Compton Scattering and maximum energy

[typeinnocent]

Homework Statement

If the maximum kinetic energy given to the electrons in a Compton scattering experiment is 10 keV, what is the wavelength of the incident X-Rays?

Homework Equations

[tex]\Delta[/tex][tex]\lambda[/tex] = (h/mc)*(1-cos[tex]\theta[/tex])
E = hc/[tex]\lambda[/tex]
[tex]\Delta[/tex][tex]\lambda[/tex] = [tex]\lambda[/tex]scattered - [tex]\lambda[/tex]incident

[c]3. The Attempt at a Solution [/b]
I think I made this question more complex than it is...
So I know that [tex]\Delta[/tex][tex]\lambda[/tex] = .00243 nm and Einitial=Ephoton + Eelectron, and Eelectron =10 keV.
I made [tex]\lambda[/tex]scattered = hc/Ephoton.
This equals [tex]\Delta[/tex][tex]\lambda[/tex] = hc/(Einitial - 10 keV), and then I plugged hc/[tex]\lambda[/tex]incident for Einitial.

My final equation is .00243 - [tex]\lambda[/tex]incident = 1240 eVnm/(hc/[tex]\lambda[/tex]incident - 10 kEv).

The solution is .0239 nm, however the correct answer in the book is .022 nm.
Although the solution is close, I feel that it's more due to luck than to actually doing the correct methodology. Could anyone help me solve this problem? Thanks!

 

[lanedance]

so if you're happy with the wavelength shift to find the energy shift knowing

[tex] E = \frac{hc}{\lambda} [/tex]

differentiating and assuming only small changes, which probably isn't too far form the truth..
[tex] dE = -\frac{hc}{\lambda^2}d\lambda [/tex]

if you want the exact answer you need to solve:
[tex] \Delta E = \frac{hc}{\lambda} -\frac{hc}{\lambda + \Delta \lambda}[/tex]
you should be able to solve for lambda, by multiplying through by the denominators & rearranging to give a quadratic in terms of lambda