Appendix 1 The steps in derivation of the probability p(n,x) for the production of an avalanche of n electrons at the distance x from the cathode are as follows: __________________________ Let N(x) is the number of electrons emitted from the cathode which pass the distance x' from the cathode without any ionizing collision. Then dN(x') = -a.N(x)dx´ N(x') = .exp(-ax') where = N(x'=0). As a consequence of this p(1,x') = N(x')/ = exp(-ax') (1) _________________________ Let the probability that the avalanche contains n-1 at x' is p(n-1,x') (2) The probability that one and only one of these electrons will ionize in the region between x' and x'+dx' can be found from the binomial distribution considering that W(k,l) = p(n-1, 1), and y = a.dx .Thus p(n-1,1) = (n-1).a.dx'(1-a.dx') ^n-2 for dx' » 0 p(n-1,1) @ (n-1).a.dx' (3) _________________________ The number of electrons in the avalanche has now increased from n-1 to n. The probability that none of these electrons will ionize in the region between x'+dx'(@ x') and x is: [p(1,x-x´)]n = [ exp { - a. (x-x´)} ] n = exp{ - n.a. (x-x´)} (4) where p(1,x-x') is the probability that a single electron will not ionize between x' and x. If we take the product of expressions (2),(3) and (4), and integrate over x ,for n>1, we get The solution of the equation (5) is: