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

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

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

My final equation is .00243 - $$\lambda$$incident = 1240 eVnm/(hc/$$\lambda$$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

$$E = \frac{hc}{\lambda}$$

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

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