The de - Broglie wavelength of a proton accelerated by 400 V400V is ??

1 Answer
Oct 17, 2017

The de Broglie wavelength would be 1.43xx10^(-12)m1.43×1012m

Explanation:

I would approach the problem this way:

First, the de Broglie wavelength is given by lambda=h/pλ=hp

which can be written as

lambda=h/(mv)λ=hmv

Now, we need the velocity of the proton that has passed through 400V. The work done by the electric field increases the kinetic energy of the proton:

qV = 1/2 mv^2qV=12mv2

which becomes v=sqrt((2qV)/m)v=2qVm

This gives v=sqrt((2*1.6xx10^(-19)xx400)/(1.67xx10^(-27)))=2.77xx10^5v=21.6×1019×4001.67×1027=2.77×105m/s

Back to the wavelength

lambda=h/(mv)=(6.63xx10^(-34))/((1.67xx10^(-27))(2.77xx10^5))=1.43xx10^(-12)mλ=hmv=6.63×1034(1.67×1027)(2.77×105)=1.43×1012m

This is quite a large wavelength when compared to the diameter of the proton at approx 10^(-15)1015 m