Compton Scattering Homework: Wavelength & Angle Calculation

[romsoy]

Homework Statement

In a Compton scattering experiment, an x-ray photon scatters through an angle of 17.40 from a free electron that is initially at rest. The electron recoils with a speed of 2180 km/s. Calculate (a) the wavelength of the incident photon and (b) the angle through which the electron scatters.

Homework Equations

$$\Delta \lambda = \lambda^{‘} − \lambda_{0} = \frac{h}{mc} (1−cos\theta)$$
$$K = \frac{1}{2}mv^{2}$$
$$E = hc(\frac{1}{\lambda^{‘}} − \frac{1}{\lambda_{0}})$$

The Attempt at a Solution

For (a), I set the kinetic energy gained by the electron equal to the energy lost by the scattered photon - so I basically equated my second and third equations above. Then I managed to get it into a quadratic form where I solve for $\lambda _{0}$, which I got by eliminating $$\lambda'$$ with $$\lambda' = \Delta \lambda + \lambda_{0}$$. I ended up getting $$\frac{−A\Delta \lambda \pm \sqrt{(A\Delta \lambda)^2 − 4A\Delta \lambda}}{2A}$$ where $$A = \frac{mv^{2}}{2hc}$$
But turns out the term under the square root is negative :/. Can anyone help me out please?

 

[TSny]

Hello and welcome to PF!

$$E = hc(\frac{1}{\lambda^{‘}} − \frac{1}{\lambda_{0}})$$

Does E represent a positive number? Does the right hand side of the equation yield a positive number?

 

[romsoy]

Hello and welcome to PF!
Does E represent a positive number? Does the right hand side of the equation yield a positive number?

Ah...E would be the energy lost by the photon so it'd be negative... so the A term will be negative. Turns out that results nicely in 1nm, thanks!

 

[TSny]

Yes. Good work.