3.7. Sources of Error in Gaseous Detectors 199
Hereαis called the recombination coefficient andSrepresents the source of charges.
The above two equations can be combined to give
d(n−−n+)
dt=0
⇒n− = n++C 1 , (3.7.3)whereC 1 is the constant of integration and depends on the initial difference be-
tween the number of positive and negative charges. Substituting equation 3.7.3 into
equation 3.7.2 gives
dn−
dt
=S−α
(
n−) 2
+αC 1 n−. (3.7.4)This is a first order linear differential equation with a solution of
n−=r 1 −r 2 C 2 exp(√
C 12 +rS/αt)
1 −C 2 exp(√
C^21 +rS/αt). (3.7.5)
Herer 1 andr 2 are the roots of the quadratic equation on the right side of equation
3.7.4 given by
r 1 ,r 2 =1
2
[
C 1 ±
√
C 12 +4S/α]
. (3.7.6)
Similarly the solution for positive charges can be obtained from equation 3.7.1. The
roots in this case are given by
r 1 ,r 2 =1
2
[
−C 1 ±
√
C 12 +4S/α]
(3.7.7)
The constantsC 1 andC 2 can be determined by using the boundary conditions:
n=n 0 att=0andn=natt=t. Instead of solving this equation for a particular
case, we note that these solutions represent complex transcendental behavior, which
eventually reach the steady state value ofr 2 ,thatis
n∞→r 2 as t→∞.For a special case when the initial concentrations of positive and negative charges
are equal, the constantC 1 assumes the value zero and consequently the steady state
charge concentration becomes
n∞=√
S
α. (3.7.8)
This shows that the equilibrium or steady state charge concentration is completely
determined by the source producing electron-ion pairs and the recombination coef-
ficient.
Application of electric field forces the charges to move toward respective elec-
trodes thus reducing the recombination probability.
Example:
Estimate the steady state density of ions in a 0.5atmhelium filled ionization
chamber if the ionization rate is 1. 5 × 1011 cm−^3 s−^1. The recombination