About Compton scattering

Does the Compton scattering equation

[tex]

\lambda ' - \lambda = \frac{h}{m_{e} c} (1-cos \theta )

[/tex]

work even when [tex] \theta = 180^{\circ}[/tex]?

 

I just did a textbook problem which asked for the percentage of a photon's energy that would be passed to an electron via Compton scattering given that it was the maximum amount possible, so I assumed the angle was 180 degrees and my answer did not exactly match the book's answer. Then I wondered if perhaps "scattering" implied a kind of deflection and not a complete rebound.

Maybe the mismatch is just due to rounding..

 

The book asks, "for the Compton effect, determine the percentage of energy transferred to the recoil electrons that acquire the maximum kinetic energy when [tex] \lambda = 1.21 x 10^{-10}m[/tex]."

I computed [tex]1-\frac {\lambda}{\lambda '}[/tex]

where [tex]\lambda ' = \lambda + \frac{h}{m_0 c}(1-cos \theta )[/tex]

and [tex]\theta = 180^{\circ}[/tex] since that maximizes [tex]1-cos \theta[/tex].

My answer: 3.85% (rounded from 3.85294..)
Book's answer: 3.93%.

 

I guess that happens. Thanks for stepping in. I just looked at the Compton scattering section of another textbook and saw, "the scattered wavelength is angle-dependent and is greatest for scattering in the backward direction ([tex] \theta = 180^{\circ}[/tex])." So there you go.