# Calculating Kinetic Energy

1. Jan 28, 2016

### FlorenceC

1. The problem statement, all variables and given/known data
Calculate the deBroglie wavelength of a proton moving at 1/4 the speed of light.

How does the kinetic energy of this proton compare to its p^2 /2m?

What does p^2 / 2m anyway conceptually (isn't it one of the triangle thingies in the lower version of E=mc^2)?

2. Relevant equations
deBroglie equation

3. The attempt at a solution
hf = KE + phi
lambda = h/mv
v = 0.25c, m = proton
that calculates lambda
then lambda = c/f to find frequency ....but i don't know how to find KE. (is it just 1/2mv^2?)

2. Jan 28, 2016

### Suraj M

$\frac{P^2}{2m}$ is nothing but the kinetic energy in terms of momentum right?
I'm not sure but since they've given velocity to be in terms of "speed of light" don't you think you should consider the relativistic mass?

3. Jan 29, 2016

### IAN 25

Your equation hf = KE + phi is the photoelectric effect equation.

The de Broglie wavelength, λd = h/p, where as Suraj says you need to insert the relativistic momentum of a proton p = γmv. Where γ is the ubiquitous relativistic factor. Try this and if you get what I'm saying let me know.

4. Jan 29, 2016

### Ray Vickson

I think you are being asked to compare classical kinetic energy $(1/2) m v^2$ with relativistic kinetic energy $m c^2 (\gamma - 1)$, where $\gamma = 1/\sqrt{1-(v/c)^2}$.