about Compton scattering

snoopies622

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]?

malawi_glenn

yeah sure, why shouldn't it?

snoopies622

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..

malawi_glenn

Maximum transfer to electron is when E_prime is minimum

lightarrow

At 180°, (E-E')/E = a/(1+a)
where:
E = photon's energy before scattering
E' = photon's energy after scattering
a = 2hv/mc^2 (v = photon's frequency before scattering).

The maximun value of (E-E')/E is 100% for: a = +oo, that is for an infinite energy of the photon before scattering.

What did you and your book get, instead?

snoopies622

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%.

jtbell

For what it's worth, I get 3.86% (rounded from 3.8558%). I used five significant figures for all constants, and did not round off any intermediate steps. I suspect that whoever calculated the book's answer either used less-precise values for the constants, or rounded off one or more intermediate steps.

snoopies622

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.

lightarrow

I get 3.86% using three significant figures for all constants:

lambda = 1.21*10^(-10) m
m =9.11*10^(-31) kg
c = 3.00*10^8 m/s
h = 6.63+10^(-34) J*s

I get 3.90% if I take m = 9.00*10^(-31) kg