1.2 Rutherford Scattering

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1 Rutherford Scattering Rutherford Scattering The series of measurements performed by Hans Geiger and Ernest Marsden under Rutherford's direction at Manchester provide a classic example of a "fixed target" experiment. The target was a thin metal foil of relatively large atomic number, while the projectiles consisted of a collimated beam of low energy a-particles, which, as we will see in the next chapter, are nothing more than the nuclei of helium atoms. The basic outcome of these experiments was that most of the a-particles went straight through the foil with very little angular deviation. Occasionally, however, the deflections were quite large. A detailed analysis of these observations revealed the structure of the target, which ultimately led to the nuclear model of the atom. To fully appreciate the beauty of these experiments, it is essential to analyze the results in their proper historical context. Prior to this work, the only popular model of the atom was due to Joseph Thomson, who visualized the electrically neutral atom as a "plum pudding" where negatively charged electrons were embedded, like raisins, within a uniform distribution of positive charge. If this model were correct, one would expect only small deviations in the a-particles' trajectories (primarily due to scattering from the electrons), unlike what was found by Geiger and Marsden. To see this, let us do a few simple kinematic calculations. Because the velocities of the a-particles in these experiments were well below 0.1c (where c refers to the speed of light), we will ignore relativistic effects. Let us assume that an a-particle with mass ma and initial velocity vo collides head-on with a target particle of mass mt, which is initially at rest (see Fig. 1.1). After the collision, both particles move with respective velocities va and vt- Assuming that the collision is elastic (namely, that no kinetic energy is converted or lost in the process), momentum and energy conservation yield the following relations. Momentum conservation: mav0 mava + mtvt, or v0 = va H vt. (1.1)

2 4 Nuclear and ^ m,, w, Fig. 1.1 Collision of a particle of mass ma and velocity vo with a target particle of mass mt. Energy conservation: - mavl = - mavl + - mtv\, or vl = vl + ^ v l (1.2) where we have labeled (Hi)2 = Vi-Vi as vf, for i = 0, a and i. Squaring the relation in Eq. (1.1) and comparing with Eq. (1.2), we obtain or»,2(l-^)=2i?a.s1. (1.3) It is clear from this analysis that, if mt <C ma, then the left hand side of Eq. (1.3) is positive and, consequently, from the right hand side we conclude that the motion of the a-particle and the target must be essentially along the incident direction. In other words, in such a case, one would expect only small deviations in the trajectory of the a-particle. On the other hand, if mt» ma, then the left hand side of Eq. (1.3) is negative, which implies large angles between the trajectories of the a-particle and the recoiling nucleus, or large-angle scattering. To get a feeling for the magnitude of the numbers, let us recall that the masses of the electron and the a-particle have the following approximate values

3 Rutherford Scattering 5 me «0.5MeV/c2, ma «4 x 103 MeV/c2. (1.4) Therefore, if we identify then, m( = me, TTl a 10~4. (1.5) Now, from Eq. (1.3) it follows that ve = vt < 2va, and then Eq. (1.2) yields va & v0. Therefore, meve = ma ^f- ve < 2 x 10~4 mava «2 x 10~4 mavo, and the magnitude of the momentum transfer to the electron target is therefore < 10~4 of the incident momentum. Consequently, the change in the momentum of the a-particle is quite small and, in the framework of the "plum pudding" model of the atom, we would expect only slight deviations in the a-trajectory after scattering from atomic electrons; thus, the outcome of the experiments, namely the occasional scatters through large angles, would pose a serious puzzle. On the other hand, if we accept the nuclear model, wherein the atom has a positively charged core (the nucleus) containing most of the mass of the atom, and electrons moving around it, then the experimental observations would follow quite naturally. For example, setting the mass of the target to that of the gold nucleus yields mt = mau «2 x 105 MeV/c2, (1.6) i«50. (1.7) ma A simple analysis of Eq. (1.3) gives vt < 2m7^v"., and from Eq. (1.2) we again obtain that va «VQ- Therefore, mtvt < 2mava ftj 2mavo. This means that the nucleus can carry away up to twice the incident momentum, which implies that the a-particle can recoil backwards with a momentum essentially equal and opposite to its initial value. Such large momentum

4 6 Nuclear and Particle Physics transfers to the nucleus can, therefore, provide large scattering angles. Consequently, in the Rutherford picture, we would expect those a-particles that scatter off the atomic electrons in gold to have only small-angle deflections in their trajectories, while the a-particles that occasionally scatter off the massive nuclear centers to suffer large angular deviations. The analysis of the scattering process, however, is not this straightforward, and this is simply because we have completely ignored the forces involved in the problem.1 We know that a particle with charge Ze produces a Coulomb potential of the form U[f) = ^. (1.8) We also know that two electrically charged particles separated by a distance r = \f\ experience a Coulomb force giving rise to a potential energy V(r) = ^ -. (1.9) Here Ze and Z'e are the charges of the two particles. An important point to note about the Coulomb force is that it is conservative and central. A force is said to be conservative if it can be related to the potential energy through a gradient, namely and it is denned to be central if F{r) = -V^(r), (1.10) V(f) = V(\f\) = V(r). (1.11) In other words, the potential energy associated with a central force depends only on the distance between the particles and not on their angular coordinates. Because the description of scattering in a central potential is no more complicated than that in a Coulomb potential, we will first discuss the general case. Let us consider the classical scattering of a particle from a fixed center. We will assume that the particle is incident along the z-axis with an initial xwe have also tacitly assumed, in the context of the Thomson model, that contributions to large-angle scattering from the diffuse positively charged nuclear matter can be ignored. This is, in fact, the case, as discussed by Thomson in his historic paper.

5 Rutherford Scattering 7 velocity vo- (It is worth noting that, outside the foil, the incident and the outgoing trajectories are essentially straight lines, and that all the deflection occurs at close distances of the order of atomic dimensions, where the interaction is most intense.) If we assume that the potential (force) falls off at infinity, then conservation of energy would imply that the total energy equals the initial energy E = - mvl = constant > 0. (1-12) Equivalently, we can relate the incident velocity to the total energy v0 = \. (1.13) V m Let us describe the motion of the particle using spherical coordinates with the fixed center as the origin (see Fig. 1.2). If r denotes the radial coordinate of the incident particle, and \ the angle with respect to the z-axis, then the potential (being central) would be independent of x- Consequently, the angular momentum will be a constant during the entire motion. (That is, since r and F are collinear, the torque r x F vanishes, and the angular momentum r x mv cannot change.) For the incident particle, the angular momentum is clearly perpendicular to the plane of motion and has a magnitude = mvob, where b is known as the impact parameter. The impact parameter represents the transverse distance that the incident particle would fly by the source if there was no force acting. Using Eq. (1.13), we can obtain the following relation [2E I = m\ b = b v2me, V m 1 2mE or.fc2 =~p-- (L14) From its definition, the angular momentum can also be related to the angular frequency, x, as follows r + r-f- x ) I = mr2 - = mr2x, (1-15) at at J at

6 8 Nuclear and Particle Physics L i_il_r^ra ^z Fig. 1.2 The scattering of a particle of mass m, with initial (asymptotic) velocity vo, from a center of force at the origin. where, as usual, we have defined a unit vector x perpendicular to r = rf, with v(r) = rf + rxx expressed in terms of a radial and an angular component of the velocity, and the dot above a variable stands for differentiation with respect to time. Equation (1.15) can be rewritten as (1.16) at mr2 The energy is identical at every point of the trajectory, and can be written as *-H )>+Mt)"+ - Hi)'=*->?-"< > or *._f»(js_v(r)-5ft)]i. (I.X7) dt [m \ 2mr2) J The term ^~s is referred to as the centrifugal barrier, which for I ^ 0 can be considered as a repulsive contribution to an overall effective potential yeff(r) = V{r) + 2^J- Both positive and negative roots are allowed in Eq. (1.17), but we have chosen the negative root because the radial coordinate decreases with time until the point of closest approach, and that is the time

7 Rutherford Scattering 9 domain we will be examining.2 Rearranging the factors in Eq. (1.17) and using Eq. (1.15), we obtain dr \2 P_ f 2mEr2 / _ V )\ _ 11 * dt ~ [m 2mr2 \ (? \ E ) J J --ilh1- )-*}'- (1.18) Prom Eqs. (1.16) and (1.18), we now obtain A l A* l dl A d-x = 2 dt = _- dr I dr or dx = r- (1-19) r[,2(l--m)-6f Integrating this between the initial point and the point of closest approach, we obtain 2 The motion is completely symmetric about the point of closest approach (r = ro), and consequently the positive and negative roots provide identical information. In fact, if the a-particle approached the target with the velocity vo along the exiting trajectory in Fig. 1.2, it would then emerge on the entering trajectory, with the same asymptotic velocity. A simple way to see that this is true is to imagine the collision as observed from both above and below the plane of scattering shown in Fig Viewed from these two perspectives, the motion in Fig. 1.2 appears as the mirror image of the reversed trajectory. This symmetry is a consequence of time-reversal invariance of the equations of motion, a concept that will be discussed in Chapter 11.

8 10 Nuclear and Particle Physics fx, fro bdr / d\ = - r> Jo Joo r[r2(i_y±riyb2y / dr -r. (1.20). o r^(i-yg.yh2y The point of closest approach is determined by noting that, as the particle approaches from infinity, its velocity decreases continuously (assuming the repulsive potential for the case of an a-particle approaching a nucleus), until the point of closest approach, where the radial velocity (^) vanishes and subsequently changes sign. That is, beyond this point, the velocity of the particle increases again. Therefore, at the distance of closest approach, when r = ro, both the radial and the absolute velocities attain a minimum, and we have - =0 which, from Eqs. (1.17) and (1.18), means that or rg(l-! o>)-* = o. (1.21) Thus, given a specific form of the potential, we can determine r0, and therefore xo> as a function of the impact parameter b.3 Defining the scattering angle 6 as the change in the asymptotic angles of the trajectory, we get r e = n-2xo=7t-2b -r. (1.22) ^ r[r2(l-vjrl)-b*y 3We note that, in general, with i ^ 0 and E > 0, that is, for 6^0, -^ is maximum at r = ro (see Eq. (1.16)). Also, for I ^ 0, even for an attractive Coulomb potential, there will be a finite result for ro as determined from Eq. (1.21). This is because the centrifugal barrier for I 7^ 0 acts as a repulsive potential that dominates over Coulomb attraction at small distances. dr

9 Rutherford Scattering 11 Consequently, given an impact parameter b, and a fixed energy E, the scattering angle of a particle in a potential can, at least in principle, be completely determined. As an application of the general result, let us now return to the scattering of a charged particle from a repulsive Coulomb potential, for which the potential energy is given by Eq. (1.9) V(r) = ^, (1.23) where Z'e represents the charge of the incident particle and Ze the charge of the scattering center. (The scattering of an a-particle from a nucleus would then correspond to Z' = 2, with Ze representing the nuclear charge.) The distance of closest approach can be obtained from Eq. (1.21) 2 ZZ'e2 l2 ro ^ ro-b2= 0, ^*\/(y)2+4fc2 n,. or r0 = 2 (1.24) Since the radial coordinate can by definition only be positive, we conclude that (1.25) Consequently, from Eq. (1.22), we obtain f dr e = ir-2b r. (1.26) Jr0 r[r2(1_l^ i)_62]i Let us define a new variable which gives x=1-, (1.27)

10 12 Nuclear and Particle Physics 1 2E [ 4b2E2 \ (1.28) Prom Eq. (1.27), we obtain dr, dx dx = 7, or dr = -, r x and, in terms of this new variable, we can write «,. f ( dx\ x r dx = 7r + 26/ r. (1.29) Now, using the following result from the integral tables we obtain I, ^ = 4= cos"1 (- P + 2lX ), (1-30),1 i I T + 2*>2x \ 6 = IT + 2b x - cos"1. E b W(V)2+4&v,0 o l( 1 + ^ X \ = 7T + 2 COS"1. ZZ e = Equivalently, we can write = 7r + 2cos"1 (. 1 = I -2cos"1(l) I /i I 4^g2~ / V ^ Vy1 + (ZZ'e!)5 / = 7T + 2cos~1 (. 1 I. (1.31) \V i + (ZZ>e*)*J

11 Rutherford Scattering 13 1 (9 TT\ A, 46^~COSl2~2J' V (ZZ'e2)2 1, /0 7r\., 6» 1 or..,, = cos = sin - = ^,!+#fr V2 2/ 2 cosec2f 26 0 r ZZ^=COt2' or b=-^- cot -. (1.32) This relates the scattering angle, which is a measurable quantity, to the impact parameter which cannot be observed directly. Note that, for fixed b, E and Z\ the scattering angle is larger for a larger value of Z. This is consistent with our intuition in that the Coulomb potential is stronger for larger Z, and leads to a larger deflection. Similarly, for a fixed b, Z and Z', the scattering angle is larger when E is smaller. Qualitatively, we can understand this as follows. When the particle has low energy, its velocity is smaller and, therefore, it spends more time in the potential and suffers a greater amount of scattering. Finally, for fixed Z, Z1 and E, the scattering angle is larger for smaller b. Namely, when the impact parameter is small, the particle feels the force more strongly and hence the deflection is larger. Equation (1.32) therefore incorporates all the qualitative features that we expect of scattering in the Coulomb field. 1.3 Scattering Cross Section As we have seen, the scattering of a particle in a potential is completely determined once we know the impact parameter and the energy of the particle; and, for a fixed incident energy, the deflection is therefore defined by just the impact parameter. To perform an experiment, we prepare an incident flux of beam particles of known energy, and measure the number of particles scattered out of the beam at different 6. Because this number is determined entirely by the impact parameters involved in the collisions,