- #1

satchmo05

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## Homework Statement

Suppose a 0.511[MeV] photon from a positron-electron annihilation scatters at 110 degrees from a free electron. What are the energies of the scattered photon and the recoiling electron? Relative to the initial direction of the 0.511[MeV] photon, what is the direction of the recoiling electron's velocity vector?

## Homework Equations

λ

_{2}- λ

_{1}= (h/mc)(1-cos(theta))

change in energy = (hc/(delta(λ)) = h*delta(f)

## The Attempt at a Solution

~ λ

_{2}- λ

_{1}= (h*c)/(m*c

^{2})*(1 - cos(110)) = 0.00326[nm] = delta(λ)

~ variating change in energy formula --> delta(f) = c/(delta(λ)) = 9.2025e

^{19}[Hz]

~ Multiplying frequency by Planck's constant, I get the change in energy = 6.1e

^{-19}[J] = 0.380732094 [MeV]

I am confused at what I do from here to determine the energies of either the scattered photon and the recoiling electron. Since the photon's initial energy is 0.511[MeV], and using change in energy answer from above, would this mean that the energy of recoiling electron be equal to (0.380732094 + 0.511) [MeV]?

Thank you for all help in advance. Hopefully this work I have shown in clear!