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!

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!

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