


The way to obtain the desired
equation shall only be sketched. 


First, the coupled differential
equation from above need to be solved for the initial condition 


n_{1}(t = 0) + n_{2}(t = 0) =
n_{0} 




n_{0} is the number of
thermalized positrons in the crystal at the beginning of the experiment: i.e.
at t = 0 

This is easy to do since the first
differential equation does not contain n_{2}. 


The solution must be 


n_{1}(t) 
= 
A · exp–(l_{1} +
n · c_{V})t 



Insertion in the second differential
equation and using the initial condition yields 


n(t) 
= 
n_{0} · 
l_{1} – l_{2}
l_{1} – l_{2} + n ·
c_{V} 
exp–(l_{1} + n
· c_{V})t 
+ 
n_{0} · 
n · c_{V}
l_{1} – l_{2} + n ·
c_{V} 
exp–l_{2} t 



We have two components decaying with
two lifetimes, t_{1} and t_{2}, given by 


t_{1} 
= 
1
l_{1} + n ·
c_{V} 



t_{2} 
= 
1
l_{2} 



Measurements usually only yield an
average lifetime <t>. 


The average
lifetime is not simply the average of t_{1} and t_{2} because we need weighted averages, i.e. 


<t> 
= 
1_{ }
n_{0} 

¥
ó
õ
0 
t 
dn(t)
dt 
dt 
<t> 
= 
t_{2} · 
1 + t_{2}n
· c_{V}
1 + t_{1}n ·
c_{V} 



Doing the integral takes a few lines,
but it is not too difficult  try it! 

