Compton Scattering/Conservation Of Energy

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Homework Statement


An x-ray photon of wavelength 0.025 nm collides elastically with an electron and scatters through an angle of 90 degrees. How much energy did the electron acquire in this collision and in what way did the x-ray change?

Homework Equations


E=pc
E=hc/lambda

The Attempt at a Solution


E=hc/lambda to get energy, which didn't work
(6.63e-34)(3.0e8)/(2.5e-11) = 7.96e-15 J

Then I read the whole chapter on it and could not find out how, exactly, to relate the two together, thus allowing me to figure out individual energy for the electron. The answer is 7.0e-16, which is pretty irrelevant because I want to know how to do it. Thanks for any input.
 
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You are finding the initial energy of the photon, not the energy lost in the collision.

When the x-ray photon hits the electron it loses some energy to the electron and thus it's wavelength will change. Thus, you need to find this change in wavelength to find the energy gained by the electron.

HINT:

There is a special formula for change in wavelength due to Compton Scattering, and if you text has a section on the Compton Effect, I am sure its in there.
 
G01 said:
You are finding the initial energy of the photon, not the energy lost in the collision.

When the x-ray photon hits the electron it loses some energy to the electron and thus it's wavelength will change. Thus, you need to find this change in wavelength to find the energy gained by the electron.

HINT:

There is a special formula for change in wavelength due to Compton Scattering, and if you text has a section on the Compton Effect, I am sure its in there.

I'm going to go ahead and assume the formula in which you speak is lambda = h/mc(1-Costheta), but it doesn't make any sense because you can't get energy from it.

That's where I'm lost.
 
Remember, the energy of a photon is related to its frequency (and therefore wavelength). So if you know the change in the wavelength of the photon, you can figure out what the wavelength is after the collision (since you know its initial wavelength) and what its new energy is.