De Broglie wavelength of particle in potential V

[yxgao]

The initial kinetic energy of a free particle is E and has wavelength \lambda. What is the de Broglie wavelength of a particle in a potential V?

This is what I have so far:

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

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

So that the new wavelength is:
<br /> \frac{\lambda}{\sqrt{1-\frac{V}{E}}}<br />


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:
<br /> \frac{\lambda}{\sqrt{\frac{V}{E}-1}}<br />


Thanks for any help!

 

[Nugatory]

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.