# Finding the deBroglie wavelength: conceptual issue

Hello all! I am new to this forum, though I have been lurking for a long time. I intend to fill out my profile and introduce myself, but finals leave me with little time to spare at the moment. I have this final next week and would like to sort this out beforehand, hence the hasty post!

1. Homework Statement

An electron has kinetic energy 3.50eV. Find it's wavelength.

(1) p=mv
(2) K=½mv2
(3) p=E/c=h/λ

(4) Er=mc2
(5) Etotal=K+Er

## The Attempt at a Solution

Note: my prof said not to use relativistic equations, as the energy is not high enough, but we have covered relativity in this class.
I now know how to solve this problem:
I solved for the speed of the electron with the kinetic energy formula, and put it into the following:
λ=h/(mv)
I got 0.656nm, which is correct.

My problem is that I did not initially solve the problem this way. What I did first was calculate the rest energy of the electron (Er=mc2) and add it to the kinetic energy to get the "total" energy. This was based off an equation which I was given by my prof: Etotal=K+Er. I then took that "total" energy and solved for the wavelength by deriving the equation: λ=hc/E from equation (3) above. I got the wrong answer, 0.00242nm, with this method.
What I don't understand is why my original method does not work. I'm assuming my fundamental understanding of these equations is wrong. Is the total energy of the electron not the rest energy plus the kinetic energy? Is there something I'm missing? My professor was not helpful this time around, so I'm turning to you guys!

Welcome to PF!

This was based off an equation which I was given by my prof: Etotal=K+Er. I then took that "total" energy and solved for the wavelength by deriving the equation: λ=hc/E from equation (3) above. I got the wrong answer, 0.00242nm, with this method.

In getting the equation λ=hc/E, it looks like you assumed that the momentum is related to energy by p = E/c. This is not correct for electrons. (But it is correct for photons.)