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De Broglie wavelength of particle in potential V

  1. Dec 15, 2004 #1
    The initial kinetic energy of a free particle is E and has wavelength [tex]\lambda[/tex]. What is the de Broglie wavelength of a particle in a potential V?

    This is what I have so far:

    Since [tex]k=\frac{\sqrt{2mE}}{\hbar}[/tex],
    the original de Broglie wavelength of the a free particle is:
    [tex]\lambda = \frac{h}{p} = \frac{h}{\hbar k} = \frac{h}{\hbar} \frac{\hbar}{\sqrt{2mE}}=\frac{h}{\sqrt{2mE}}[/tex]

    When the particle enters the region of potential V,
    we solve Schrodinger's equation to get
    [tex]
    \lambda_{new}=\frac{\hbar}{\sqrt{2m(E-V)}}
    [/tex]

    So that the new wavelength is:
    [tex]
    \frac{\lambda}{\sqrt{1-\frac{V}{E}}}
    [/tex]


    This kind of confusing because it means that the potential energy can only be as large as the kinetic energy to give real values of the new wavelength - why is this? Or should the wavelength be:
    [tex]
    \frac{\lambda}{\sqrt{\frac{V}{E}-1}}
    [/tex]


    Thanks for any help!
     
    Last edited: Dec 15, 2004
  2. jcsd
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