
#1
Dec2308, 12:34 AM

P: 611

Does the Compton scattering equation
[tex] \lambda '  \lambda = \frac{h}{m_{e} c} (1cos \theta ) [/tex] work even when [tex] \theta = 180^{\circ}[/tex]? 



#3
Dec2308, 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.. 



#4
Dec2708, 04:52 AM

Sci Advisor
HW Helper
P: 4,739

about Compton scattering
Maximum transfer to electron is when E_prime is minimum




#5
Dec2708, 08:26 AM

P: 1,504

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 (EE')/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? 



#6
Dec2708, 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}(1cos \theta )[/tex] and [tex]\theta = 180^{\circ}[/tex] since that maximizes [tex]1cos \theta[/tex]. My answer: 3.85% (rounded from 3.85294..) Book's answer: 3.93%. 



#7
Dec2708, 11:39 AM

Mentor
P: 11,255

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 lessprecise values for the constants, or rounded off one or more intermediate steps.




#8
Dec2708, 02:21 PM

P: 611





#9
Dec2708, 02:26 PM

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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