- #1
Feodalherren
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Homework Statement
A photon with energy 3.98E-19 J hits a stationary electron. The photon is deflected 90 degrees and the electron recoils back and down. At what angle is the electron deflected?
Hint: the electron will be going at << .10c
Homework Equations
The Attempt at a Solution
Skipping some simple plug and chug steps this is what I found:
λ(photon) = 499nm
P(photon) and therefore also Pi of the system = 1.328E-27 kgm/s.
This is where I get very confused.
Since the photon is deflected 90 degrees the Compton effect is negligible:
[itex]\Delta \lambda = \frac{h}{(9.11E-31kg)c} = 2.426E-12m[/itex].
So there is essentially no change in lambda. Therefore the energy of the photon is internally conserved. This means that the energy of the photon is essentially the same before and after it hits the photon. If the energy is conserved before and after, how on Earth can the electron gain any kinetic energy!? This seems to violate the conservation of energy.
Yet when I do the momentum part I get
[itex]P_{photon}=1.328E-27kgm/s = P_{Xe-}=(9.11E-31kg)(V_{Xe-})[/itex]
[itex]V_{Xe-}=V_{Ye-}=1458kgm/s[/itex] in the positive X and negative Y direction.
Therefore the angle is 45 degrees.
But as I said. Out of an energy point of view I can't make sense of it.