? asked in Science & MathematicsPhysics · 10 years ago

Find the frequency of revolution of an electron?

a) Find the frequency of revolution of an electron with an energy of 100 eV in a uniform magnetic field of 22.0 µT.

in Hz

(b) Calculate the radius of the path of this electron if its velocity is perpendicular to the magnetic field.

in m

1 Answer

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  • Mohasa
    Lv 4
    10 years ago
    Favorite Answer

    The electron has energy of 100 eV

    So its KE is 0.5mv² = 100 eV...(i)

    where m = 9.109 × 10^(-31) kg is the electron mass, and v is its speed

    1 eV = 1.6 * 10^(-19) J

    So 0.5mv² = 100 * 1.6 * 10^(-19) J

    0.5mv² = 1.6 * 10^(-17) J

    And v = √[2 * 1.6* 10^(-17) / {9.109 × 10^(-31)} J/kg]

    giving v = 5.9271 * 10^6 m/s....(ii)

    The magnetic force = the centrifugal force in the magnetic field:

    qvB = mv²/r

    where q = 1.602 × 10^(-19) coulombs (C) is the electron charge,

    v = electron speed, m its mass, B = 22.0 µT is the magnetic field strength

    and r is the radius of the circular path.

    So qB = mv/r

    and v/r = qB/m...(iii)

    The period T = 2πr/v

    and the frequency f is

    f = 1/T = v/(2πr)

    or f = qB / (2πm) where we used equation (iii) for v/r

    Solve, with B = 22.0 * 10(-6)T

    f = 1.602 × 10^(-19) C * 22.0 * 10(-6)T / (2*π*9.109 × 10^(-31) kg)

    giving f = 6.158 * 10^5 Hz

    or f = 615.8 kHz <=====

    b) r = mv/(qB)

    r = {9.109 × 10^(-31) kg * 5.9271 * 10^6 m/s} / {1.602 × 10^(-19) C * 22.0 * 10^(-6)T}

    r = 1.532 m <====

    Hope this helps.

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