# Wavelength of an electron

1. Jun 9, 2009

### Jules18

1. The problem statement, all variables and given/known data

-Electron has 3.00 MeV (or 4.8*10^-13 Joules)
-it's relativistic
-finding λ.

2. Relevant equations

h=6.63*10^-34

λ=h/p (obviously)

And I'm not sure if they're needed, but the relativistic eq's are:

KE = mc^2/sqrt(1-(v/c)^2)
p = mv/sqrt(1-(v/c)^2)

I'm not sure if this one applies to relativistic speeds:

E = hc/λ

3. The attempt at a solution

Attempt 1:

E = hc/λ

4.8E-13 = (6.63E-34)(3E8)/λ
λ = (6.63E-34)(3E8)/(4.8E-13)
λ = 4.14E-13 m

If you could help, that would be great.
Sorry if it's too long, and I'm a little unfamiliar with relativistic eqn's so forgive me if I screwed up on them.

2. Jun 9, 2009

### diazona

The equation you used, $E = hc/\lambda$, only applies to photons (or massless particles in general). So you're not going to need that one here. If you're familiar with the equation
$$E^2 = p^2c^2 + m^2c^4$$
I'd use that. If not, you can get the velocity from
$$E = \frac{mc^2}{\sqrt{1 - v^2/c^2}}$$
(note that that's total energy, not kinetic energy) and compute the momentum from that.

3. Jun 10, 2009

### Jules18

Thanks very much for the help.