1. Dec 23, 2008

### snoopies622

Does the Compton scattering equation

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

work even when $$\theta = 180^{\circ}$$?

2. Dec 23, 2008

### malawi_glenn

yeah sure, why shouldn't it?

3. Dec 23, 2008

### 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..

4. Dec 27, 2008

### malawi_glenn

Maximum transfer to electron is when E_prime is minimum

5. Dec 27, 2008

### 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.

Last edited: Dec 27, 2008
6. Dec 27, 2008

### 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 $$\lambda = 1.21 x 10^{-10}m$$."

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..)

Last edited: Dec 27, 2008
7. Dec 27, 2008

### Staff: Mentor

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.

8. Dec 27, 2008

### 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 ($$\theta = 180^{\circ}$$)." So there you go.

9. Dec 27, 2008

### 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