One electron collides elastically with a second electron initially at rest. After the collision, the radii of their trajectories are 1.00 cm

Question

One electron collides elastically with a second electron initially at rest. After the collision, the radii of their trajectories are 1.00 cm and 1.90 cm. The trajectories are perpendicular to a uniform magnetic field of magnitude 0.0500 T. Determine the energy (in keV) of the incident electron.

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Ladonna 1 week 2021-07-22T06:48:32+00:00 1 Answers 1 views 0

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    2021-07-22T06:50:03+00:00

    Answer:

    101.19keV

    Explanation:

    r1 = 1.0cm = 0.01m

    r2 = 1.90cm = 0.019m

    B = 0.050T

    q = 1.60*10^-19C

    m = 9.11 * 10^-31 kg

    Mv /r = qB

    v = rqB / m

    v1 = (0.01 * 1.60*10^-19 * 0.05) / 9.11 * 10^-31

    v1 = 8.78 * 10⁷ m/s

    V2 = (0.019 * 1.60*10^-19 * 0.05) / 9.11 * 10^-31

    V2 = 1.67 * 10⁸ m/s

    Kinetic energy of the system K.E = K.E₁ + K.E₂

    K.E₁ = ½mv₁² = ½ * 9.11 *10⁻³¹ * (8.78*10⁷)²

    k.e₁ = 3.511 * 10⁻¹⁵ J

    k.e₂ = 1/2 mv₂² = 1/2 * 9.11*10⁻³¹ * (1.67*10⁸)²

    k.e₂ = 1.27*10⁻¹⁴J

    K.E = k.e₁ + k.e₂

    K.E = 3.511*10⁻¹⁵ + 1.27*10⁻¹⁴

    K.E = 1.6211*10⁻¹⁴J

    1eV = 1.602 * 10⁻¹⁹J

    xeV = 1.6211*10⁻¹⁴J

    x = 101,192.23eV = 101.192keV

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