# 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
$$\lambda_{new}=\frac{\hbar}{\sqrt{2m(E-V)}}$$

So that the new wavelength is:
$$\frac{\lambda}{\sqrt{1-\frac{V}{E}}}$$

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:
$$\frac{\lambda}{\sqrt{\frac{V}{E}-1}}$$

Thanks for any help!

Last edited:
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#### Nugatory

Mentor
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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