# Wave-Particle Duality

1. Oct 30, 2014

### bennyq

1. The problem statement, all variables and given/known data
Calculate the de Broglie wavelengths of an electron with energy 120 eV ...

2. Relevant equations
lambda = h\p where p = sqrt(2*Me*E)

3. The attempt at a solution
E=1.6E-19*120ev..
Then sub into equation and I get 1.1E-10m for the wavelength, which is the answer quoted.

The question that concerns me is why can you not use E=hf, where rearranged gives lambda=(hc/E) which gives
Thanks

2. Oct 30, 2014

### ehild

E=hf is valid if E is the total energy, (E=mc2). The formula p = sqrt(2*Me*E) is valid for speeds much less than c, and E means the kinetic energy.

3. Oct 30, 2014

### Orodruin

Staff Emeritus
Note that you can only use this for the de Broglie wavelength the way done in the OP if the particle is relativistic and thus has momentum essentially equal to its energy. In other words, when the velocity is close to c - otherwise the relation between wavelength and frequency changes.