What Is the Kinetic Energy of a Proton at 1/4 the Speed of Light?

[FlorenceC]

Homework Statement

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)?

Homework Equations

deBroglie equation

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?)

 

[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?

 

[IAN 25]

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

The de Broglie wavelength, λ_d = 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.

 

[Ray Vickson]

Homework Statement

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)?

Homework Equations

deBroglie equation

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?)

Homework Statement

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)?

Homework Equations

deBroglie equation

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?)

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 $\gamma = 1/\sqrt{1-(v/c)^2}$.