Finding KE when given de Broglie wavelength

[vspectra]

[SOLVED] Finding KE when given de Broglie wavelength

Homework Statement

What is the kinetic energy, in MeV, of a proton with a de Broglie wavelength of 10fm?
1 MeV = 10^6 eV
1 fm = 10^-15m
1 eV = 1.602 x 10^-19 J
h = 6.63 x 10^-34 Js
wavelength = 10 x 10^-15 m = 10^-5 nm
Mass proton = 1.673 x 10^-27 kg

Homework Equations

wavelength = h/mv
KE = (1/2)(m)(v^2)

The Attempt at a Solution

I feel like I did this right, but I still got the incorrect answer and I can't seem to notice what I did wrong.

v = (6.63 x 10^-34) / (10^-5)(1.673 x 10^-27) = .03962 m/s
K = .5(1.673 x 10^-27)(.03862^2) = 1.316 x 10^-30 J = 8.2 x 10^-12 eV
= 8.2 x 10^-18 MeV

 

[Hootenanny]

Welcome to PF,

v = (6.63 x 10^-34) / (10^-5)(1.673 x 10^-27) = .03962 m/s
K = .5(1.673 x 10^-27)(.03862^2) = 1.316 x 10^-30 J = 8.2 x 10^-12 eV
= 8.2 x 10^-18 MeV

Shouldn't that be 10*10^-15?

 

[vspectra]

Welcome to PF,

Thanks.

Shouldn't that be 10*10^-15?

I tried that the first time I did the problem, and realized 10 x 10^-15 is in meters. Wouldn't I need the wavelength to be in nm? So it would be 1 x 10^-5 nm.

 

[Hootenanny]

I tried that the first time I did the problem, and realized 10 x 10^-15 is in meters. Wouldn't I need the wavelength to be in nm? So it would be 1 x 10^-5 nm.

Why would you want it in nm? If you want your output velocity to be in m/s, then all your inputs must be in SI units.

 

[vspectra]

Ah, you're right, silly me. I was looking at another example which was basically the same question, and to me it looked like they converted from angstroms to nanometers for wavelength. It was actually converted to meters.

Thanks!

 

[bharatxxx007]

Hey thanks both of u :) i made the same mistake :p