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yxgao
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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!
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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