How Does a Particle's Kinetic Energy Affect Its Wave-Particle Duality?

[hxcguitar101]

Homework Statement

At what Kinetic energy will a particle's debroglie wavelength equal its Compton Wavelength

Homework Equations

DeBroglie
λ = h/mv

Compton
λ = h/mc

The Attempt at a Solution

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

 

[vela]

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.

 

[hxcguitar101]

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

 

[vela]

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.