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21 February, 02:13

An alpha particle (charge + 2e) travels in a circular path of radius. 5m in a magnetic field of 1.0 T. Find the (a) period, (b) speed, and (c) kinetic energy (in electron volts) of the alpha particle. Take m = 6.65 x 10 - 27 kg for the mass of the alpha particle.

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  1. 21 February, 05:40
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    Given Information:

    Radius = r = 0.5 m

    Magnetic field = 1.0 T

    Required Information:

    Period = T = ?

    Speed = v = ?

    Kinetic energy = KE = ?

    Answer:

    Period = 0.13x10⁻⁶ seconds

    speed = 24.16x10⁶ m/s

    Kinetic energy = 12.11 MeV

    Explanation:

    (a) period

    The time period of alpha particle is related to its orbital speed as

    T = 2πr/v eq. 1

    According to newton's law

    F = ma

    Force due to magnetic field is given by

    F = qvB

    qvB = ma

    qvB = m (v²/r)

    qB = mv/r

    v = qBr/m eq. 2

    substitute the eq. 2 in eq. 1

    T = 2πr/qBr/m

    r cancels out

    T = 2π/qB/m

    T = 2πm/qB

    T = 2π*6.65x10⁻²⁷/2*1.602x10⁻¹⁹*1

    T = 0.13x10⁻⁶ seconds

    (b) speed

    From equation 1

    T = 2πr/v

    v = 2πr/T

    v = 2π*0.5/0.13x10⁻⁶

    v = 24.16x10⁶ m/s

    (c) kinetic energy (in electron volts)

    Kinetic energy is given by

    KE = 0.5mv²

    KE = 0.5*6.65x10⁻²⁷ * (24.16x10⁶) ²

    KE = 1.94x10⁻¹² J

    since 1 electron volt has 1.602x10⁻¹⁹ J

    KE = 1.94x10⁻¹²/1.602x10⁻¹⁹

    KE = 12.11 MeV
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