1000 Solved Problems in Modern Physics

(Tina Meador) #1

238 3 Quantum Mechanics – II


∑R/λ
l= 0
(2l+1)=(R/λ-)^2

∴σr=σs=πλ-^2

(

R

λ-^2

)

=πR^2

The total cross-section
σt=σr+σs= 2 πR^2
which is twice the geometrical cross-section

3.111 The potential which an electron sees as it approaches an atom of a monatomic
gas can be qualitatively represented by a square well. Slow particles are con-
sidered.
V(r)=−V 0 ;r≤R
=0; r>R
corresponding to an attractive potential. Scattering of slow particles for
whichkR<<1, is determined by the equation
(
∇^2 +k^2 −


2 μV
^2

)

ψ 2 =0 (inside the well) (1)

withk^2 = 2 μE/^2 , and the wave numberk=p/
Outside the well the equation is
(∇^2 +k^2 )ψ 1 =0(2)
Further writing
k^21 =k^2 +k^20
wherek^20 =^2 μ 2 V
andV=−V 0
The solutions are found to be
ψ 2 =Asink 1 r (3)
ψ 1 =Bsin(kr+δ 0 )(4)
ψ 1 (r) is the asymptotic solution at large distances with the boundary condi-
tion
ψ 1 (0)= 0
Matching the solutions (3) and (4) atr = Rboth in amplitude and first
derivative,
Asink 1 R=Bsin(kR+δ 0 )(5)
Ak 1 cosk 1 R=Bkcos(kR+δ 0 )(6)
Dividing one equation by the other, and settingk 1 cotk 1 R= D^1 , and with
simple algebraic manipulations we get
tanδ 0 =(kD−tankR)(1+kDtankR)−^1 (7)
The phase shiftδ 0 determined from (7) is a multivalued function but we are
only interested in the principle value lying within the interval−π 2 ≤δ 0 ≤π 2.
For small values of the energy of the relative motion
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