## X-Ray Spectra and Photon Energy

1. The problem statement, all variables and given/known data
In a particluar x-ray tube, an electron approaches the target moving at 2.35 108 m/s. It slows down on being deflected by a nucleus of the target, emitting a photon of energy 40 keV. Ignoring the nuclear recoil, but not relativity, compute the final speed of the electron.

2. Relevant equations
Equation for photon Energy: Ephoton = Ei - Ef

3. The attempt at a solution
Ephoton = Ei - Ef

Using the emitted photon energy as the change in energy (40 KeV = 6.409x10-15 Joules)
and the relativistic equations for energy: E = pc = mevc.
We have: 6.409x10-15 = mec(vi - vf)
Where mec = (9.109-31)(3x108)
Then we have: 2.345x107 = vi - vf
We can use 2.35x108 as the electron's initial velocity:
so vf = 2.35 108 - 2.345x107 = 2.1155x108.
However, the answer is 2.27x108. So I am off by a small amount.

What am I doing wrong?
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 Try the relativistic formulation of momentum p = $\gamma$mov where $\gamma=\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

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