De Broglie wavelength of electron

[lampshade]

Homework Statement

Show that the de Broglie wavelength of an electron of kinetic energy E (eV) is

\lambda = \frac{12.3*10^{-8}}{E^{1/2}}

Homework Equations

\lambda = \frac{h}{p}
E = \frac{p^2}{2m}

The Attempt at a Solution

I've played around with substituting and things like that, but I can't seem to find that 12.3 number anywhere. I feel like I must be missing something simple, but I'm not sure what.

 

[CompuChip]

First convert the second equation to the right units and solve it for p, then plug it into the first equation and work out all the things you have numerical values for. Note that h is actually \hbar = h / (2\pi).

I then get 12.26 \times 10^{-9}.