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21 October, 09:44

n alpha particle (q = + 2e, m = 4.00 u) travels in a circular path of radius 5.94 cm in a uniform magnetic field with B = 1.10 T. Calculate (a) its speed, (b) its period of revolution, (c) its kinetic energy, and (d) the potential difference through which it would have to be accelerated to achieve this energy.

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  1. 21 October, 10:03
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    a). V = 3.13*10⁶ m/s

    b). T = 1.19*10^-7s

    c). K. E = 2.04*10⁵

    d). V = 1.02*10⁵V

    Explanation:

    q = + 2e

    M = 4.0u

    r = 5.94cm = 0.0594m

    B = 1.10T

    1u = 1.67 * 10^-27kg

    M = 4.0 * 1.67*10^-27 = 6.68*10^-27kg

    a). Centripetal force = magnetic force

    Mv / r = qB

    V = qBr / m

    V = [ (2 * 1.60*10^-19) * 1.10 * 0.0594] / 6.68*10^-27

    V = 2.09088 * 10^-20 / 6.68 * 10^-27

    V = 3.13*10⁶ m/s

    b). Period of revolution.

    T = 2Πr / v

    T = (2*π*0.0594) / 3.13*10⁶

    T = 1.19*10⁻⁷s

    c). kinetic energy = ½mv²

    K. E = ½ * 6.68*10^-27 * (3.13*10⁶) ²

    K. E = 3.27*10^-14J

    1ev = 1.60*10^-19J

    xeV = 3.27*10^-14J

    X = 2.04*10⁵eV

    K. E = 2.04*10⁵eV

    d). K. E = qV

    V = K / q

    V = 2.04*10⁵ / (2eV) ... 2e-

    V = 1.02*10⁵V
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