Photo-electric effect, Compton Scattering

mm2424

1. The problem statement, all variables and given/known data
What is the maximum kinetic energy of electrons knocked out of a thin copper foil by Compston scattering of an incident beam of 17.5 KeV rays? Assume the work function is negligible.


2. Relevant equations
Δλ = h/mc (1-cosθ)


3. The attempt at a solution

I reasoned that the greatest energy transfer to an electron will occur when the x-ray rebounds at 180 degrees, in which case the change in wavelength is 4.85 x 10^-12 m.

I figured that the wavelength of the x-rays increases by this amount, thereby decreasing in energy. I thought I could therefore take this change in wavelength and calculate the energy associated with it using E = hc/Δλ, and I got E = 256 KeV.

However, the answer key requires that you calculate the wavelength of the original x-ray, add Δλ, then calculate the energy and deduct the original energy. It yields a different answer, 1.1 KeV.

I don't understand why you can't say that the energy loss associated with the increase in the wavelength of the x-ray is completely transferred to the electron and then be done with it. What am I missing? Thanks!!!!

mm2424

Does anybody have any insight into what I'm doing wrong here? I'd appreciate any and all help :).

Redbelly98

So far, so good.
Not quite. First, note that it's impossible for a 17.5 keV photon to lose 256 keV of energy.

You have two energies, E₁ = hc/λ₁ and E₂ = hc/λ₂.

The energy difference E₁-E₂ is the difference between the hc/λ terms, not hc/Δλ. What you did is equivalent to saying

(1/5) - (1/3) = 1/(5-3) = 1/2,

which is not true.

Can you take it from here?

Yes, that is the basic idea.

mm2424

Thanks, that makes sense!