Potential difference to acquire momentum

[desmond iking]

Homework Statement

Electron diffracation experiment show that the wavelength of a electron is 0.15nm. Find the momentum of the electron in the beam. ii) Through what the potential diffrenece should the electron accelerated from rest to acquire this momentum?

Homework Equations

The Attempt at a Solution

λ= h/p , p=momentum
= (6.63x10^-34) / 550x10^-9
= 4.42x10^-24kgm/s
1/2 mc2
= eV

pc = 2eV
V= pc/2e

= (4.42X10^-24) ( 3x10^8) / (2 x 1.6x10^-19)
= 4143 V

but the ans is 67V ... why i am wrong?[/B]

 

[Orodruin]

The electron will not be traveling at the speed of light (quite far from it). If it was anywhere close, you would also have to use relativistic expressions. As it is, the speed is relatively low and classical expressions should be fine.

 

[gneill]

The electrons are not moving at the speed of light; they are not photons. Your solution should not involve c.

Classically, the momentum is mv. You should know the value of the mass of the electron. So find v (and check to make sure it's well less that c so that relativity doesn't come into play!).

edit: Ha! Orodruin beat me to it!

 

[nasu]

The wavelength of the electron is not 550 nm. Where did you get that value from?

 

[desmond iking]

so

The wavelength of the electron is not 550 nm. Where did you get that value from?

sorry it's a typo... it should be 0.15x10^-9