Rutherford’s formula 53 on returning to the original variables r and P, the following two equivalent forms of the final result: ===== (dx/dp) (4) This formula determines implicitly the function w(r) (and therefore U(r)) for all r > rmin, i.e. in the range of r which can be reached by a scattered particle of given energy E. §19. Rutherford’s formula One of the most important applications of the formulae derived above is to the scattering of charged particles in a Coulomb field. Putting in (18.4) U = a/r and effecting the elementary integration, we obtain whence p2 = (a2/m2voo4) tan2oo, or, putting 00 = 1(-) from (18.1), p2 = (19.1) Differentiating this expression with respect to X and substituting in (18.7) or (18.8) gives do = (a/v2cosxdx/sin31 (19.2) or do = (19.3) This is Rutherford’s formula. It may be noted that the effective cross-section is independent of the sign of a, so that the result is equally valid for repulsive and attractive Coulomb fields. Formula (19.3) gives the effective cross-section in the frame of reference in which the centre of mass of the colliding particles is at rest. The trans- formation to the laboratory system is effected by means of formulae (17.4). For particles initially at rest we substitute X = 77 - 202 in (19.2) and obtain do2 = 2n(a/mvoo2)2 sin de2/cos302 = (19.4) The same transformation for the incident particles leads, in general, to a very complex formula, and we shall merely note two particular cases. If the mass M2 of the scattering particle is large compared with the mass M1 of the scattered particle, then X 2 O1 and m 22 M1, SO that do1 = = (a/4E1)2do1/sin4301 (19.5) where E1 = 1M1U..02 is the energy of the incident particle. 54 Collisions Between Particles § 19 If the masses of the two particles are equal (m1 = M2, m = 1M1), then by (17.9) X = 201, and substitution in (19.2) gives do1 = 2(/E1)2 cos 01 d01/sin³01 = (19.6) If the particles are entirely identical, that which was initially at rest cannot be distinguished after the collision. The total effective cross-section for all particles is obtained by adding do1 and do2, and replacing A1 and O2 by their common value 0: do = do. (19.7) Let us return to the general formula (19.2) and use it to determine the distribution of the scattered particles with respect to the energy lost in the collision. When the masses of the scattered (m1) and scattering (m2) particles are arbitrary, the velocity acquired by the latter is given in terms of the angle of scattering in the C system by V2’ = [2m1/(m1+m2)]%“00 sin 1x; see (17.5). The energy acquired by M2 and lost by M1 is therefore E = 1M2U2’2 = (2m2/m2)/0002 sin21x. Expressing sin 1x in terms of E and substituting in (19.2), we obtain do = de/e2. (19.8) This is the required formula: it gives the effective cross-section as a function of the energy loss E, which takes values from zero to Emax = 2m2voo2/m2. PROBLEMS PROBLEM 1. Find the effective cross-section for scattering in a field U = a/r2 (a > 0). SOLUTION. The angle of deflection is The effective cross-section is do sin PROBLEM 2. Find the effective cross-section for scattering by a spherical”potential well” of radius a and “depth” U0 (i.e. a field with U = 0 for r > a and U = - U0 for r < a). SOLUTION. The particle moves in a straight line which is “refracted” on entering and leav- - ing the well. According to §7, Problem, the angle of incidence a and the angle of refraction B (Fig. 21) are such that sin x/sin B = n, where n = W(1+2U0/mVo2). The angle of deflection is X = 2(a-B). Hence = Eliminating a from this equation and the relation a sin a p, which is evident from the diagram, we find the relation between P and X: cos 1x