X-Ray Spectra and Photon Energy

[kachilous]

Homework Statement

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.

Homework Equations

Equation for photon Energy: E_photon = E_i - E_f

The Attempt at a Solution

E_photon = E_i - E_f

Using the emitted photon energy as the change in energy (40 KeV = 6.409x10^-15 Joules)
and the relativistic equations for energy: E = pc = m_evc.
We have: 6.409x10^-15 = m_ec(v_i - v_f)
Where m_ec = (9.109^-31)(3x10⁸)
Then we have: 2.345x10⁷ = v_i - v_f
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?

 

[Basic_Physics]

Try the relativistic formulation of momentum

p = \gammam_ov
where
\gamma=\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}}