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Perfectly inelastic collision - energies

  1. Apr 2, 2012 #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 [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.

    2. Relevant equations
    Show that the minimum kinetic energy [tex]K_{initial}[/tex] that the electron must have in order to excite the atom is given as:
    [tex]K_{initial} = \frac{(M+m_e)E}{M}[/tex] and derive a formula for the associated minimum kinetic speed [tex]v_{0min}[/tex]

    3. Additional information
    From conservation of momentum [tex]v_{final} = \frac{m_e v_0}{m_e +M}[/tex] and so [tex]KE_{final}=1/2 \frac{(m_e*v_0)^2}{m_e +M}[/tex]
    which can be written [tex]KE_{final}=\frac{K_{initial}}{M+m_e}[/tex]
    4. The attempt at a solution

    [tex]minimum KE_{initial} = KE_{final}+ E = \frac{KE_{initial}}{M+m_e} + E[/tex]
    so [tex]KE_{initial}(1-1/(M+m_e))= E [/tex]
    [tex]KE_{initial} = \frac{E}{1-1/(M+m_e)}[/tex]

    which is different from the correct answer.
    What am I doing wrong?
     
  2. jcsd
  3. Apr 2, 2012 #2

    NascentOxygen

    User Avatar

    Staff: Mentor

    Check that last equation.
     
  4. Apr 2, 2012 #3
    I got the mistake, the answer should be:
    [tex]KE_{final}=1/2 (\frac{m_e*v_0}{m_e +M})^2[/tex]
    which can be written [tex]KE_{final}=\frac{K_{initial}m_e}{M+m_e}[/tex]
    Hence
    [tex]minimum KE_{initial} = KE_{final}+ E = \frac{K_{initial}m_e}{M+m_e} + E[/tex]
    so [tex]KE_{initial}(1-{m_e}/(M+m_e))= E [/tex]
    [tex]KE_{initial} = \frac{E(M+m_e)}{M}[/tex]

    Thank you for your reply!
     
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