- #1
CoffeeCrow
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Homework Statement
X-rays with wavelength 55pm are scattered from a graphite target. Consider an x-ray which is scattered from a valence electron at an angle of 180 degrees (back towards the x-ray source).
a. Is the wavelength of the X-ray greater or less than before the collision?
b. What is the momentum of the recoiling electrons?
Homework Equations
##\Delta\lambda=\frac {h} {mc}(1-cos(\theta))##
##p=\frac {h} {\lambda}##
The Attempt at a Solution
Part a. seems obvious. The X-ray collides with an electron, transfers some kinetic energy to it, losing some energy in the process. Less energy, longer wavelength, right?
Part b however, is giving me a few issues.
I start by finding the Compton shift for 180 degrees, which gives \Delta\lambda=4.85pm.
This means that the scattered X-ray should have a wavelength of 55+4.85=59.85pm, correct?
This is where things get a bit hairy. The solution for this problem however, states that the recoiling X-ray has a wavelength of 50.14 pm. This doesn't seem to make sense, as the energy of the X-ray has increased, despite transferring energy to the electron (these solutions have had issues in the past). Anyway,they then simply use p=\frac{h}{\lambda} to calculate the momentum of the recoiling electron, yielding (with their wavelength), 1.32E-23 Ns.
I however have tried the following:
##p_{i}=p_{f}##
##\frac{h}{\lambda_{i}}=p_{electron}-\frac{h}{\lambda_{f}}##
##p_{electron}=h(\frac{1}{\lambda_{i}}+\frac{1}{\lambda_{f}})##
##p_{electron}=2.31E-23Ns## (using my values for wavelength, 55pm and 59.85pm)
OR:
##p_{electron}=2.53E-23Ns##(using their values)
Which of course, doesn't agree with the solutions.
So, I was wondering, am I missing something fundamental and conceptual in regards to energy and wavelength shift and, is my momentum approach valid?
