P27_009

# P27_009 - 10 − 27 kg) = 6 . 68 × 10 − 27 kg, so v = s...

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9. (a) The charge that strikes the surface in time ∆ t is given by ∆ q = i t ,where i is the current. Since each particle carries charge 2 e , the number of particles that strike the surface is N = q 2 e = i t 2 e = (0 . 25 × 10 6 A)(3 . 0 s) 2(1 . 6 × 10 19 C) =2 . 3 × 10 12 . (b) Now let N be the number of particles in a length L of the beam. They will all pass through the beam cross section at one end in time t = L/v ,where v is the particle speed. The current is the charge that moves through the cross section per unit time. That is, i =2 eN/t =2 eNv/L .T h u s N = iL/ 2 ev . To Fnd the particle speed, we note the kinetic energy of a particle is K =20MeV=(20 × 10 6 eV)(1 . 60 × 10 19 J / eV) = 3 . 2 × 10 12 J . Since K = 1 2 mv 2 , then the speed is v = p 2 K/m . The mass of an alpha particle is (very nearly) 4 times the mass of a proton, or m =4(1
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Unformatted text preview: 10 − 27 kg) = 6 . 68 × 10 − 27 kg, so v = s 2(3 . 2 × 10 − 12 J) 6 . 68 × 10 − 27 kg = 3 . 1 × 10 7 m / s and N = iL 2 ev = (0 . 25 × 10 − 6 )(20 × 10 − 2 m) 2(1 . 60 × 10 − 19 C)(3 . 1 × 10 7 m / s) = 5 . × 10 3 . (c) We use conservation of energy, where the initial kinetic energy is zero and the Fnal kinetic energy is 20 MeV = 3 . 2 × 10 − 12 J. We note, too, that the initial potential energy is U i = qV = 2 eV , and the Fnal potential energy is zero. Here V is the electric potential through which the particles are accelerated. Consequently, K f = U i = 2 eV = ⇒ V = K f 2 e = 3 . 2 × 10 − 12 J 2(1 . 60 × 10 − 19 C) = 10 × 10 6 V ....
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## This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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