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