# Wave-Particle Duality

1. Oct 19, 2011

### hxcguitar101

1. The problem statement, all variables and given/known data
At what Kinetic energy will a particle's debroglie wavelength equal its Compton Wavelength

2. Relevant equations
DeBroglie
λ = h/mv

Compton
λ = h/mc

3. The attempt at a solution

Setting the two equations equal to eachother, I got v = c, then said KE = (1/2)mc^2, but somehow that just doesn't sound right. What do you think?

2. Oct 19, 2011

### vela

Staff Emeritus
Nope, that's not right, as you suspected. The DeBroglie wavelength is actually $\lambda = h/p$ where p is the momentum of the particle. You need to use the relativistic expression for the kinetic energy to get the correct answer.

3. Oct 19, 2011

### hxcguitar101

So, it's when p = mc, so would you use E = Sqrt((pc)^2+(mc^2)^2) = sqrt((mc)^2+(mc^2))?

4. Oct 19, 2011

### vela

Staff Emeritus
Essentially, yes, but you need to get the algebra right. You have $p = mc$ so $pc = mc^2$ and
$$E=\sqrt{(pc)^2+(mc^2)^2} = \sqrt{(mc^2)^2+(mc^2)^2} = \cdots$$Also, remember E gives the total energy, not the kinetic energy.