Solution to Exercise 4.1-1 "Lifetime of Positrons"

Show that the solution of the differential equations for the positron concentrations n1 and n2
 =  – (l1  +  n · cV) · n1
 =  l2 · n2 + n · cV · n1
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
n1(t = 0) + n2(t = 0) = n0
n0 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 n2.
The solution must be
n1(t)  =  A · exp–(l1  +  n · cV)t
Insertion in the second differential equation and using the initial condition yields
n(t)  =  n0 · l1l2
l1l2 + n · cV
exp–(l1  +  n · cV)t  +   n0 · n · cV
l1l2 + n · cV
exp–l2 t
We have two components decaying with two lifetimes, t1 and t2, given by
t1  =  1
l1 + n · cV
t2  =  1
Measurements usually only yield an average lifetime <t>.
The average lifetime is not simply the average of t1 and t2 because we need weighted averages, i.e.
<t>  =  1 
t dn(t)

<t>  =  t2 · 1 + t2n · cV
1 + t1n · cV
Doing the integral takes a few lines, but it is not too difficult - try it!

With frame With frame as PDF

go to Exercise 4.1-1 Lifetime of Positrons

© H. Föll (Defects - Script)