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Chapter 2. Classical 2-body collision and kinematical relations. We will show later that a collision between two atoms with a total kinetic energy in the keV-MeV region can be described to good approximation with classical theory. Hereby the electronic excitations are not taken into account. At much lower or higher energies quantum-mechanical or relativistics effects have to be taken into account.

Fig. 2.1. Two particles with position vectors r1 and r2 w.r.t. the origin O. Consider the collision between two particles with masses m1 en m2 (see fig. 2.1) with a central interaction potential: r rr (2.1) V (r ) = V (r 2 - r 1) = V (r ) The position of the centre of mass is given by: r r r m1r 1 + m 2 r 2 R= m1 + m 2 We assume there are no external forces or force momenta: r r Next = Fext = 0 (2.2) For this isolated system conseravation laws apply in any inertial system, including the centre of mass system. Total energy: E = constant (elastic collision) (2.3) r r r d (m1r + m 2 r 2 ) = Pcm = constant Momentum: (2.4) dt Angular momentum: (Central field of force) r r rr Ltot = Lcm + RcmxPcm = constant (2.5) r r Here Rcm x Pcm is the impulse-momentum of the centre of gravity with respect to an arbitrary origin (see fig. 2.1) There are no external forces, so: rr v Rcm xPcm = constant L cm = constant (2.6) r r Here Lcm is the angular momentum with respect to the centre of gravity. That Lcm = constant follows also from the fact that the momenta of the particles with respect to the centre of mass = 0: the force r vector passes through the centre of gravity. From Lcm = 0 follows that the trajectories of m1 and m2 are in one plane. 2.1 Collision of a moving particle with a particle at rest. The velocity of the centre of mass is: m1 r r dr r (v2 = 0) R = v cm = v1 dt m1 + m2

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r r Here v 1 en v 2 are the velocities in the laboratory system. The velocities of the particle in the cms before the collision are: m2 r r r v 1 = v1 - v cm = v1 (2.7) m1 + m2 r r m1 r v 2 = v cm = v1 m1 + m2 The sum of E of the kinetic energies in the cm system is: m2 E= E1 (2.8) m1 + m 2 Here E1 = m1 v 1 /2 is the kinetic energy in the laboratory system. The energy E is the energy in the cms. We apply the conservation laws in the cm system: r r | v 1 | = | ~ 1 | | v 2 | = | ~ 2 | v v (2.9) Here ~ indicates the velocity after the collision. From this we get the velocities after the collision by adding the velocity of the centre of mass (see fig. 2.2). Applying the sine-rule r r v1 v 2 = in this figure, gives: sin ( - ) = (m1/m2) sin sin sin ( - ) With sin ( - ) = sin cos - cos sin we get: m1 + cos m2 For m2 >>m1: tg = sin = sin 1 + cos µ with µ = m m
2 1 2

(2.10)

Fig. 2.2. Vector diagram for the derivation of the kinematic elations. - scattering angle in laboratory system (ls); - scattering angle in cm system (cms); - recoil angle cms; - recoil angle ls .

From figure 2.2 we get for the recoil angle: =- =½ = 1 - , 22 or tg = sin 1 - cos (2.11)

From eq. 2.10 it can be derived that for µ < 1 there is a maximum of as a function of .

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Differentiate (2.11): 1 cos + cos + sin 2 µ 1 d = d cos 2 1 cos + µ 2 d 1 sin = = tg 2 + 1 = +1 2 2 d cos 1 + cos µ For = : cos*= - µ , 0<µ<1 /2<< (For µ = 1 is d/d = ½; µ = 0 is a non-physical tg = sin 1 cos + µ = 1- µ2 = 1 -µ+ µ µ 1- µ
2

cos + µ 1 µ cos + µ value of µ) + µ sin 2
2

=

cos + µ =0 1 2 cos + + µ µ (2.12)

0

(1 - µ2)1/2 As can be seen from the triangle: sin = µ 0 < < /2

(2.13)

Fig.2.3. The scattering angle (ls) as a function of the scattering angle (cm) for various values of µ. For cos = -µ we have d2/d2 < 0, so for µ<1 () has a maximum, as indicated in fig. 2.3. From this figure we see:

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For each < * and µ < 1 there are two cms scattering angles 1 and 2, with recoil angles 1 = /2 ½1 and 2 = /2 - ½2 (see eq. 2.11), and therefore two energies E1 and E2 (ls). For 1 and 2 we get:

sin (tg)2 = 1 + cos µ

2

and

- sin 2 ± cos µ 2 - sin 2 (2.14) µ For µ > 1 the - sign has no meaning because it leads to a value < -1 for cos . For µ < 1 ± and for µ 1 + ~ ~ The energies E1 and E 2 can be calculated from the cm velocities after the collision. Applying the cosine rule in fig. 2.2 we get: ~ ~ ~ ~~ (v1)2 = (v1L )2 + (v2 )2 - 2v1L v2 cos ~ ~ ~ ~ With v = v ; v = v we get: sin2 = (1 ­ cos2 ) cos =
1 1L 2L 2 2 2 ~2 ~ v1L v1L m1 m2 m1 cos + =0 -2 v1 m1 + m2 v1 m1 + m2 m1 - m2



~ v1L 1 = cos ± µ 2 - sin 2 v1 µ +1

[

]
(2.15)

2 ~ E cos ± µ 2 - sin 2 = ± for µ < 1 and < and + for µ 1 µ +1 E1 For µ >> 1 (m2 >> m1) this relation can be approximated by: 2 ~ E1 cos + µ = E1 µ + 1 We use this relation to see under which conditions the energy dispersion is large m2 and for which scattering angle . This is important to know if we want to scattering from atoms of different masses. We require:

(2.16) as a function of mass discriminate between

~ ~ d E1 d v1L maximal maximal dm2 dm2 ~ ~ d v1L d v1L dµ v d cos + µ v1 1 - cos m1 = 1 µ + 1 m (µ + 1)2 v1 m 2 (1 - cos dm2 dµ dm2 m1 dµ 1 2 mass separation for µ >> 1 increases with m1 decreases with m2 increases with v1 has a maximum for = 180°

)

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For the recoiling particle we have: = /2 ­ ½ 0 (see eq. 2.11) 90° The recoiling particle is in the ls always emitted to forward angles. For the recoil angle we get: cos = - cos 2 = 2 sin2 - 1

- sin 2 ) ± cos µ 2 - sin 2 ± if µ < 1 ; + if µ 1 (2.17) sin = 2µ Of course for µ < 1 again one scattering angle corresponds to two recoil angles. In fig. 2.4 is plotted as a function of . Also for µ < 1 the values of and are plotted as a function of µ. As can be seen in the figure, for the critical scattering angle the function = f () has a vertical asymptote: 1 d d d d d for cos = - µ and µ < 1 = =d d d d 2 d The recoil angle corresponding to the critical angle is given by: 1- µ (µ < 1) (2.18) sin = 2

(µ

Fig. 2.4. The recoil angle (ls) as a function of the scattering angle (ls) for different values of µ. For µ < 1 there is a vertical asymptote and for each (allowed) value of there are two corresponding values of . Also for µ < 1 * en * are plotted as a function of µ.

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To calculate the energy Er (ls) of the recoiling particle we use again fig. 2.2: 1~ v2 2 = v cos = 2 Er = m1 v1 cos m1 + m 2 ~ v2L = 2 v1 cos µ +1

L

4µ cos 2 E1 = Tmax cos 2 (2.19) 2 (1 + µ ) 4m1m2 (2.20) Tmax = E (m1 + m2 )2 1 Here Tmax is the maximum energy that can be transferred to a recoiling particle. This quantity plays an important role, for instance in the description of the stopping of ions in solids. As an example of application of the kinematic relations we consider the bombardment of a Cu surface with 10 keV Ne+ ions. The surface is contaminated with a very small amount (fraction of an atomic mono-layer) of H and C. Particles scattered or recoiled from the surface are detected at a forward angle = 30°. The scattered or recoiled particles are mainly from the outermost surface layers; in a spectrum (see fig.2.5) the signal from deeper layers forms a continuous background. At the angle = 30° we should detect Ne ions scattered from Cu (µ > 1) and recoiling Cu, H and C atoms. For forward angles < /2 recoiling atoms must be visible, independent of their mass. If Ne-atoms scattered from H or C (µ < 1) are visible at this angle depends on the critical scattering angle , for which we have sin = µ. In fig.2.5 a measured time-of-flight spectrum is displayed for this case. Here the yield is plotted as a function of time required for the particle to cover the distance between the Cu surface and the detector. The time scale is inversely proportional with a velocity scale. At certain flight times peaks are visible that can be identified with recoiling H, C, Cu and scattered Ne. This identification can be made using the kinematic relations (see exercise 1).

Fig. 2.5. Time of ­flight spectrum of the bombardment of a Cu-surface with 10 keV Ne+- ions. The surface is contaminated with H and C. The peaks can be identified as follows: A - recoil H, m1 = 20, m2 = 1, µ = 1/20; B - recoil C, m1 = 20, m2 = 12; C - Ne scattering from Cu, m1 = 20, m2 = 63(69%) and 65(31%); D - recoil Cu from a clean surface (measured separately).

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2.2.

Summary of kinematic relations. tg = sin m1 / m2 + cos

Scattered particle ls cm

(m1 < < m2 ) small angles m2 µ = µ +1 m1 + m 2

has a maximum for sin = µ; µ < 1 (µ = m2/m1), < /2 Scattered particle cm ls Recoiled particle ls cm Recoiled particle ls ls ~ E cos ± µ 2 - sin 2 = E1 µ +1
2

cos =

- sin 2 ± cos µ 2 - sin 2 µ

µ<1± µ1+

= (/2-1/2) = 1/2 sin =

/2 µ<1± µ1+ ± µ < 1 and < +1

(µ

- sin 2 ) ± cos µ 2 - sin 2 2µ

Energy scattered particle

For µ > Energy Energy Energy

> 1: dispersion maximal for = 180° dispersion increasing with increasing E1 and m1 dispersion decreasing with increasing m2 T Er 4µ cos 2 = max cos 2 ; = 2 E1 (1 + µ ) E1 4m1m2 Tmax = E (m1 + m2 )2 1

Energy recoiled particle

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