What is the wavelength of the x-rays in the direct beam?

[asdf1]

for the following question:
a beam of x-rays is scattered b a target. At 45 degrees from the beam derection the scattered x-rays have a wavelength of 2.2pm. What is the wavelength of the x-rays in the direct beam?

my problem:
the question doesn't specify what the target is...
so what are you supposed to plug in for the compton's wavelength if you the compton effect formula?

 

[big man]

Don't you just use the expression:

$$\lambda$$-$$\lambda_0$$ = ($$h/m_ec$$)*(1-cos($$\theta$$))

Where $$\lambda$$ is the scattered wavelength and $$\lambda_0$$ is the original wavelength. The term $$h/m_ec$$ is the comptom wavelength and you can just check that value in your book no doubt. So the target source is irrelevant for this problem (I think).

 

[asdf1]

for the term, $$h/m_ec$$,
doesn't m depend on the target?

 

[big man]

Nope it doesn't depend on the material at all. $$h$$ is Planck's constant, $$c$$ is the speed of light and $$m_e$$ is the mass of the electron, which are all independent of the target material. You can calculate its value based on that information, or you can look in your book and it should give an accepted value for the compton wavelength ($$h/m_ec$$).

 

[asdf1]

@@
i think i don't get the formula thoroughly... i thought the x-rays were supposed to hit the target, and get scattered by it~
unless the target is the electron, why's $$m_e$$ the mass of the electron?

 

[big man]

Okay you have your x-rays which are incident on a target source. The incident x-rays have a wavelength $$\lambda_0$$. Now what happens is that the x-ray is scattered from an electron in the target source. The scattered x-ray has a new wavelength $$\lambda'$$. Now in this process there is also a recoiling electron, which has the compton wavelength $$\lambda_c = h/m_ec$$

 

[asdf1]

so it doesn't matter what the target is...
whatever target the photon hits will send off an electron?

 

[big man]

Yeah if the photon does interact with the target source there will always be a recoiling electron in order to be consistent with the fact that energy and momentum are conserved.

 

[asdf1]

i understand it a whole better~
thank you very much for explaining! :)