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Kinetic Energy and Nonrelativistic Collision

  1. Oct 19, 2005 #1
    An atom of mass [tex]M[/tex] is initially at rest, in its ground state. A moving (nonrelativistic) electron of mass [tex]m_e[/tex] 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[tex] K_e[/tex] that the electron must have in order to excite the atom.
    Express your answer in terms of E, M , and [tex]m_e[/tex] .


    Ok, I've gotten this part so far: [tex] K_e = E + 1/2 * (M+m_e)v_2^2[/tex]
    But how do you get rid of [tex]v_2^2[/tex]? The computer says that my answer shouldn't depend on [tex]v_2[/tex].
    How do I get rid of v...or Is my equation wrong?
    Thanks
     
    Last edited: Oct 19, 2005
  2. jcsd
  3. Oct 19, 2005 #2

    Astronuc

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    Staff: Mentor

    I think that M is the mass of the atom and me would be the mass of the electron, but that's editorial.

    Normally in a collision of moving objects, one has an equation of energy (kinetic and potential) and an equation of momentum.

    So the moving electron has kinetic energy and momentum.

    It is absorbed by the atom, which is excited to some internal energy E above ground state, and there is some kinetic energy and momentum of the negative ion (electron + atom).

    me << M

    Nice little overview of collisions - http://www.virginia.edu/ep/Interactions/1__introduction_&_collision_kinematics.htm
     
    Last edited: Oct 19, 2005
  4. Oct 19, 2005 #3
    Hi quicknote,

    You must use the CM system (CM=center of mass). At the limit, after the collision, the system must be at rest in this system. So, if the electron collides the atom with [tex]v[/tex] in the laboratory coordinate system, in the CM coordinate system the electron is moving with a velocity given by the difference [tex]v-v_{CM}=v-\frac{mv}{M+m}[/tex]. The atom will have the velocity [tex]v_{CM}=\frac{mv}{M+m}[/tex]. The initial velocity (or its kinetic energy) of the electron can be obtained now from the conservation law of total energy in this system.

    [tex]E_{c,el}^{CM}+E_{c,at}^{CM}=E[/tex]

    Good luck!
     
    Last edited: Oct 19, 2005
  5. Oct 19, 2005 #4
    Hey Astronuc and Clive,
    thanks for taking a look at the question. I think I got it.

    To find the electron's kinetic energy it would be
    [tex]K_e=E+K_f[/tex]

    Using conservation of momentum, [tex]v_f[/tex] is [tex]v_f = m_e v_o/(m_e+M)[/tex]

    Using formula for kinetic energy, find K_f of the atom/atom is [tex] k_f = 1/2 (m_e+M) v_f^2[/tex]

    subbing in v_f gives: [tex] K_f = \frac {\ 1}{2} (m_e +M)( \frac {\ (m_e v_i)}{m_e + M})^2 [/tex]

    Using formula for kinetic energy, eliminated v_i from previous step.
    [tex] K_f = \frac {\ m_e k_i}{m_e+M} [/tex]

    Finally, subbing back into K_e = E + K_f, I get:

    [tex] k_e = \frac {\ E}{1- \frac {\ m_e}{m_e +M}}[/tex]

    *Whew*

    But now that I look at this equation...it looks familar, but I'm not totally sure. Can anyone confirm it?

    Thanks again
     
    Last edited: Oct 19, 2005
  6. Mar 12, 2008 #5
    It's correct, but...

    I'm probably not seeing something super basic, but how did you go from the velocity-less equation for K_final to that one for k_e? I think my algebra must be off since I'm having problems figuring out how to go from:

    [tex]
    K_f = \frac {\ m_e k_i}{m_e+M}
    [/tex]

    to:

    [tex]
    k_e = \frac {\ E}{1- \frac {\ m_e}{m_e +M}}
    [/tex]
     
  7. Mar 12, 2008 #6
    You need to enter K_f into the equation K_i= E + K_f
    Therefore K_i= E + m_e/(m_e+M)*K_i
    Then solve for K_i
    K_i - m_e/(m_e+M)*K_i=E
    K_i * [1-m_e/(m_e+M)]=E
    K_i=E/[1-(m_e/m_e+M)]
     
    Last edited: Mar 12, 2008
  8. Mar 12, 2008 #7
    Perfect, thanks. :)
     
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