p31318notes01a.pdf - 12 Before proceeding there are two...

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12 Before proceeding, there are two more points of slight interest. (i) The distance of closest approach for zero impact parameter is . 4 0 E Qq πε What is the distance of closest approach for a particle of impact parameter b ? (ii) What is the equation to the envelope of the trajectories - i.e. the red dashed line in the drawing of page 4? We’ll leave that for an assignment. A blank space follows in case you want to add this to your notes
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11 0 1 2 3 4 5 6 7 8 9 10 0 20 40 60 80 100 120 140 160 180 b/d 0 φ degrees ψ φ
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10 . 1 4 2 sin , 1 4 1 cos 2 2 + β β = α + β = α (16) On substitution of this into equation 15, we see, after some algebra and trigonometry, that the polar equation to the trajectories is . 1 sin 2 cos 2 2 - θ β + θ β = ρ (17) This was the equation used to generate figure 2, which is drawn for β = 0 to 1 in steps of 0.1. Of particular interest in the present context is the angle of deviation φ , which is (see drawing below) ψ - π 2 , where ψ is the semi-angle between the asymptotes, given by . 2 / cot tan 2 1 β = = φ = ψ a b Thus, for a given impact parameter, the angle of deviation is given by . cot 2 cot 2 2 1 2 1 1 φ = β β = φ - (18) This is graphed below.
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9 Figure 4 Also of interest is the semiangle between the asymptotes, which is given by 0 2 / 2 1 / tan d b e a b = - = = ψ Introduce 0 d r = ρ and 0 d b = β , and equation (14) becomes 1 ) cos( 1 4 2 2 2 - α - θ + β β = ρ . (15) The angle α is an arbitrary constant of integration, and we can choose its value so that the incoming alpha particle is moving parallel to the initial line of the polar coordinates. That means that we choose α so that ρ is infinite for θ = 0. This requires that
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8 Hence 2 2 u du d - κ = θ (11) Therefore α + κ = θ - ) / ( cos 1 u (12) ). cos( α - θ κ = u (13) Recall that 2 0 2 1 b d r u + = and 2 2 0 2 2 4 b d b + = κ , and we see that the equation is 1 ) cos( 4 2 0 2 0 2 0 2 - α - θ + = d d b d b r (14) We recall that the polar equation to a hyperbola is 1 ) cos( - α - θ = e l r , (15) and so we see that the path is a hyperbola of semi latus rectum 0 2 2 d b l = and eccentricity . 4 0 2 0 2 d d b e + = (16), (17) The angle α is an arbitrary constant of integration, and it is the angle that the axis of the hyperbola makes with the initial line of the polar coordinates. Let us pause to recall some properties of a hyperbola. We normally use the symbol a to represent the length of the semi transverse of the hyperbola , and the symbol b to represent the length of the semi transverse axis if the conjugate hyperbola. This is also equal to the perpendicular distance between a focus and an asymptote - i.e. to the impact parameter. a and b are related by 1 2 - = e a b , and the semi latus rectum is ) 1 ( 2 - = e a l . Combined with equations (16) and (17), this shows that . 2 0 a d = Recall that d 0 is the distance of closest approach for an alpha with zero impact parameter, given by . 4 0 0 E Qq d πε = . That is, all the hyperbolic trajectories in figure 2 have the same length of transverse axis. The length of the transverse axis depends only on the energy of the alpha particles, which is the same for all the alphas from the radon source.
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7 Energy: . ) ( 0 2 2 2 2 1 E r Ed r r m = + θ + & & (4) [Skip down to equation (14) if you like, and don’t bother with the derivation.]
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