- #1
MrPunk44
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1. Homework Statement
Problem 1) A photon having 44 keV scatters from a free electron at rest. What is the maximum energy that the electron can obtain?
I already used all of my attempts on this problem so I can't check whether or not any attempt on this forum is successful. Hopefully that's not a problem.
1) ##E = \frac{hc}{\lambda}##
2) ##\delta \lambda = \frac{h}{mc} (1-cosx)##
3) ##E_i = E_s + KE_e##
In order to find the maximum energy the electron can obtain I first need to find the energy of the incident photon. In order to do that I must first find the wavelength of the scattered photon.
I use equation 1 for the wavelength of scattered photon. 1240ev*nm / 44000ev = .02818 nm. Then I maximize equation 2 by allowing x to equal pi. RHS equals 4.8488E-12 m. Since the incident photon has a higher energy I know it's wavelength must be smaller than .02818 nm. Therefore, I calculate .02818E-9 m - 4.8488E-12 m = .0233nm, which is the wavelength of the incident photon. Equation 1 gives me 53150ev as the energy of incident photon. Finally, equation 3 gives me 9150 ev as the maximum energy the electron can obtain.
Where'd I go wrong?[/B]
Problem 1) A photon having 44 keV scatters from a free electron at rest. What is the maximum energy that the electron can obtain?
I already used all of my attempts on this problem so I can't check whether or not any attempt on this forum is successful. Hopefully that's not a problem.
Homework Equations
1) ##E = \frac{hc}{\lambda}##
2) ##\delta \lambda = \frac{h}{mc} (1-cosx)##
3) ##E_i = E_s + KE_e##
The Attempt at a Solution
In order to find the maximum energy the electron can obtain I first need to find the energy of the incident photon. In order to do that I must first find the wavelength of the scattered photon.
I use equation 1 for the wavelength of scattered photon. 1240ev*nm / 44000ev = .02818 nm. Then I maximize equation 2 by allowing x to equal pi. RHS equals 4.8488E-12 m. Since the incident photon has a higher energy I know it's wavelength must be smaller than .02818 nm. Therefore, I calculate .02818E-9 m - 4.8488E-12 m = .0233nm, which is the wavelength of the incident photon. Equation 1 gives me 53150ev as the energy of incident photon. Finally, equation 3 gives me 9150 ev as the maximum energy the electron can obtain.
Where'd I go wrong?[/B]
