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Unformatted text preview: Physics 31 Spring, 2007 Solution to HW #4 Problem A A parallel beam of α particles with fixed kinetic energy is normally incident on a piece of gold foil. (a) If 100 α particles per minute are detected at 20 ◦ , how many will be counted at 40, 60, 80, and 100 ◦ ? (b) If the kinetic energy of the incident α particles is dou bled, how many scattered α particles will be observed at 20 ◦ ? (c) If the original α particles were incident on a copper foil of the same thickness, how many scattered α particles would be detected at 20 ◦ ? The density of copper is 8.9 gm/cm 3 , and the density of gold is 19.3 gm/cm 3 . The formula for Rutherford scattering can be written N ( θ ) = A nZ 2 ( KE ) 2 sin 4 ( θ/ 2) , where we have lumped all the factors not involved in the present problem into a constant A . The factors we need are n , the number of atoms per unit volume in the foil; Z , the nuclear charge; and KE , the kinetic energy of the incident α particles. For part (a), note that N ( θ ) N (20 ◦ ) = sin − 4 ( θ/ 2) sin − 4 (20 ◦ / 2) Hence N ( θ ) = N (20 ◦ ) sin 4 (20 ◦ / 2) sin 4 ( θ/ 2) = 100 × . 000909 sin 4 ( θ/ 2) Substituting numbers, we get θ N ( θ ) 20 100.00 40 6.64 60 1.45 80 0.53 100 0.26 For part (b), if you double the KE , you change N by a factor of 2 − 2 = 0 . 25. So N = 25....
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 Spring '07
 Hickman
 Atom, Energy, Kinetic Energy, ev

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