- #1

jowens1988

- 15

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## Homework Statement

Compton's derivation of his scattering formula:

[tex]\delta \lambda \equiv \lambda' - \lambda = \frac{hc}{m_e c^2} (1 - cos\theta)[/tex]

assumed that that the target electrons were at rest. In reality, they are orbiting around nuclei. In a material like lead, the electrons have a kinetic energy as large as K ~ 200 eV.

Suppose that the atomic electron was heading directly toward the incoming X-ray photon. Calculate the wavelength of the scattered photon [tex]\lambda'[/tex]. By how much is this different from the expected Compton shift? Neglect terms of order K,K^2, or the electron momentum squared.

## Homework Equations

Conservation of Momentum:

x-direction: [tex] p_1 c - p_{e1}c = p_2 c cos(\theta) + p_{e2}c cos\phi[/tex]

and

y-direction: [tex] p_2 c sin\theta = p_{e2}c sin\phi[/tex]

Conservation of Energy:

[tex]p_1 c + \sqrt((p_{e1}c)^2 + (m_e c^2)^2) = p_2 c + \sqrt((p_{e2}c)^2 + (m_e c^2)^2) [/tex]

## The Attempt at a Solution

I think I am setting it up correctly, just adding the kinetic energy of the electron to the energy equations and an initial momentum to the x-direction of the momentum equations.

But if I neglect the terms that it tells me to neglect, then I get [tex] p_1 = p_2 [/tex], which would imply there is no wavelength shift at all, which doesn't seem right.

Would it make more sense to view the post collision frame in a center of momentum frame?