Scattering


Contents

  1. Introduction
  2. Scattering Fundamentals
    Kinds of Scattering Experiments
    Solid Angle
    Cross Section
    Center-of-Mass System
    Elastic and Inelastic Collisions
    Nuclear Examples
    Classical Scattering of Hard Spheres
  3. Quantum Mechanics of Scattering
    Setting Up the Time-Independent Problem
    The Method of Partial Waves
    Finding the Phase Shifts
  4. Ramsauer Scattering
  5. References

Introduction

In general terms, "scattering" means a random distribution of certain items. In physics, it has a different and quite definite meaning. The scattering of light is a concrete example. We think of a collimated beam of light from which a portion of the energy is diverted from the beam, as when the light passes through dusty air; a searchlight beam is seen from the side by such scattered light. This is extended to experiments in which a collimated beam of particles suffers a similar fate. With light, we think of particles which reflect, refract or re-radiate the light in directions other than that of the axis of the beam. With particles, we think of binary collisions of particles in the beam with other particles that knock the beam particles into different directions. In either case, we usually concentrate on the interactions of the wave with the particle, or the interactions of two particles, and this phenomenon is called scattering, though scattering is really the effect of a multitude of such processes, some more complex than simple binary collisions.

Scattering experiments were rare in physics before quantum mechanics, except perhaps in relation to the scattering of light by colloids and density fluctuations. After quantum mechanics, scattering experiments became the principal method for the study of atoms, molecules and nuclei, since the properties of the scattering system were reflected in the scattering process. The earliest were the alpha-particle scattering experiments of Geiger and Marsden that revealed the atomic nucleus. The study of electrical discharges and thermionic vacuum tubes had made available methods of creating high vacua that are necessary for scattering experiments. The collisions of electrons with atoms were studied extensively, and then the collisions of artificially accelerated nuclear particles--protons, deuterons, neutrons and so on with each other and with nuclei. Scattering is still the primary experimental resource for studying fundamental particles.

In this article, only nonrelativistic potential scattering of particles will be considered, and even then only the fundamentals. The whole subject of scattering is gigantic, and its literature is overwhelming. The size of the study of even atomic scattering is huge, as can be seen from the size of the work by Mott and Massey. We use the term "particle" only to differentiate from the scattering theory of electromagnetic and mechanical waves, which requires a very different treatment. "Particle" refers to electrons, atoms, nuclei and so on. Wave scattering can have particle aspects at high energies (photons), as in Compton scattering, but for this relativity is essential.

Scattering Fundamentals

Kinds of Scattering Experiments

The two main kinds of scattering experiments are illustrated at the right. Both have a source of particles at the left that produces a collimated beam of intensity Io particles per square cm per second, called the flux. These particles may be, for example, electrons, protons, neutrons, gamma-rays (photons), and so on. In general, the kinetic energy or velocity of the particles can be controlled, and a monoenergetic beam created. The velocity is given in terms of the energy by v = √(2qV/m), where V is the energy in electron volt. eV. The energy in joule is E = qV. The absolute value of the charge of the particle is q, and m is its mass. The momentum p = mv = √(2mqV), and the wave vector k = 2π/λ = p/h' (h' = h/2π), where λ is the deBroglie wavelength. If the density of particles in the beam is n cm-3, the flux is nv particles cm-2s-1. This is a directed flux. In some cases, an isotropic flux may be appropriate, as in a nuclear reactor. The isotropic flux (number of particles crossing a 1 cm2 area per second) is φ = nv/4.

In an experiment of type (a), the beam falls on a localized target that is here represented as a thin foil. All the scattering centres are subject to the same flux. Often, the scattered particles are received by a detector at some angle θ from the beam direction, so the numbers of particles received per second can be measured as a function of angle. The current of charged particles can be measured with a Faraday cage, and ionizing particles with a Geiger-Müller counter. This is only one possibility of many. A photographic emulsion, or a cloud or bubble chamber, or its equivalent, can make the paths of scattered particles evident. The incident particles can cause a nuclear reaction to occur in the target, creating a radioactive isotope. Measuring the radiation can tell how many reactions occurred in the time of irradiation.

In an experiment of type (b), the beam passes through a scattering volume of length L, in which it undergoes interactions, scattering particles from the beam so that the beam is attenuated. In many cases, the attenuation is exponential, so that I(x) = Io e-μx. The attenuation coefficient μ can be computed from the ratio of I(L) to Io: μ = (1/L)ln[Io/I(L)]. The target is generally a gas, though thin metal foils can also be used. By varying L and the gas pressure, the exponential law can be verified, and also that μ is proportional to the density of gas particles. The scattered particles can also be detected, and related to the probability of scattering, as well as induced radioactivity.

Solid Angle

The concept of solid angle is used to describe the angular distribution of scattered particles. A solid angle (with respect to some centre) is product of an area dA and the cosine that its normal makes with the radius to the centre, divided by the square of the radius, dω = dA cosθ/r2, which is obviously dimensionless. The area of a sphere of radius r is 4πr2, so (4πr2)(cos 0)/r2 = 4&pi. A solid angle defined this way is said to be measured in steradians, sr, and so there are 4π steradians around a point. The steradian is not, of course, a real unit, only a note saying what we are talking about. If the reference point is inside a closed surface, the integral of dA cosθ/r2 will be 4π no matter what the shape of the surface, which can be proved by Gauss's Law. If the reference point is outside the closed surface, the total solid angle is zero. In general, if we arbitrarily choose the direction of the normal to dA (say, in the direction a right-handed screw would advance if we go round the boundary keeping dA to our left), then if r is in the same direction (θ < 90°), dω > 0, and otherwise is negative.

These matters are probably quite clear, but it gives further insight to compare the case in a plane. We define the angle subtended by a line segment ds as the component of the segment normal to the radius r divided by r, or dθ = ds cosθ / r. This dimensionless quantity is the angle in radians. If the reference point is at the centre of a circle, then the total angle is 2πr/r = 2π radians. The total angle subtended by any closed curve with respect to a point within it is 2&pi. If the point is outside, the angle subtended is zero. Angles can be considered to be positive and negative depending on the direction of description of the line segment. For a circle, anticlockwise is chosen to give positive angles.

In terms of the diameter d, the circumference of a circle is C = πd and the area of a sphere is A = πd2. This gives the remarkable result that C/d = A/d2 = π. Once you have squared the circle, you have also squared the sphere! The area of a sphere is just the circumference of a circle equal to its diameter times the diameter, or the area of a circular cylinder that just fits over the sphere, and is as high as the sphere.

Very often there is symmetry about the axis, so that the solid angle element at a polar angle θ is conical, as shown in the diagram. This is the solid angle between θ and θ + dθ. If it is integrated from 0 to π, the total solid angle Ω = 4π is obtained.

Cross Section

Let's concentrate on a particular outcome of a scattering encounter. This can be anything, from absorption of the particle to sending it off into a solid angle dω in the direction (θ,φ) given by spherical coordinate angles. The rate at which this particular outcome occurs is proportional to the incident flux nv (twice as many chances, twice as many outcomes), and also to the number of scattering centres N on which the flux falls. Both these conditions have nothing to do with the actual process, so we would like to have a measure of the probability of the process that is independent of them. If we let n be the beam particle density in cm-3, and N the number of centres per cm3, then the rate of outcomes will be per cubic centimetre per second. We do not have to have a square centimetre of beam, nor a cubic centimetre of target, but can adjust the answer to fit whatever beam flux and target geometry we happen to have.

The ratio (outcomes per cc per s)/Nnv has the dimensions of cm2, an area, and is called a cross section. However, it is not basically an area, but a probability, and reference to this will resolve many paradoxes. If I is the number of outcomes per cc per s, then I = Nnvσ, where σ is the cross section for the process. In many cases, the cross section can be represented by an actual area, making it quite clear what is going on. In the diagram at the left, a typical square centimetre of beam is shown, with the dots perhaps representing the points at which beam particles were directed. They are uniform over the surface, as they will be on average. The small circles are drawn at the position of scattering centres, of which there are 15 in this square centimetre. A smaller or larger target will have a proportionally smaller number of scattering centres. It will be seen that 7 of the centres have been "hit" out of 289 incident beam particles, or a sample probability of 7/289 = 0.024. The fifteen small circles have a total area of 180 pixel2, while the whole square has an area of about 4624 pixel2, so the population probability is 180/4624 = 0.039. We have been a bit unlucky with hits, but the result is within the difference expected from chance.

All the scattering centres have been projected onto a plane in this case. If N is the density of scattering centres, and t the thickness of the target, then Σ = Ntσ is the total projected cross section. Since the probability Ntσ/1 is small, there is little chance of overlap. Also, the scattering centres act independently of each other, which is another condition for the validity of the analysis. In the more general case of a thick target, we consider a thin slice of thickness dx, which gives a total cross section of Nσdx. A beam of intensity I will scatter dI = -INσdx particles in distance dx, and become weaker by this amount. Integrating dI/I = -Nσdx from x = 0 to x = L, we have I(x) = Ioe-NσL, an exponential attenuation of the beam. This expression holds well even for beams of particles, as it would for a wave, since in quantum mechanics a scattering centre does not take an infinite hole out of the beam, as it would in a classical case. This deserves a little thought, though it is perfectly all right for NσL to be greater than 1, when either the beam would be completely blocked (if the scattering centres were cunningly arranged not to overlap), or else many scattering centres would be shielded by those in front and not feel the full beam flux. Strange things can happen with thick targets. An initially homogeneous beam can become nonhomogeneous, and worse.

Let's now look at what the numbers are like in practice. The number density N for a solid or liquid can be found from Avogadro's Number NA, the density d, and the atomic weight A. In fact, N = NAd/A. For Al, d = 2.7 and A = 27, so N = 6.02 x 1022 cm-3. For Cd, d = 8.65 and A = 112.4, so N = 2.32 x 1022 cm-3. For Au, d = 19.32 and A = 197, giving N = 5.90 x 1022. For metals at least, we estimate N is between 1022 and 1023 cm-3. For a gas, a mol at STP occupies 22,400 cm3, so N = 2.69 x 1019 cm-3, a factor of 1000 fewer than for a metal.

An atom is less than 0.2 nm in diameter, so its projected area will not exceed about 10-16 cm2. If we consider atoms colliding with each other, classically, the cross section will be four times this. We see that Nσ will be on the order of 3000 cm-1, which means that a beam of particles will not get very far at atmospheric pressure. At a vacuum of 1 μmHg, however, N will be only about 3.5 x 1013, and Nσ will be only 3.5 x 10-3 cm-1. 1/Nσ is the average distance a particle travels before colliding, the mean free path, which here will be 286 cm. This illustrates clearly why scattering experiments must be done in a high vacuum.

A scattering target is often a thin metal foil, thin enough for substantial penetration. Foil targets are usually specified by weight, not thickness, though the weight is related to the thickness by w = td. An Al foil of 10 mg/cm2 has a thickness of t = 0.01/2.7 = 0.0037 cm. The number of atoms per cm2 will be 2.23 x 1020 cm-2. This is NAw/A. A gold foil of 10 mg/cm2 has 3.05 x 1019 atoms per cm2.

Center-of-Mass System

The arbitrary inertial system in which we consider the collision is called the laboratory system. For example, one kind of particle may be projected in a beam toward another kind at rest in a target. This system is not the best for analyzing the collision, since what happens changes if we use a system moving with constant relative velocity with respect to the first. That is, the results will depend on this relative velocity in addition to the dynamics of the collision. To remove this dependence, we transform to an inertial system in which the center of mass does not move, called the center-of-mass or CM system.

The transformation r' = r - vt to a system moving with constant relative velocity v is called a Galilean transformation. It leaves the Newtonian equations of motion d2r/dt2 = f unchanged, and, in particular, does not affect the forces. Suppose we have two particles of masses m1 and m2, with the second exerting a force f on the first, and the first exerting a force -f on the second, in accordance with the Third Law. The equations of motion are m1d2r1/dt2 = f and m2d2r2/dt2 = -f. If we add the two equations, we find d2(m1r1 + m2r2)/dt2 = 0. If we define R = (m1r1 + m2r2)/(m1 + m2), then this equation is d2R/dt2 = 0, which integrates to R = a + bt, where a and b are vector constants. R is clearly the position of the center of mass. The velocity of the center of mass is v = (m1v1 + m2v2)/(m1 + m2). If we transform to an inertial system moving with constant relative velocity -v, and let the center of mass be at the origin of this system, then R = 0 in this system. We have taken care of 6 of the 12 arbitrary constants that enter in a complete specification of the problem.

If we divide each equation of motion by the corresponding mass and subtract, we find d2(r1 - r2)/dt2 = (1/m1 + 1/m2)f. This is the equation of motion for the relative position r. The reduced mass μ is defined as 1/μ = (1/m1 + 1/m2), or μ = m1 m2/( m1 + m2). It is less than either of the masses, and if the masses are equal, it is half. Study of the relative motion is quite convenient, especially if the force depends only on the relative separation r of the two masses. Then, μd2r/dt2 = f(r), in which the angular and r variables are easy to separate. Orbital motion is analyzed in this way.

The center of gravity lies on the line joining the masses. m1 is a distance r' = [m2/(m1 + m2)]r from the center of mass. The equation of motion for motion relative to the fixed center of mass is then m1d2r'/dt2 = f([(m1 + m2)/m2]r). The mass is the same, but the force must be corrected for the different distance. The result is the same as when the relative motion is considered, but the two masses describe similar trajectories about the center of mass.

Often m1 is projected with velocity v onto m2 at rest. The center of mass then moves with velocity v' = [m1/( m1 + m2)]v. The kinetic energy E in the laboratory system is E = m1v2/2, while in the CM system it is E' = m1(v - v')2/2 + m2v'2/2 = μv2/2. E' is smaller than E in the ratio μ/m1 = m2/( m1 + m2). If m2 is much larger than m1, then E' is about equal to E. If the two masses are equal, then E' = E/2. Finally, if m2 is much less than m1, E' is close to zero. E' is the maximum energy available for causing an endothermic reaction during the collision, a matter of considerable importance in nuclear collisions. All this depends on the target's being at rest. If it is in motion, then an appropriate analysis should be made. For example, if two equal masses are projected in opposite directions with the same speed, then E' is the sum of their kinetic energies in the laboratory system.

Elastic and Inelastic Collisions

Collisions are usefully distinguished as elastic and inelastic. In an elastic collision, total kinetic energy is conserved. This is very common and exact in quantum-mechanical collisions, but is only approximate in classical collisions. In an elastic collision of a light particle with a heavy one, most people seem to think that this means that the velocity of the light particle is only changed in direction, but not in magnitude, so its kinetic energy is conserved. This, however, is not true when the heavy particle is in motion, in which case the velocity of the light particle is increased or decreased up to a maximum equal to the velocity of the heavy particle. A tiny fractional change in the kinetic energy of the heavy particle can cause a considerable change in the kinetic energy of the light particle.

The collision problem is solved most conveniently by transforming to the center of mass system. In this system, the result will be that the velocities have changed direction by some angle Θ, which is the sole result of an elastic collision in the CM system. Afterwards, transformation back to the laboratory system gives the desired solution.

In an inelastic collision, there is a decrease or increase in total kinetic energy that comes from the internal energy of the colliding partners. This may be rotation or vibration, or a change in structure, or even the disappearance of one particle. If the kinetic energy before collision is E, after the collision it is E' = E + Q, where Q is the decrease in energy of the internal coordinates. In the CM system, the final momentum must be zero, so the final velocities are in the ratio of the masses, and can be found from E'. In either case, the description of the collision is much simpler in the CM system, and the final velocities can be determined by the conservation of energy and momentum, and the CM scattering angle Θ. Finally, the original velocity of the center of mass is added to all velocities to find the result of the collision in the laboratory system.

The diagrams usually drawn in connection with the analysis of a collision are hodograms, not trajectories. They show all velocities relative to a common origin, which is not the origin of our reference systems. The colliding particles in the CM system do not, in general, collide head-on, as so often seen depicted in diagrams. Classically, we can identify an impact parameter p, which is the distance of closest approach if the particles did not interact, and moved in straight lines. This distance is not changed by transformation to the CM system, which affects velocities only. In quantum mechanics, the role of the impact parameter is assumed by the relative angular momentum of the particles.

Nuclear Examples

An atomic nucleus is very small. The nuclear radius is well-defined, and is about R = (1.5 x 10-13)A1/3 cm, where A is the mass number, the total number of nucleons in the nucleus, and close to the atomic weight. The cross-sectional area of the nucleus is πR2 = 7.07 x 10-26A2/3. For Al27, this is 5.88 x 10-25 cm2. For Cd113, it is 1.65 x 10-24 cm2, and for Au197 it is 2.40 x 10-24 cm2. Nuclear cross sections are typically around 10-24 cm2, so this area is called a barn ("big as a barn") and is used to specify cross sections. Cross sections run from zero to several hundred thousand barns, but a few barns is typical.

A nucleus of atomic number Z has a charge of Ze C, and strongly repels any other positively charged particle, such as a proton. It is very difficult to get a proton to react with a nucleus except at very high energies, since it cannot get close enough for the nuclear forces to act. The Coulomb scattering, however, is very easy to see. Although it was used to establish the existence of the small nucleus, not much can be learned about the nucleus from it. It is very different for the uncharged neutron, which can get close enough to interact strongly with the nucleons in a nucleus by means of the strong, but short-range, nuclear force.

If Al27 is bombarded with slow neutrons, the most common reaction by far is elastic scattering, or an (n,n) reaction. The cross section is about four times the geometrical cross-section of the Al nucleus, about 2-3 b. This result is expected on the basis of quantum mechanics, as will be shown later. It also shows that the nucleus behaves more or less like a rigid sphere with respect to the neutron, which finds no place there and just bounces off. At certain energies, however, the neutron can dance with the neutrons and protons in the Al nucleus to form what is almost, but not quite, a stationary state, called the compound nucleus. Around these energies, the cross section is much higher, up to 12 b, and makes a peak in the plot of the cross section, called a resonance. Mostly the neutron eventually leaves again elastically. However, about one in a thousand times a photon is emitted, carrying off the energy the neutron added, and leaving the neutron bound in an Al28 nucleus. This is called radiative capture, a (n,γ) reaction. Al28 is radioactive with a half-life of 2.3 minutes, emitting an electron to form stable Si28, the most common isotope of silicon. The 2.87 MeV beta ray and 1.78 MeV gamma ray can easily be detected to give evidence that the reaction has occurred.

Cadmium isotope Cd113 represents 12.26% of natural cadmium, and has a thermal neutron absorption cross section of 20,000 b. It emits a gamma ray in the (n,γ) reaction to form stable Cd114, the most common isotope. Cadmium rods are used to control nuclear reactors, since they drink up neutrons and lower the reactivity. Practically every neutron that hits them is absorbed. When the Cd113 at the surface is depleted, the neutrons can then penetrate deeper, always finding Cd113 at some depth.

Au197 is the only naturally-occurring isotope of gold. Its cross section for the (n,γ) reaction is 98 b. When it absorbs a neutron, it becomes an excited state of Au198, which emits a gamma ray promptly, then decays by beta emission with a half-life of 64.8 hr to Hg198, a stable isotope of mercury. This is an excellent reaction for determining neutron flux, since the reaction is probable, and the half-life is convenient for measurement. A foil is exposed to the neutron flux for a certain time, then the activity is measured and extrapolated back to the end of the exposure.

Classical Scattering of Hard Spheres

This problem is interesting to compare with the quantum-mechanical results we shall derive later. An example often adduced is that of billiards, but the collision of billiard balls is really quite a complex problem, since billiard balls rotate, and roll on a rough surface. We consider an ideal billiard ball whose surface is ideally smooth, so the ball is not made to rotate. Suppose the balls have radius s/2. Let p be the impact parameter, the distance between the parallel paths of two balls in the CM system. When p = 0 the balls collide head-on. When p = s the balls just graze one another, and the scattering angle is zero. The scattering angle θ will be a function of the impact parameter p, varying from 0 when p = s to 180° when p = 0.

Conditions are as shown in the figure at the right. The centre of mass is located at the middle of the line joining the centres of the two balls when they collide, at the point of collision. It is equivalent to think of a point particle bouncing off of a sphere of radius s, called the collision diameter. The angle α is given by p = s sin α, then the angle of scattering θ = π - 2θ. This is the relation between scattering angle and impact parameter.

The area πp2 is the area within which a collision will result in scattering through an angle greater than θ. By differentiation, we can find the dp corresponding to scattering between θ and θ + dθ. The solid angle corresponding is dω = 2π sinθ dθ. The area corresponding to dp is dA = 2πpdp = (π/2)s2sinθ dθ. The ratio |dA/d&omega)| is the probability for scattering into unit solid angle, called the differential cross section. The ratio is s2/4. This is independent of angle, so the scattering is isotropic in the CM system. The total cross section is obtained by integrating over all solid angles, or σ = 4π x (s2/4) = πs2, which we certainly know must be the case. When we add the velocity of the CM, the scattering will become nonisotropic in the laboratory system, favoring the forward direction.

Quantum Mechanics of Scattering

Setting Up the Time-Independent Problem

At first sight, the scattering problem would appear to be a very difficult time-dependent problem involving the coordinates of two particles that initially are separated wave packets, then they move into the same region and interact, and finally separate into two wave packets moving relatively in a new direction. This is, indeed, a very difficult problem, but it is possible to cast the problem into a time-independent, stationary-state problem that is much easier to solve. In this view, we consider a state of one particle corresponding to a plane wave at large distances that interacts with a fixed potential at the origin. Outgoing waves are identified, and they represent the scattered particle. The process is conceived to be a steady one, which is very well represented by the usual interpretations of the wave function. The coordinate system is that described in connection with the CM system, in which a particle of mass m moves in the field of a potential at the origin. The potential must be corrected for the difference between the distance from the origin and the relative distance of the two particles (but this is seldom mentioned).

The wave function ψ(r) is the solution of the Schrödinger equation (-h'2/2m)div grad ψ + V(r')ψ = E'ψ, where V(r') is the potential causing the scattering, and E' is the total energy in the CM system. It must satisfy the usual conditions of continuity and integrability appropriate to such solutions. If z is the coordinate in the direction of the initial velocity through the center of mass, then as z → -∞ ψ must look like eikz, a wave function normalized to unity in a cube 1 cm on a side, and corresponding to a momentum p = h'k in the + z direction. The corresponding flux is h'k/m particles per cm2 per second.

To this we must add a term representing outgoing spherical waves at a large distance, such as f(θ,φ)eikr/r, which depends on the spherical coordinate angles θ and φ. The flux passing through a small area dA of solid angle dω = dA/r2 will then be h'k/m times |f(θ,φ)|2 dω Comparing incident and scattered fluxes, σ(θ,φ) = |f(θ,φ)|2. The differential cross section can be found from the solution of the wave equation that behaves asymptotically as eikz + f(θ,φ)eikr/r. This is really an extraordinary simplification of a difficult problem. Note carefully that this is the asymptotic form of the solution, not the actual solution for any particular case.

The Method of Partial Waves

Now we go back to the Schrödinger equation and see if we can find the actual form of the solutions. First of all, we write ψ = R(r)Y(θ,φ), and separate the radial and angular dependence. Y(θ,φ) is a linear combination of spherical harmonics Ylm(θ,φ). Ylm corresponds to a total angular momentum L of L2 = l(l + 1)h'2 and z-component Lz = mh', with z ranging from -l to l. Since V(r) does not depend on the angles, we can set m = 0, and use only the Yl0, which are proportional to the Legendre polynomials Pl(cosθ). We'll consider, then, solutions of the form Rl(r)Pl(cosθ). Rl satisfies the radial equation, which in the absence of a scattering potential V(r) is exactly that arising in the solution of the three-dimensional wave equation, and in particular the Helmholtz equation (div grad + k2)ψ = 0. This is a form of Bessel's equation, of which the solutions are linear combinations of Jl + 1/2(kr)/r1/2 and Nl + 1/2(kr)/r1/2. Multiplied by √(π/2), these are the spherical Bessel functions jl(kr) and nl(kr). jl is finite at r = 0, but nl blows up there. At large kr, they behave like (1/kr) sin(kr - lπ/2) and -(1/kr) cos(kr - lπ/2). The general radial function for V = 0 is Ajl(kr) + Bnl(kr), where A and B are arbitrary constants.

By substitution, we can find that eikz = eikrcos(θ) is a solution of the Schrödinger equation for V = 0, and must have this form. Since it is finite at the origin, it must be ΣAljl(kr)Pl(cosθ). To find the constants Al, expand both sides in powers of kr cosθ and equate the coefficients of equal powers. This is easy to do using the expansion of jl(x) in powers of x, xl/(2l + 1)!!, and noting that the coefficient of coslθ in Pl is (2l)!/2ll!. The result is Al = (2l + 1)il, so that finally eikz = Σ (2l + 1)iljl(kr)Pl(cosθ).

Now let's see what the radial function Rl(kr) looks like when V ≠ 0, but we are far enough away so that the scattering potential has no influence. It might seem that this is always possible, but it is not. In the case of the Coulomb potential 1/r, there is an effect on the wave function even at large distances, and this must receive special consideration. However, let' put such cases aside, and study potentials of limited range, where there is an asymptotic region where V(r) has no effect. In this region, then, Rl(kr) = Al[cos δl jl(kr) - sin δl nl(kr)]. We have represented the two constants in terms of other constants δl, called the phase shifts, and an amplitude Al. The asymptotic form of Rl is then (Al/kr) sin(kr - lπ/2 + δl).

Now we are lucky enough to have two asymptotic forms. One, eikz + f(θ)eikr/r, defines the scattering amplitude f(θ) from which we can find the differential cross section, and the other, (Al/kr) sin(kr - lπ/2 + δl) expressed in terms of phase shifts that we can calculate in terms of V(r). By comparing the two asymptotic forms, we can find the scattering amplitude f(θ) as well as the coefficients Al. Simply set the two expressions equal, using the expansion of eikz in terms of the jl, and equate coefficients of corresponding terms. The algebra I shall leave for the pleasure of the reader. It is not difficult, using the expansion of the sine in terms of complex exponentials. The results are Al = el/k and f(θ) = (1/2ik)Σ(2l + 1)(e2iδl - 1)Pl(cosθ).

Since σ(θ) = |f(θ)|2, we can integrate over dω to find the total cross section: σ = (4π/k)Σ(2l + 1)sin2δl. Moreover, f(θ = 0) = (1/2ik)Σ(2l + 1)(e2iδl - 1). The imaginary part, Im f(0) = (1/k)Σ(2l + 1)sin2δl = kσ/4&pi. This can be written σ = (4&pi/k) Im f(0), a special case of the optical theorem. This is a consequence of the removal of particles from the beam due to scattering, which decreases the intensity of the beam at θ = 0. It turns out to be quite a general result, and gives the total cross section in most cases.

This method of solving the scattering problem is called the method of partial waves. It is particularly applicable to atomic and sub-atomic scattering, when only a few values of l contribute. It is not convenient for the classical limit, when there are many partial waves. To use it, we must be able to find the phase shifts in terms of the scattering potential.

Finding The Phase Shifts

Let's consider the problem analogous to the classical scattering of hard spheres, which we solved above. We suppose V(r) → ∞ at a distance a from the origin. (We assume this has been corrected for the relative masses). For r > a, Rl(kr) is just the V = 0 solution in terms of jl and nl. The boundary condition is Rl(ka) = 0, so cosδljl(ka) = sinδlnl(ka), or tanδl = jl(ka)/nl(ka). For l = 0, we can use the explicit closed forms for j and n to find that tanδ0 = -sin(ka)/cos(ka) = - tan(ka), or δ0 = -ka. For ka << 1, we can use the expansions of j and n for small x to find tanδ1 = -(ka)3, tanδ2 = -(ka)5/45, and so on. It is clear that these are not only small, but approach zero as ka → 0. However, the 0th partial wave does give a finite contribution, σ = 4πa2, because of the k2. The total cross section, therefore, approaches four times the geometric cross section πa2. This prediction has been amply confirmed by experiment. Scattering is an excellent proof of the wave nature of matter.

Now we turn to a more general problem. The radial equation can be integrated for each value of l that is of interest, numerically if necessary, starting at r = 0 and proceeding outward through the region in which V(r) ≠ 0. This equation is (1/r2)(d/dr)(r2dR/dr) + {2m[E - V(r)]/h'2 - l(l + 1)/r2}R = 0. (As usual, h' = h/2π) Eventually, we will reach a point where V(r) can be considered zero and no longer affects the solution. Let's call this r = a. Now we have, on one hand, our solution coming out from r = 0, and on the other, the general form of the radial function when V = 0, Rl(kr) = Al[cos δl jl(kr) - sin δl nl(kr)]. Note that this is not the asymptotic solution at this point, but the phase shifts are included in the coefficients. We must fit the two solutions together at r = a to find the asymptotic behavior of the wave function. The value and slope must be continuous at r = a, the usual joining conditions for wave functions.

We are not really interested in the value of Al, only the phase shift. If u(r) is a wave function of the form of Rl, then the value of (1/u)du/dr will be continuous at a point where wave functions are joined, since both u and du/dr are continuous. However, this expression does not depend on Al. The continuity of (1/u)du/dr, the logarithmic derivative, should then give us the phase shifts without involving Al. Then we can determine Al from, say, the continuity of the wave function. We need not go this far, since we already have the phase shifts, which is what we desire.

Let γl be the value of the ratio (1/u)du/dr at r = a for the wave function for r < a. Then, the continuity of the logarithmic derivative gives k[cos δl jl'(ka) - sin δl nl'(ka)]/[cos δl jl(ka) - sin δl nl(ka)] = γl, where the ' indicates the derivative with respect to the argument. Dividing numerator and denominator by cos δl and solving for tan δl, we determine the phase shifts in terms of the γl. The result is tan δl = [kjl'(ka) - γljl(ka)]/[knl'(ka) - γlnl(ka)]. This is all we need to find the cross section. It is now possible to solve a great number of scattering problems, and indeed that has been done.

Ramsauer Scattering

An excellent example of particle scattering theory is offered by the scattering of low-energy electrons by atoms and molecules. The speed of an electron accelerated through a potential of V volt is v = √(2eV/m) = 5.931 x 107 √V cm/s. A 1 keV electron moves at 1.88 x 109 cm/s, or 0.06c. The relativity factor 1/√[1 - (v/c)2] = 1.0020, so an electron of this energy is certainly nonrelativistic, and the formula for the speed that we have given is valid. The rest energy mc2 of an electron is 0.51 MeV, so energies of a few hundred keV are necessary before relativity must be taken into account. Let's call electrons of an energy of 1 keV or lower "low energy" electrons.

Atoms are the usual target, since their spherical symmetry makes theoretical treatment easier. The Bohr radius ao = 0.0529 nm is a typical atomic distance, and atoms have a diameter on the order of 2ao = 0.106 nm. An atom has a small nucleus whose radius is on the order of 10-13 cm, and whose charge is +Ze. This charge is neutralized by the electrons, so there is no electric field outside the atom. Inside, however, there is a field that increases as the nucleus is approached. This field gives rise to a scattering potential like V(r) = -Ze2(1/r - 1/ro) for r < ro. This means that the interior of an atom attracts electrons rather strongly. An extra electron can even form a stable negative ion in many cases.

Several outcomes are possible when a low-energy electron is incident on an atom. First, and most probably, it can be elastically scattered by the attractive potential. If its energy is just right, it may knock the atom into a higher stationary state, losing the corresponding amount of kinetic energy. Finally, it can eject an electron from the atom, producing a positive ion. The lowest excitation and ionization energies are in the range of a few eV. Electrons, therefore, are attractive probes for investigating the structure of atoms.

Most of this was already obvious in 1920, when the Bohr-Sommerfeld atom was the best model then available, and electrons had been studied for two decades, but wave mechanics was not yet known. Ramsauer thought it would be interesting to study the scattering of electrons by noble gas atoms. Atoms collide very much like hard spheres, or perhaps fuzzy hard spheres, and these collisions had been studied in kinetic theory since Maxwell. The sizes of the atoms were rather well known, therefore. Argon had a radius of 0.191 nm, so its classical collision cross section was 1.146 x 10-15 cm2. From this, the mean free path 1/√2Nσ could be calculated. The mean free path of an electron can be estimated at 4√2 times this, the electron being lighter and moving so rapidly the the atoms could be regarded as still. We'll see that this reasoning is fallacious.

Ramsauer's apparatus is shown schematically at the right. It was quite small in size, and evacuated by a diffusion pump to a low pressure, so the electron mean free path would be much greater than the dimensions of the apparatus. The chamber in the lower left-hand corner was fed with argon gas through a capillary, so the argon pressure in this region could be controlled. Free electrons were produced at a photocathode, illuminated from the outside. By controlling the wavelength of the light, the energy of the emitted electrons could be made very small. The electrons were then accelerated through a potential difference between the photocathode and the first slit. The body of the apparatus was grounded, and the photocathode made negative the required amount.

It was necessary to control the electron energy much more closely than could be done by simply changing the accelerating potential, so a magnetic velocity selector was used. The whole apparatus was placed in a uniform magnetic field directed toward the observer in the diagram. Only a rather weak field was required, and the earth's field had to be compensated for. An electron of speed v describes a circular trajectory in a uniform magnetic field, whose radius is given by mv2/r = Bev in SI units. The velocity v = (e/m)Br. A 1 V electron has v = 5.931 x 105 m/s, so Br = 3.372 x 10-6 T-m, or in kinder units, 3.372 gauss-cm. The diameter of the circle would be about 6.7 cm for a field of 1 gauss. 100-V electrons would then require B = 10 gauss. This is all quite reasonable. A nearly monoenergetic beam enters the scattering volume. Electrons that are scattered along the approximate quadrant of a circle are collected by the walls of this volume, and their current can be measured with an electrometer. Those that are not scattered continue into the final chamber, where they are collected and their current measured. The sum of the currents is the initial beam intensity, and the current from the final chamber is the final beam intensity. The attenuation coefficient μ = (1/L)ln[(I1 + I2)/I2, where L is the length of the arc, and I1,I2 are the measured currents. Then, the electron mean free path is 1/μ, and from the density of the argon, the total cross section can be determined.

Ramsauer must have been astonished by the results. At the lowest energies he could study, the cross section was small, and at about 0.7 V it all but vanished! After that it rose to a maximum of 25 (in units of πao2 = 0.88 x 10-16 cm2) at around 12 V, then declined steadily as the voltage increased. All the noble gases gave similar results. A constant value equal to the geometrical cross-section of the argon atom was expected, but was definitely not observed. When the differential cross-section was investigated, it showed a strong dependence on angle, with peaks and valleys that changed with the electron energy, not the expected isotropic scattering. There was no relationship to the gas-kinetic cross section. An argon atom appeared very differently to an electron than to another argon atom. The Ramsauer Effect was a complete mystery.

Quantum mechanics, which arrived in the course of the next decade, explained everything. The deBroglie wavelength of an electron is given by λ = 1.226/√V nm. This is easily found from p = mv = h'k, where k = 2π/λ. For low energies, the deBroglie wavelength of the electrons is comparable to the sizes of atoms, so wave effects can be expected. The scattering potential could be the one given above, or Morse's (Ze2/r)exp(-2r/ro). The size parameter ro can be taken as the radius at which the electron density is a maximum. For argon, ro = 1.3ao. Analytic or numerical methods lead to the phase shifts, and from them the cross section is found. At low energies, the l = 0 partial wave is the most important. The strong attractive potential "sucks in" the wave function enough that the phase shift becomes π radians. Then, σ = (4π/k)sin2δo = 0! This cannot occur for a repulsive interaction, since the phase shift cannot rise to such a value without bringing in contributions from higher partial waves. In this case, ka is small enough that the contributions of higher partial waves is small. The observed cross sections can be very well explained along these lines.

The Ramsauer Effect is very strong evidence for the wave nature of the electron, and one of the earliest. In 1923, Davisson and Germer published their first results on the reflection of 50-V electrons from the surface of a nickel crystal, showing that they interfered like X-rays of a similar wavelength. This is a more difficult experiment, since the crystal surface must be extremely clean, but is clear evidence of wave properties that is easy to understand through the analogy with X-rays. Electron diffraction has become an important tool in surface studies. Textbooks tend to make much of the Davisson and Germer result, but fail to mention the Ramsauer Effect at all. Townsend and Bailey published similar observations a year later, so Townsend's name is tacked on by many people so an Englishman would figure in the discovery. Bailey probably did the work. One thing we learn from all this is that electron mean free paths cannot be simply deduced from the gas-kinetic cross sections, as implied in so many sources, and that they can vary widely.

References

L. I. Schiff, Quantum Mechanics, 3rd ed. (New York: McGraw-Hill, 1968). Chapter 5.

N. F. Mott and H. S. W. Massey, The Theory of Atomic Collisions, 3rd ed. (Oxford: Clarendon Press, 1965). A large and comprehensive treatise.

I. Kaplan, Nuclear Physics (Cambridge, MA: Addison-Wesley, 1955). Chapter 16, Nuclear Reactions.


Return to Physics Index

Composed by J. B. Calvert
Created 15 May 2003
Last revised