# 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 4.2-1 Diffusion During Cooling

Exercise 4.1-1 Lifetime of Positrons

© H. Föll (Defects - Script)