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Homework Help: Inelastic collision energy relationship

  1. Oct 21, 2007 #1
    1. The problem statement, all variables and given/known data
    An atom of mass M is initially at rest, in its ground state. A moving (nonrelativistic) electron of mass me collides with the atom. The atom+electron system can exist in an excited state in which the electron is absorbed into the atom. The excited state has an extra, "internal," energy E relative to the atom's ground state.

    Find the kinetic energy that the electron must have in order to excite the atom.
    Express your answer in terms of E, me, and M.

    2. Relevant equations


    Inelastic collision:
    m1*v1 + m2*v2 = (m1+m2)Vf

    and possibly:

    [tex]\Delta[/tex]K = Ki - Kf

    3. The attempt at a solution

    What i've got so far:
    m1*v1 + m2*v2 = (m1+m2)Vf -->

    me*ve + (0) = (me + M)*Vf -->

    Vf = (me*ve)/(me + M)

    [tex]\Delta[/tex]K = Ki - Kf -->

    Kf = (1/2)(me + M) * [(meve)/(me + M)]^2 --> simplified -->


    Ki = just the Kinetic energy of the electron = (1/2)(me*ve^2 --> so..

    (me*ve)^2/2(me+M) - (1/2)(me*ve^2) = E(?.. an assumption) --->

    so K_e = (m_e * v_e)^2/2(m_e + M) - E --->

    there's the problem, i have v_e in my solution. But first i need to know i'm on the right track which i think i am. Any advice?
    Last edited: Oct 21, 2007
  2. jcsd
  3. Oct 21, 2007 #2

    Doc Al

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    Staff: Mentor

    You want K_i - K_f = E, since kinetic energy is lost to internal energy.

    You are exactly on track, except for the error that I pointed out. Correct that, of course.

    Then: Eliminate the v_e in your solution by expressing it in terms of K_e, and then solving for K_e in terms of E and the masses.
  4. Oct 21, 2007 #3
    I'm having trouble isolating K_e when i substitute V_e = (2k_e)/m_e

    Here's my new eq. as K_i - K_f and V_e in terms of K_e:

    K_e =[(m_e *(2K_e/m_e)]^2/[2(m_e + M)] - E ---> i'm having trouble isolating K_e. Is this setup correct? --->

    4(K_e)^2/[2(m_e + M)] - E = K_e
    Last edited: Oct 21, 2007
  5. Oct 21, 2007 #4

    Doc Al

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    Staff: Mentor

    That's not quite right. K_e = 1/2 m_e v_e^2 (Don't forget that v_e is squared!)
  6. Oct 21, 2007 #5
    wow, i keep making dumb mistakes.

    I've ended up with this:

    K_e = E(m_e + M)/2m_e + M --->
  7. Oct 21, 2007 #6
    nevermind ^ that.

    I got K_e = E(m_e + M)/-M

    I don't think the negative sign is correct. I gotta check my algebra. What do you think?
  8. Oct 22, 2007 #7

    Doc Al

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    Staff: Mentor

    The negative sign is not correct. Did you correct the sign error I pointed out in post #2? (Other than that, it's good.)
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