Solution to Exercise 4.1-1 "Lifetime of Positrons"

	Show that the solution of the differential equations for the positron concentrations n1 and n2
	dn1 dt   =  – (l1  +  n · cV) · n1       dn2 dt   =  – 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 · l1 – l2 l1 – l2 + n · cV exp–(l1  +  n · cV)t  +   n0 · n · cV l1 – l2 + 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 l2
	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  n0    ¥ ó õ 0 t dn(t) dt dt <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

Exercise 4.1-1 Lifetime of Positrons

© H. Föll (Defects - Script)