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PENELOPE 2003 - OECD Nuclear Energy Agency

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2.4. Electron-positron pair production 59<br />

where<br />

v = (0.2840 − 0.1909a) ln(4/κ) + (0.1095 + 0.2206a) ln 2 (4/κ)<br />

+ (0.02888 − 0.04269a) ln 3 (4/κ) + (0.002527 + 0.002623a) ln 4 (4/κ). (2.89)<br />

Then, the single quantity η ∞ characterizes the triplet production for each element.<br />

The approximation given by eq. (2.77) with the fitted value of the screening radius,<br />

fails at low energies where it systematically underestimates the total cross section (it<br />

can even become negative). To compensate for this fact we introduce an empirical<br />

correcting term F 0 (κ, Z), which acts in a way similar to the Coulomb correction. To<br />

facilitate the random sampling, the Bethe-Heitler DCS, eq. (2.77), including this lowenergy<br />

correction and a high-energy radiative correction, is written in the form<br />

[ ( )<br />

dσ pp<br />

= r 2 2 1 2<br />

]<br />

dɛ<br />

eαZ[Z + η] C r 2<br />

3 2 − ɛ φ 1 (ɛ) + φ 2 (ɛ) , (2.90)<br />

where<br />

with<br />

φ 1 (ɛ) = g 1 (b) + g 0 (κ)<br />

φ 2 (ɛ) = g 2 (b) + g 0 (κ) (2.91)<br />

g 1 (b) = 1 2 (3Φ 1 − Φ 2 ) − 4 ln(Rm e c/¯h) = 7 3 − 2 ln(1 + b2 ) − 6b arctan(b −1 )<br />

− b 2 [ 4 − 4b arctan(b −1 ) − 3 ln(1 + b −2 ) ] ,<br />

g 2 (b) = 1 4 (3Φ 1 + Φ 2 ) − 4 ln(Rm e c/¯h) = 11 6 − 2 ln(1 + b2 ) − 3b arctan(b −1 )<br />

+ 1 2 b2 [ 4 − 4b arctan(b −1 ) − 3 ln(1 + b −2 ) ] ,<br />

g 0 (κ) = 4 ln(Rm e c/¯h) − 4f C (Z) + F 0 (κ, Z). (2.92)<br />

C r = 1.0093 is the high-energy limit of Mork and Olsen’s radiative correction (Hubbell<br />

et al., 1980).<br />

The correcting factor F 0 (κ, Z) has been determined by requiring that the total cross<br />

section for pair production obtained from the expression given in eq. (2.90) (with η = 0)<br />

coincides with the total cross sections for pair production in the field of the nucleus tabulated<br />

by Hubbell et al. (1980). By inspection and numerical fitting, we have obtained<br />

the following analytical approximation<br />

F 0 (κ, Z) = (−0.1774 − 12.10a + 11.18a 2 )(2/κ) 1/2<br />

+ (8.523 + 73.26a − 44.41a 2 )(2/κ)<br />

− (13.52 + 121.1a − 96.41a 2 )(2/κ) 3/2<br />

+ (8.946 + 62.05a − 63.41a 2 )(2/κ) 2 . (2.93)

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