Compton Scattering Homework: Wavelength & Angle Calculation

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SUMMARY

The discussion focuses on a Compton scattering homework problem involving the calculation of the wavelength of an incident x-ray photon and the scattering angle of a free electron. The relevant equations include the change in wavelength formula, kinetic energy equation, and energy conservation equation. The user initially encountered a negative term under the square root while solving for the initial wavelength, which was resolved by recognizing that the energy lost by the photon is negative. The final wavelength calculated is 1 nm.

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romsoy
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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?
 
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Hello and welcome to PF!
romsoy said:
$$E = hc(\frac{1}{\lambda^{‘}} − \frac{1}{\lambda_{0}})$$
Does E represent a positive number? Does the right hand side of the equation yield a positive number?
 
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TSny said:
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!
 
Yes. Good work.
 

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