Find σ_{k}(θ)
and σ_{k} for the scattering of a particle
from a perfectly rigid sphere (an infinitely repulsive potential) of radius a.
Choose the energy of the particle such that ka << 1.

Solution:

- Concepts:

Elastic scattering, the method of partial waves, s-wave scattering. - Reasoning:

For very slow particles or very short-range potentials the method of partial waves is the preferred method of calculating the scattering cross section, because only s-waves need to be considered. - Details of the calculation:

V(r) = 0, for r > a, and V(r) = ∞ for r < a. Since ka << 1, ka << (l(l + 1))^{½}for all l except l = 0. We can neglect all phase shifts except that of the s-wave. T

hen σ_{k}(θ) = |f_{k}(θ)|^{2}= (1/k^{2})sin^{2}δ_{0}and σ_{k}= (4π/k^{2})sin^{2}δ_{0}. To calculate δ_{0}we must solve the radial equation for l = 0.

(∂^{2}/∂r^{2}+ k^{2})u_{k0}(r) = 0, u_{k0}(a) = 0

yields u_{k0}(r) = Csin(kr - ka), r > a and u_{k0}(r) = 0, r < a.

As r --> ∞ we expect u_{k0}(r) to be of the form Csin(kr + δ_{0}). We therefore have δ_{0}= -ka. The phase shift is negative, the wave function is "pushed out".

Since ka << 1 we now have (1/k^{2})k^{2}a^{2}= a^{2}and σ_{k }= 4πa^{2}. The total cross section is independent of energy (as long as ka << 1) and equal to four times the geometrical cross section of the hard sphere. Classical mechanics would have yielded πa^{2}. Low energy scattering means very large wavelength scattering, and we do not necessarily expect a classically reasonable result.

[But for high energy scattering off a hard sphere we might expect the classical result. However we obtain σ_{k }= 2πa^{2}twice the geometrical cross section. This is called**shadow scattering**. If the wavelength of the incident particle is very small compared to a, then the sphere will cast a shadow. Directly behind the sphere we will find no scattered particles, but the shadow will extend only up to a finite distance. Very far away from the sphere we will not see the shadow at all, so it must get filled in by scattering of some of the waves at the edges of the sphere. The scattered flux must have the same magnitude as the flux that is taken out of the incident beam by the geometrical cross section of the sphere in order to fill in the shadow. The total scattering cross section therefore must have twice the magnitude of the geometrical cross section.]

A slow particle is scattered by a spherical potential well of the form V(r) =
-V_{0}, for r < a, V(r) = 0, for r > a.

(a) Write down the radial wave equation for this potential and boundary
conditions that apply at r = 0, r = a, and r = ∞.

(b) Assume that the de Broglie wavelength exceeds the dimension of the well, so
that s-wave scattering dominates and write solutions both inside and outside the
r = a sphere. Using the continuity conditions at r = a, calculate the phase
shift that occurs at this boundary.

Solution:

- Concepts:

Elastic scattering, the partial wave method - Reasoning:

For low energy scattering from a finite-range, central potential, the partial wave method is the preferred method for calculating the scattering cross section.

We find the scattering cross section by calculating the phase shifts of the partial waves. - Details of the calculation:

(a) (∂^{2}/∂r^{2}+ k^{2}- U(r) - l(l + 1)/r^{2})u_{kl}(r) = 0 is the radial equation.

Here E = ħ^{2}k^{2}/(2m), V(r) = ħ^{2}U(r)/(2m).

We need u_{kl}(0) = 0, u_{kl}(r) and ∂u_{kl}(r)/∂r^{ }to be continuous at r = a, and u_{kl}(r) to stay finite at infinity.

(b) To calculate δ_{0}we must solve the radial equation for l = 0.

(∂^{2}/∂r^{2}+ k^{2}+ U_{0})u_{k0}(r) = 0, u_{k0}(0) = 0, for r < a.

Therefore u_{k0}(r) = C_{1}sin((k^{2}+ U_{0})^{½}r).

(∂^{2}/∂r^{2}+ k^{2})u_{k0}(r) = 0, for r > a.

u_{k0}(r) = C_{2}sin(kr + δ_{0}).

We need u_{k0}(r) and ∂u_{k0}(r)/∂r^{ }to be continuous at r = a. This yields

C_{1}sin((k^{2}+ U_{0})^{½}a) = C_{2}sin(ka + δ_{0})

and

(k^{2}+ U_{0})^{½}C_{1}cos((k^{2}+ U_{0})^{½}a) = kC_{2}cos(ka + δ_{0})

or

tan(ka + δ_{0}) = (k/(k^{2}+ U_{0})^{½})tan((k^{2}+ U_{0})^{½}a) = C.

We have

δ_{0}= nπ + tan^{-1}(C) - ka.

A slow particle is scattered by a spherical potential well of the form V(r) =
-V_{0}, for r < a, V(r) = 0, for r > a.

What must V_{0}a^{2} be for a 3-dimensional square well
potential in order that the scattering cross section be zero at zero bombarding
energy (Ramsauer-Townsend effect)?

Solution:

- Concepts:

Elastic scattering, the partial wave method - Reasoning:

As E --> 0, k --> 0, ka --> 0. If ka << 1 then the scattering is dominated by s-wave scattering and we use the partial wave method to calculate the scattering cross section.

Then σ_{k}(θ) = (1/k^{2})sin^{2}δ_{0}and σ_{k}= (4π/k^{2})sin^{2}δ_{0}. - Details of the calculation:

When δ_{0}= π, then σ_{k}(θ) = σ_{k}= 0.

To find what V_{0}a^{2}has to be for δ_{0}= π, we need to find the expression for a 3-dimensional square-well potential.

To calculate δ_{0}we must solve the radial equation for l = 0.

Let E = ħ^{2}k^{2}/(2m), V_{0}= ħ^{2}U_{0}/(2m). Then

(∂^{2}/∂r^{2}+ k^{2}+ U_{0})u_{k0}(r) = 0, u_{k0}(0) = 0, for r < a.

Therefore u_{k0}(r) = C_{1}sin((k^{2}+ U_{0})^{½}r).

(∂^{2}/∂r^{2}+ k^{2})u_{k0}(r) = 0, for r > a.

u_{k0}(r) = C_{2}sin(kr + δ_{0}).

We need u_{k0}(r) and ∂u_{k0}(r)/∂r^{ }to be continuous at r = a. This yields

C_{1}sin((k^{2}+ U_{0})^{½}a) = C_{2}sin(ka + δ_{0})

and

(k^{2}+ U_{0})^{½}C_{1}cos((k^{2}+ U_{0})^{½}a) = kC_{2}cos(ka + δ_{0})

or

tan(ka + δ_{0}) = (k/(k^{2}+ U_{0})^{½})tan((k^{2}+ U_{0})^{½}a).

We want δ_{0}= π. Then

tan(ka) = (k/(k^{2}+ U_{0})^{½})tan((k^{2}+ U_{0})^{½}a).

As ka --> 0 this becomes

ka = (k/U_{0}^{½})tan(U_{0}^{½}a) or U_{0}^{½}a = tan(U_{0}^{½}a),

U_{0}^{½}a = nπ + b.

We are not interested in the solution n = 0, b = 0 for the free particle.

n = 1, b = 1.352, U_{0}^{½}a = 4.494, tan(U_{0}^{½}a) = 4.497.

V_{0}a^{2}= ħ^{2}U_{0}a^{2}/(2m) = 20.2 ħ^{2}/(2m).

The Ramsauer-Townsend effect is analogous to the perfect transmission found at particular energies in a one dimensional problem. It is the explanation for the extremely low minimum observed in the scattering cross section of electrons by rare gas atoms at about 0.7 eV bombarding energy. The Ramsauer-Townsend effect cannot occur with a repulsive potential, since ka would have to be at least π to make |δ_{0}|_{ }= π. The perfectly rigid sphere produces the largest |δ_{0}| for any repulsive potential with range a, |δ_{0}| = ka. A potential of this large range produces higher l phase shifts.

Determine the differential scattering cross section σ(θ) in units of cm^{2}/sr
for a particle of mass m = 9.1*10^{-31}kg incident on a spherically
symmetric potential

V(r) = 0, 0 < r < a, V(r) = V_{0}, a
< r < b, V(r) = 0, r > b,

with a = 0.05 nm and b = 0.1 nm. Let E = 1 eV and V_{0 }= 0.8
eV.

Solution:

- Concepts:

Elastic scattering, the method of partial waves, s-wave scattering. - Reasoning:

E = ħ^{2}k^{2}/(2m), k^{2}= 2mE/ħ^{2}, k = 5.1*10^{9}/m, kb = 0.51 < 1.

Only s-wave scattering is important. We have σ_{k}(θ) = (1/k^{2})sin^{2}δ_{0}. - Details of the calculation:

To calculate δ_{0}we must solve the radial equation for l = 0.

(∂^{2}/∂r^{2}+ k^{2}- U(r))u_{k0}(r) = 0, V(r) = ħ^{2}U(r)/(2m).

r < a: U(r) = 0, u_{k0}(r) = C_{1}sin(kr).

a < r < b: U(r) = U_{0}, V_{0 }= ħ^{2}U_{0}/(2m), E > V_{0}.

Therefore u_{k0}(r) = C_{2}sin((k^{2}- U_{0})^{½}r + δ') = C_{2}sin((k_{1}r + δ').

r > a: U(r) = 0, u_{k0}(r) = C_{3}sin(kr + δ_{0}).

We need u_{k0}(r) and ∂u_{k0}(r)/∂r^{ }to be continuous at r = a and r = b.

This yields at r = a

C_{1}sin(ka) = C_{2}sin(k_{1}a + δ'),

kC_{1}cos(ka) = k_{1}C_{2}cos(k_{1}a + δ'),

or

δ' = tan^{-1}((k_{1}/k)tan(ka)) - k_{1}a.

k_{1}/k = ((E-V_{0})/E)^{½}= 0.447, ka = 0.256, δ' = 0.002.

At r = b we have

C_{2}sin((k_{1}b + δ') = C_{3}sin(kb + δ_{0}),

k_{1}C_{2}cos((k_{1}b + δ') = kC_{3}cos(kb + δ_{0}),

or

δ_{0}= tan^{-1}((k/k_{1})tan(k_{1}b + δ')) - kb.

δ_{0}= -2.8*10^{-2}.

σ_{k}(θ) = (1/k^{2})sin^{2}δ_{0}= 3*10^{-19}cm^{2}/sr.

Calculate the differential and total elastic scattering cross section for an
electron of energy 0.25 eV incident on a spherical positive potential of form
V(r) = 2 eV (r < 4a_{0}) and V(r) = 0 (r > 4a_{0}).
Justify your method. Give numerical answers.

Solution:

- Concepts:

Elastic scattering, the partial wave method - Reasoning:

E = ħ^{2}k^{2}/(2m), k^{2}= 2mE/ħ^{2}, k = 2.56*10^{9}/m, k4a_{0}= 0.54 < 1.

Only s-wave scattering is important. We have σ_{k}(θ) = (1/k^{2})sin^{2}δ_{0}. -
Details of the calculation:

To calculate δ_{0}we must solve the radial equation for l = 0.

(∂^{2}/∂r^{2}+ k^{2}- U(r))u_{k0}(r) = 0, V(r) = ħ^{2}U(r)/(2m).

Let a = 4a_{0}.

r < a: U(r) = U_{0}, V_{0}= ħ^{2}U_{0}/(2m), E = ħ^{2}k^{2}/(2m), E < V_{0}.

Therefore u_{k0}(r) = C_{1}sinh((U_{0}-k^{2})^{½}r).

r > a: U(r) = 0, u_{k0}(r) = C_{2}sin(kr + δ_{0}).

We need u_{k0}(r) and ∂u_{k0}(r)/∂r^{ }to be continuous at r = a.

This yields at r = a

C_{1}sinh((U_{0}-k^{2})^{½}a) = C_{2}sin(ka + δ_{0}),

(U_{0}-k^{2})^{½}C_{1}cosh((U_{0}-k^{2})^{½}a) = kC_{2}cos(ka + δ_{0}),

or

tan(ka + δ_{0}) = (k/(U_{0}-k^{2})^{½})tanh((U_{0}-k^{2})^{½}a) = C.

δ_{0}= tan^{-1}(C) - ka.

U_{0}= 2mV_{0}/ħ^{2}= 5.24*10^{19}m^{-2}.

k^{2}= 6.55*10^{18}m^{-2}.

ka = 0.54.

(U_{0}-k^{2})^{½}a = 1.43, k/(U_{0}-k^{2})^{½}= 0.376.

C = 0.335, tan^{-1}(C) = 0.323, δ_{0}= -0.21.

σ_{k}(θ) = (1/k^{2})sin^{2}δ_{0}= 7*10^{17}cm^{2}/sr.

σ_{k}= 8.85*10^{-16}cm^{2}.

Determine the total cross section for the scattering of slow particles (ka < 1) by a potential V(r) = Cδ(r-a).

Solution:

- Concepts:

Elastic scattering, the method of partial waves, s-wave scattering. - Reasoning:

For very slow particles or very short-range potentials the method of partial waves is the preferred method of calculating the scattering cross section, because only s-waves need to be considered. - Details of the calculation:

σ_{k}= (4π/k^{2})sin^{2}δ_{0}.

We need to find the phase shift δ_{0}for V(r) = Cδ(r-a).

To calculate δ_{0}we must solve the radial equation for l = 0.

(∂^{2}/∂r^{2}+ k^{2}- 2mV(r)/ħ^{2})u_{k0}(r) = 0.

Region 1: V(r) = 0, u^{1}_{k0}(r) = C_{1}sin(kr).

Region 2: V(r) = 0, u^{2}_{k0}(r) = C_{2}sin(kr + δ_{0}).

Boundary conditions:

u^{1}_{k0}(a) = u^{2}_{k0}(a), C_{1}sin(ka) = C_{2}sin(ka + δ_{0}).

∂u^{2}_{k0}(r)/∂r|_{a+ε}- ∂u^{1}_{k0}(r)/∂r|_{a-ε}= (2m/ħ^{2})Cu(a).

kC_{1}cos(ka) + (2mC/ħ^{2})C_{1}sin(ka) = kC_{2}cos(ka + δ_{0}).

kcot(ka) + 2mC/ħ^{2}= kcot(ka + δ_{0}).

δ_{0}= cot^{-1}(cot(ka) + 2mC/(kħ^{2})) - ka.