II above. Using the methodology described in NCRP Report # 51 (Ref. 14, p. 49), the dose rate D_x through the shield at point r was determined from the expression

D_x = D_o B_x / r² (5)

where D_o is dose rate from the unshielded target calculated at 1 m (see Table II); B_x is the transmission through a given distance obtained from Figure 9 for concrete or Figure 10 for lead; and r is the distance in meters from the target to the location where the dose rate is being calculated. For calculations where the incident electron energy is higher than those provided in the figures, the highest energy provided in the figure was used. These values are acceptable since the transmission is controlled by the Compton attenuation minimum which does not change for increasing electron energies.

For neutron attenuation, the formulation described in NCRP Report # 51 (Ref. 14, p. 55) was used to calculate H_n where

H_n = F_o B_n / r² (6)

Where F_o is the unshielded fluence at one meter from the source, B_n is the transmission through a given distance obtained from Figure 13 for concrete; and r is the distance in meters from the target to the location where the dose rate is being calculated. No credit for neutron attenuation in lead was taken in the calculations. The dose equivalent calculated in this manner explicitly includes the additional dose associated with photons generated from neutron capture.

IV. Examples of Shielding Calculations

Example 1

Calculate the photon dose rate at 90^o at a distance of 3.2 m through 4 feet of concrete for a 20 nA 300 MeV electron striking a thick copper flange. The beam pipe is also surrounded by 2" of lead shielding in the horizontal direction. Using equation (5)

D_x = D_o B_x / r² (7)

D_o = 7.7 R/hr at 1 m from Table II and r = 3.2 meters. The thickness of 4 feet of ordinary concrete expressed in units of g/cm² is

4 ft x 30.48 cm/ft x 2.35 g/cm³ = 286.5 g/ cm² (8)

From Figure 9 B_x,c = 8 x 10^-3. In addition, we calculate the attenuation from lead in the same manner.

2" x 2.54 cm/inch x 11.35 g/cm³ = 57.7 g/ cm² (9)

From Figure 10 B_x,l = 1.5 x 10^-1. Therefore D_x is

D_x = 7.7 R/hr x 1.5 x 10^-1 x 8 x 10^-3 / (3.2)² = 0.9 mR/hr (10)

Example 2

At a distance of 3.2 meters, calculate the neutron dose equivalent rate at 90^o through 4 feet of concrete for a 20 nA 300 MeV beam striking a thick copper target. Using equation (6),

H_n = F_o B_n / r² (11)

We need to calculate the dose equivalent rate for both GRN and HEN and add to get the total.

From Table 3 F_o(GRN) = 4.78 x 10⁴ n/cm²- s. For 4 feet of ordinary concrete or 286 g/cm²

We get B_n(GRN) = 5 x 10^-13 rem- cm² from Figure 13 (using an average energy of 1.5 MeV for the GRN). Subbing these values into eq. (11) we get

H_n (GRN) = 4.8 x 10⁴ x 5 x 10^-13 x 3.6 x 10³ / (3.2) ² = 8.4 x 10^-6 rem/hr (12)

From Table 4 F_o(HEN) = 9 x 10² n/cm²- s and B_n(HEN) = 3 x 10^-9 rem-cm² from Figure 13 (using an average energy of 50 MeV for the HEN). 50 MeV is a conservative value for the HEN as the total HEN fluence is dominated by neutrons in the energy range from 20 - 50 MeV. Substituting these values into eq. (11) we get

H_n (HEN) = 9 x 10² x 3 x 10^-9 x 3.6 x 10³ / (3.2) ² = 0.95 x 10^-3 rem/hr (13)

Therefore

H_n (Total) = H_n (GRN) + H_n (HEN) = 0.96 x 10^{-3 rem}/hr (14)

The following table provides the values of B_n and B_x used in this report for common thicknesses of shield material.

Table V

Examples of Values of B_n and B_x taken from Figures 9, 10, & 13

	2" Lead	3" lead	4" lead	5" lead	32" concrete	48" Concrete
Bx	1.5 x 10^-1	4 x 10^-2	1x 10^-2	3x 10^-3	5 x 10^-2	8 x 10^-3
Bn(GRN)	-	-	-	-	1 x 10^-11	5 x 10^-13
Bn(HEN)	-	-	-	-	1.2 x 10^-8	3 x 10^-9

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