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:
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   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') = -.N(x)dx´
                    N(x') =  .exp(-x')
where  = N(x'=0). As a consequence of this
              p(1,x') = N(x')/  = exp(-x')                                                      (1)
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   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 = .dx .Thus
                p(n-1,1) = (n-1)..dx'(1-.dx') n-2   
for dx'   0
                     p(n-1,1)  (n-1)..dx'                                                             (3)
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   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: