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3 August, 22:30

An alpha particle (q = + 2e, m = 4.00 u) travels in a circular path of radius 6.56 cm in a uniform magnetic field with B = 1.70 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. 3 August, 22:39
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    dа ta:

    Charge of alpha particle, q = 2e

    = 2 * 1.6 x 10^{-19 }

    = 3.2 x 10^{-19} C

    Radius of circular path, r = 4.5 cm

    = 0.045 m

    Magnetic field, B = 1.8 T

    Mass of alpha particle, m = 4 u

    = 4*1.67 x 10^{-27 }

    = 6.68 x 10^{-27} kg

    Solution:

    (a) Centripetal force = Force due to magnetic field

    m v^2 / r = q v B

    m v / r = q B

    v = r q B / m

    = 0.045*3.2 x 10^{-19}*1.8 / 6.68 x 10^{-27 }

    = 3.88 x 10^6 m/s

    Speed of alpha particle, v = 3.88 x 10^6 m/s

    (b)

    Time period, T = 2 π r / v

    = 2 * π * 0.045 / 3.88 x 10^6

    = 7.2 x 10^{-8} s

    Time period, T = 7.2 x 10^-8 s

    (c) Kinetic energy, KE = (1/2) m v^2

    = 0.5*6.68 x 10^{-27} * (3.88 x 10^6) ^2

    = 5.02 x 10^{-14}J

    = (5.02 x 10^{-14} / 1.6 x 10^{-19}) eV

    = 3.14 x 10^5 eV

    Kinetic energy, KE = 3.14 x 10^5 eV

    (d) Potential energy = Kinetic energy

    q V = KE

    2 e V = 3.14 x 10^5 e V

    2V = 3.14 x 10^5 V

    V = 1.57 x 10^5 V

    Potential difference, V = 1.57 x 10^5 V
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