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eep
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I'm working out a problem from a text concerning the scattering of monochromatic light by free electrons (Compton effect) which asks me to derive expressions for the wavelength shift, electron momentum, and electron scattering angle in terms of the photon scattering angle assuming that the electron is initially at rest.
I've managed to derive the standard wavelength shift formula no problem, it depends only on the scattering angle of the photon. However, I can't manage to derive anything for the momentum of the electron without it also depending on the initial wavelength of the photon. Is there an expression for the momentum of the electron which ONLY depends on the scattering angle of the photon?
My hunch is no, since if one looks at the following formula:
[tex]\Delta E = h(\nu - \nu^{\prime}) = h(\frac{c}{\lambda} - \frac{c}{\lambda^{\prime}}) = hc(\frac{\lambda^{\prime} - \lambda}{\lambda\lambda^{\prime}})[/tex]
The change in energy (which will lead you to the electron momentum) depends on the product of [itex]\lambda[/itex] and [itex]\lambda^{\prime}[/itex]. Am I missing something obvious?
I've managed to derive the standard wavelength shift formula no problem, it depends only on the scattering angle of the photon. However, I can't manage to derive anything for the momentum of the electron without it also depending on the initial wavelength of the photon. Is there an expression for the momentum of the electron which ONLY depends on the scattering angle of the photon?
My hunch is no, since if one looks at the following formula:
[tex]\Delta E = h(\nu - \nu^{\prime}) = h(\frac{c}{\lambda} - \frac{c}{\lambda^{\prime}}) = hc(\frac{\lambda^{\prime} - \lambda}{\lambda\lambda^{\prime}})[/tex]
The change in energy (which will lead you to the electron momentum) depends on the product of [itex]\lambda[/itex] and [itex]\lambda^{\prime}[/itex]. Am I missing something obvious?
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