# Difficulty with relativistic units, eV/c^2

1. Mar 27, 2012

### yoshtov

I'm having some difficulty in working with units of mass at the quantum level. This difficulty most clearly manifests itself when I'm doing a Compton scattering problem.

Recall that Compton scattering is given by
$$\Delta \lambda =\frac{h}{m_{e}c}(1-cos(\theta ))$$
and that the rest mass of an electron is 5.11 x 10^5 eV/c^2

My confusion seems to come about in the denominator. I find myself unsure of whether to multiply by c, c^2, or maybe something else entirely. In any case, for my problem, I keep getting the wrong answer, regardless of my method of attack.

If anyone could shed any insight on my problem, and also give me a big picture of relativistic units (i.e., here is how to handle relativistic units in calculations), it would be very much appreciated.

Last edited: Mar 27, 2012
2. Mar 27, 2012

### fzero

To convert from mass to length, we use

$$\hbar c \sim 197~ \mathrm{MeV\cdot fm}.$$

$$\frac{h}{m_e c} = \frac{2\pi \hbar c}{m_e c^2},$$

which is easily computed from the information given.

3. Mar 27, 2012

### Staff: Mentor

I like to think not in terms of $m = 5.11 \times 10^5$ eV/c2, but rather $mc^2 = 5.11 \times 10^5$ eV. Before plugging this into an equation, I insert c's so that I have this combination to substitute for. In your example:

$$\Delta \lambda =\frac{h}{m_{e}c}(1- \cos \theta) = \frac{hc}{m_e c^2}(1 - \cos \theta)$$
$$\Delta \lambda = \frac{1.24 \times 10^{-6} eV \cdot m}{5.11 \times 10^5 eV} (1 - \cos \theta)$$

Most tables of constants in textbooks list values for hc in various units, or of course you can construct them yourself from values for h and c. hc comes up often enough that I've memorized it naturally.

Similarly for momentum: p = 100 keV/c means pc = 100 keV. An electron with that momentum has energy

$$E = \sqrt{(mc^2)^2 + (pc)^2} = \sqrt{(511 keV)^2 + (100 keV)^2} = 520.7 keV$$

Last edited: Mar 27, 2012