- #1
VenomHowell15
- 13
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Pretty easy one, I'd imagine. Just want to check it over.
lamba = (h*c)/(E)
(lamba' - lambdao) = (h/mc) (1 - cos (theta))
I've figured out that I'm looking at a 0 degree deflection here, with the photon bouncing back at 180 degrees if there is to be a maximum energy transfer. I'm just wondering if I have to simply plug 50 keV into lamba = (h*c)/(E), or is there some factor to take into account? I'm wondering if not all of the photons energy could be transferred to the electron and if there's any kinks involving that.
The thing is, looking at the equations above, if cos is 180, that means lambda' = lambda o... Which doesn't make sense, because that would mean the wavelengths are equal and no energy transfer could possibly have happened, since wavelength is inversely proportional to energy.
Homework Statement
If the maximum energy transferred to an electron during Compton Scattering is 50keV, what is the wavelength of the incident photon?Homework Equations
lamba = (h*c)/(E)
(lamba' - lambdao) = (h/mc) (1 - cos (theta))
The Attempt at a Solution
I've figured out that I'm looking at a 0 degree deflection here, with the photon bouncing back at 180 degrees if there is to be a maximum energy transfer. I'm just wondering if I have to simply plug 50 keV into lamba = (h*c)/(E), or is there some factor to take into account? I'm wondering if not all of the photons energy could be transferred to the electron and if there's any kinks involving that.
The thing is, looking at the equations above, if cos is 180, that means lambda' = lambda o... Which doesn't make sense, because that would mean the wavelengths are equal and no energy transfer could possibly have happened, since wavelength is inversely proportional to energy.
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