3.3 Solutions
239
tan k R ≈ k R +
(k R)
3
3
+ · · ·
We therefore have
tan δ 0 ≈
k
D − R −
(k R)
3
3k
1 + k 2 D R
If the inequalities kR << 1 and k
2 DR << 1 are satisfied we can still
further simplify the expression for tan δ 0 .
tan δ 0 ≈ k(D − R) = k R
tan k 1 R
k 1 R
− 1
(8)
The total cross-section is then
σ =
4π
k 2 sin
2
δ 0 ≈ 4π(D − R)
2
= 4π R
2
1 −
tan k 1 R
k 1 R
2
(9)
It follows that if the condition
tan k 1 R = k 1 R
(10)
is satisfied, the phase shift and the scattering cross-section both vanish. This
phenomenon is known as the Ramsauer-Townsend effect. The field of the
inert gas atoms decreases appreciably faster with distance than the field of
any other atom, so that to a first approximation, we can replace this field by a
rectangular spherical well with sharply defined range and use Equation (10)
to evaluate the cross-section for slow electrons.
Physically, the Ramasuer – Townsend effect is explained as the diffraction
of the electron around the rare-gas atom, in which the wave function inside
the atom is distorted in such a way that it fits on smoothly to an undistorted
function outside.
Here the partial wave wave with l = 0 has exactly a half cycle more of
oscillation inside the atomic potential then the wave in the force-free field,
and the wavelength of the electron is large enough in comparision with R so
that higher l phase-shifts are negligible.
3.112 In order that the Schrodinger equation is reduced to the given form is that the
potential V (r ) does not depend on time. From Problem 3.104 the total wave
function
ψ = Σi
l (2l + 1)
e
iδl
kr
sin
kr −
1
2
πl + δ l
p l (cos θ )
For slow neutrons only the first term (l = 0) in the summation is important.
As p 0 (cos θ) = 1
ψ =
exp(iδ 0 )
kr
sin(kr + δ 0 )
u = ψr =
exp(iδ 0
k
sin(kr + δ 0 ) = const. sin(kr + δ 0 )
We assume that the wave function inside the well is identical with that
in the deuteron problem. This is justifiable since the total energy inside the
potential well is raised by little over 2 MeV corresponding to the
239
tan k R ≈ k R +
(k R)
3
3
+ · · ·
We therefore have
tan δ 0 ≈
k
D − R −
(k R)
3
3k
1 + k 2 D R
If the inequalities kR << 1 and k
2 DR << 1 are satisfied we can still
further simplify the expression for tan δ 0 .
tan δ 0 ≈ k(D − R) = k R
tan k 1 R
k 1 R
− 1
(8)
The total cross-section is then
σ =
4π
k 2 sin
2
δ 0 ≈ 4π(D − R)
2
= 4π R
2
1 −
tan k 1 R
k 1 R
2
(9)
It follows that if the condition
tan k 1 R = k 1 R
(10)
is satisfied, the phase shift and the scattering cross-section both vanish. This
phenomenon is known as the Ramsauer-Townsend effect. The field of the
inert gas atoms decreases appreciably faster with distance than the field of
any other atom, so that to a first approximation, we can replace this field by a
rectangular spherical well with sharply defined range and use Equation (10)
to evaluate the cross-section for slow electrons.
Physically, the Ramasuer – Townsend effect is explained as the diffraction
of the electron around the rare-gas atom, in which the wave function inside
the atom is distorted in such a way that it fits on smoothly to an undistorted
function outside.
Here the partial wave wave with l = 0 has exactly a half cycle more of
oscillation inside the atomic potential then the wave in the force-free field,
and the wavelength of the electron is large enough in comparision with R so
that higher l phase-shifts are negligible.
3.112 In order that the Schrodinger equation is reduced to the given form is that the
potential V (r ) does not depend on time. From Problem 3.104 the total wave
function
ψ = Σi
l (2l + 1)
e
iδl
kr
sin
kr −
1
2
πl + δ l
p l (cos θ )
For slow neutrons only the first term (l = 0) in the summation is important.
As p 0 (cos θ) = 1
ψ =
exp(iδ 0 )
kr
sin(kr + δ 0 )
u = ψr =
exp(iδ 0
k
sin(kr + δ 0 ) = const. sin(kr + δ 0 )
We assume that the wave function inside the well is identical with that
in the deuteron problem. This is justifiable since the total energy inside the
potential well is raised by little over 2 MeV corresponding to the