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Unformatted text preview: 1 Problem 13.9 Using Jackson's equation 13.50 and the fact that n = p " ( ! ) yields: cos c = 1 n (1) Now, we know that K = ( 1) mc 2 = 1 p 1 2 1 mc 2 . Solving this expression for yields: = p K 2 + 2 Kmc 2 K + mc 2 (2) Plugging this into equation (1) yields: cos c = K + mc 2 n p K 2 + 2 Kmc 2 Jackson's equation 13.48 gives us: dE dx = Z z 2 e 2 c 2 ! 1 1 n 2 2 d! A single energy quantum (i.e., a photon) radiated will have energy ~ ! . Thus, the above equation can be rewritten to express the number of quanta emitted: dN dx = Z z 2 e 2 ~ c 2 1 1 n 2 2 d! = z 2 e 2 ~ c 2 1 1 n 2 2 [ ! max ! min ] = z 2 e 2 ~ c 2 1 1 n 2 2 2 c n min 2 c n max = 2 z 2 e 2 n ~ c 1 1 n 2 2 1 min 1 max For z = 1 (since we're dealing with isolated particles), n = 1 : 5, and min = 4000 A, this equation becomes: dN dx = 283 1 1 1 : 5 2 2 cm 1 (3) Plugging K = 1 MeV and mc 2 = 0 : 511 MeV into equation (2) yields = 0 : 941. Plugging this into equation (3) yields dN=dx = 149 photons per cm. Plugging K = 500 MeV and mc 2 = 938 MeV into equation (2) yields = 0 : 758. Plugging this into equation (3) yields dN=dx = 64 photons per cm. Plugging K = 5000 MeV and mc 2 = 938 MeV into equation (2) yields = 0 : 987. Plugging this into equation (3) yields dN=dx = 154 photons per cm. 1 2 Problem 14.4 2.1 Part a ~ z = a cos ( ! t ) ^ z ~v = a! sin ( ! t ) ^ z = ) ~ = a! c sin ( ! t ) ^ z _ ~ = a! 2 c cos ( ! t ) ^ z The observer is located at the zenith angle from the zaxis. Thus, the angle between ^ n and _ ~ is . Equation 14.20 becomes: dP d = e 2 4 c ^ n ^ n _ ~ 2 = e 2 4 c ^ n _ ~ 2 = e 2 4 c _ ~ 2 sin 2 = e 2 a 2 !...
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This note was uploaded on 12/08/2011 for the course PHYSICS 505 taught by Professor Liu during the Winter '08 term at University of Michigan.
 Winter '08
 LIU

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