De Broglie wavelength of particle in potential V

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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:

Nugatory

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The first expression, ##(1-V/E)^{-1/2}## is correct; you can get that result from the de Broglie wavelength formula if you take the kinetic energy to be ##E-V##, consistent with the kinetic energy being ##E## when there is no potential.

The result you get for ##V\gt{E}## is just telling you that the particle cannot be found in regions where the potential is greater than the initial kinetic energy.
 

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