# The Compton Effect: Find the angles of scattered photon and electron

 P: 939 The energy you are given is the kinetic energy (call it K). So you have $E = K + mc^2$. Oh, and stick to electron volts in a calculation like this. There is no reason to fool around with kilograms.
 Emeritus Sci Advisor HW Helper Thanks PF Gold P: 11,687 It's useful to know that ##hc = 1240\text{ nm eV}##. Don't plug in a value for ##c##. Pull the factors of ##c## into the units. For example, the 120-keV photon has a momentum of 120 keV/c, and the mass of the electron is 511 keV/c2. Sometimes, you'll want to introduce factors of ##c## to make formulas work out more easily. For instance, you found ##E_\text{before} = 160\text{ keV}##. The momentum of the photon is then ##p = 160\text{ keV/}c##. To find its wavelength, you used ##\lambda = h/p##. If you throw in some factors of ##c##, everything works out nicely: $$\lambda = \frac{h}{p} = \frac{hc}{pc} = \frac{1240\text{ nm eV}}{160\text{ keV}} = 7.75\times 10^{-3}\text{ nm}.$$ It's worth taking a little time to learn how to work in these units. It'll save you from a lot of tedious unit conversions.