One electron collides elastically with a second electron initially at rest. After the collision, the radii of their trajectories are 0.00 cm

Question

One electron collides elastically with a second electron initially at rest. After the collision, the radii of their trajectories are 0.00 cm and 2.30 cm. The trajectories are perpendicular to a uniform magnetic field of magnitude 0.0370 T. Determine the energy (in keV) of the incident electron.

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Kiệt Gia 4 years 2021-08-06T22:12:37+00:00 1 Answers 172 views 0

Answers ( )

  1. trucchi
    0
    2021-08-06T22:14:00+00:00
    Reply

    Answer:

    63.750KeV

    Explanation:

    We are given that

    Initial velocity of second electron,u_2=0

    Radius,r_1=0

    r_2=2.3 cm=\frac{2.3}{100}=0.023 m

    1 m=100 cm

    Magnetic field,B=0.0370 T

    We have to determine the energy of the incident electron.

    Mass of electron,m=9.1\times 10^{-31} kg

    Charge on an electron,q=-1.6\times 10^{-19} C

    Velocity,v=\frac{Bqr}{m}

    Using the formula

    Speed of electron,v_1=\frac{Bqr_1}{m}=\frac{0.0370\times 1.6\times 10^{-19}\times 0}{9.1\times 10^{-31}}=0

    Speed of second electron,v_2=\frac{0.0370\times 1.6\times 10^{-19}\times 0.023}{9.1\times 10^{-31}}

    v_2=1.5\times 10^8 m/s

    Kinetic energy of incident electron=\frac{1}{2}mv^2_1+\frac{1}{2}mv^2_2

    Kinetic energy of incident electron=0+\frac{1}{2}(9.1\times 10^{-31})(1.5\times 10^8)^2=1.02\times 10^{-14} J

    Kinetic energy of incident electron=\frac{1.02\times 10^{-14}}{1.6\times 10^{-19}}=63750eV=\frac{63750}{1000}=63.750KeV

    1KeV=1000eV

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