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About Compton scattering

by snoopies622
Tags: compton, scattering
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snoopies622
#1
Dec23-08, 12:34 AM
P: 611
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]?
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malawi_glenn
#2
Dec23-08, 03:16 AM
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yeah sure, why shouldn't it?
snoopies622
#3
Dec23-08, 07:18 AM
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..

malawi_glenn
#4
Dec27-08, 04:52 AM
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About Compton scattering

Maximum transfer to electron is when E_prime is minimum
lightarrow
#5
Dec27-08, 08:26 AM
P: 1,521
Quote Quote by snoopies622 View Post
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.

What did you and your book get, instead?
snoopies622
#6
Dec27-08, 09:45 AM
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 [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
#7
Dec27-08, 11:39 AM
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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
#8
Dec27-08, 02:21 PM
P: 611
Quote Quote by jtbell View Post
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 ([tex] \theta = 180^{\circ}[/tex])." So there you go.
lightarrow
#9
Dec27-08, 02:26 PM
P: 1,521
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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