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17 April, 21:06

A steady beam of alpha particles (q = + 2e, mass m = 6.68 * 10-27 kg) traveling with constant kinetic energy 22 MeV carries a current of 0.27 µA. (a) If the beam is directed perpendicular to a flat surface, how many alpha particles strike the surface in 2.8 s? (b) At any instant, how many alpha particles are there in a given 16 cm length of the beam? (c) Through what potential difference in volts is it necessary to accelerate each alpha particle from rest to bring it to an energy of 22 MeV?

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  1. 17 April, 21:22
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    q = 2e = 3.2 x 10^-19 C

    mass, m = 6.68 x 10^-27 kg

    Kinetic energy, K = 22 MeV

    Current, i = 0.27 micro Ampere = 0.27 x 10^-6 A

    (a) time, t = 2.8 s

    Let N be the alpha particles strike the surface.

    N x 2e = q

    N x 3.2 x 10^-19 = i t

    N x 3.2 x 10^-19 = 0.27 x 10^-6 x 2.8

    N = 2.36 x 10^12

    (b) Length, L = 16 cm = 0.16 m

    Let N be the alpha particles

    K = 0.5 x mv²

    22 x 1.6 x 10^-13 = 0.5 x 6.68 x 10^-27 x v²

    v² = 1.054 x 10^15

    v = 3.25 x 10^7 m/s

    So, N x 2e = i x t

    N x 2e = i x L / v

    N x 3.2 x 10^-19 = 2.7 x 10^-7 x 0.16 / (3.25 x 10^7)

    N = 4153.85

    (c) Us ethe conservation of energy

    Kinetic energy = Potential energy

    K = q x V

    22 x 1.6 x 10^-13 = 2 x 1.5 x 10^-19 x V

    V = 1.17 x 10^7 V
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