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Posted by on Monday, September 29, 2008 at 1:15am.

An alpha particle, the nucleus of a helium atom, is at rest at the origin of a Cartesian
coordinate system. A proton is moving with a velocity of v towards the alpha particle
from the positive x axis direction. If the proton is initially far enough away to have no
potential energy, how close does the proton get to the alpha particle? Your answer
should be in terms of v, mp (mass of a proton), e (charge of a proton).

  • physics - , Monday, September 29, 2008 at 1:24am

    I shall assume that the alpha particle, which has a charge of 2e, cannot move.

    The proton will stop and turn around when all if its initial kinetic energy, (1/2) mp V^2, is converted to electrostatic potential energy. The electrostatic potential energy when they are a distance r apart is
    2 k e^2/r
    because the alpha particle has a charge of 2e and the proton has a charge of e. k is the Coulomb constant. You will have to look that up.

    When the particle stops,
    (1/2) mp * v^2 = 2 k e^2/r

    v is the initial velocity, when they are very far apart and the electrostatic potential energy is zero.

    Solve for r

  • physics - , Thursday, September 26, 2013 at 7:32pm

    r = (4ke^2)/(mpV^2)

  • University of Rochester Physics - , Thursday, September 26, 2013 at 7:37pm

    it stops when kinetic energy is equal to electrostatic energy. therefor:
    .5mpv^2= kqq/r
    .5mpv^2=(2ke^2)/(mpv^2)
    solve for r: (4ke^2)/(mpv^2)

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