by snoopies622
Tags: compton, scattering
 P: 611 Does the Compton scattering equation $$\lambda ' - \lambda = \frac{h}{m_{e} c} (1-cos \theta )$$ work even when $$\theta = 180^{\circ}$$?
 Sci Advisor HW Helper P: 4,739 yeah sure, why shouldn't it?
 P: 611 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..
HW Helper
P: 4,739

Maximum transfer to electron is when E_prime is minimum
P: 1,504
 Quote by 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..
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.

 P: 611 The book asks, "for the Compton effect, determine the percentage of energy transferred to the recoil electrons that acquire the maximum kinetic energy when $$\lambda = 1.21 x 10^{-10}m$$." I computed $$1-\frac {\lambda}{\lambda '}$$ where $$\lambda ' = \lambda + \frac{h}{m_0 c}(1-cos \theta )$$ and $$\theta = 180^{\circ}$$ since that maximizes $$1-cos \theta$$. My answer: 3.85% (rounded from 3.85294..) Book's answer: 3.93%.
 Mentor P: 11,224 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.
P: 611
 Quote by jtbell 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.
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 ($$\theta = 180^{\circ}$$)." So there you go.
 P: 1,504 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

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