Exam draft for Atomic Physics in Fusion, ED2235, suggested solutions
December 9, 2009
Part A
1.
v  2.5 106 m/s . [Can be derived e.g. by classical mechanics for a circular orbit, if one
remembers, say, that the orbit radius is about 0.5 Å, or use that the kinetic energy is about
the same as the ionization energy for hydrogen, 13.6 eV]
2.
kT
3.
E 2
4. Elastic binary collisions and continuous braking due to electronic excitations
5. V / (2 )3 where V is the volume in real space, or twice that density due to spin
Part B
6. The boundary condition at infinity should be that the wave function vanishes sufficiently fast
for normalization. The boundary conditions at x=0 should be firstly that the wave function is
continuous, secondly a condition on the derivative, which is found from the Schrödinger
equation. Integrate both sides of 
2
2m
 2  V ( x)  E from x   to x   and take
the limit   0 , remembering that the wave function is continuous. This gives
 '( )  '( )  2m /
2
as boundary condition for the derivative. For x>0 the
Schrödinger equation gives  ( x)  Ae
2 mE x /
and for x<0  ( x)  Ae
2 mE x /
, continuous at
2mE
. The allowable energy is determined
2
from the boundary condition on derivative: E    , there is only one bound state.
x=0. A is determined from normalization: A 
7. Let the CM-system be moving at speed VCM . The velocities of particles 1 and 2 in the CM
system are v1  VCM and VCM respectively. The condition that the total momentum is zero in
the CM system gives m1 (v1  VCM )  m2VCM 
VCM 
m1
v1
m1  m2
while
v1  VCM 
m2
v1
m1  m2
so that   VCM / (v1  VCM )  m1 / m2 . From vector addition like in the figure
vL sin  L  v sin 
vL cos  L  VL  v cos 
and consequently:
tan  L 
sin 
  cos 
which is a relation of the desired kind.
8. We can use the result that for classical diffusion, a random walk with step length rL 
me v
,
eB
the diffusivity is D  rL2 /  . Only a rough estimate is required. Setting v  kT / me gives
D  3.5 103  m 2 /s . For the confinement time a simple dimensional argument or some
more elaborate procedure, e.g. in slab geometry gives t  L2 /4 D0.7s .
Part C
9. The Thomas-Fermi model of the atom assumes a spherically symmetric electron charge
distribution. Each volume element is associated with a potential energy density as for the
classical electrostatic charge density and a kinetic energy density as for a free electron gas
with the given local electron density. Minimizing the total energy of the system gives a nonlinear differential equation for the potential distribution in the atom. The solution is a
screened potential V (r ) 
Ze2
(r / aTF ) , from which the charge density distribution can
4 0 r
also be calculated. The Thomas-Fermi model of the atom can be used for calculating
approximate interatomic potentials when two atoms collide, if the charge distributions of
both atoms are taken to be unperturbed by the collision process and the total energy of the
system of both atoms is calculated from the overlap of the two charge distributions.
10. The ionization cross section can be calculated classically by assuming that the projectile is
effectively colliding with a single bound electron of the target atom. It is reasonable to
assume that ionization results if the projectile imparts more energy to the target electron
than its binding energy. The total cross section for ionization is obtained by summing the
cross sections for all target electrons, at all binding energies. The result is a cross section that
increases with energy, starting from the threshold at the ionization energy, reaches a
maximum and decreases again at high impact energies as E 1 .