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

Question

6727 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-10-17T01:15:18

The de Broglie wavelength would be #1.43xx10^(-12)m#

I would approach the problem this way:

First, the de Broglie wavelength is given by #lambda=h/p#

which can be written as

#lambda=h/(mv)#

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^2#

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

This gives #v=sqrt((2*1.6xx10^(-19)xx400)/(1.67xx10^(-27)))=2.77xx10^5#m/s

Back to the wavelength

#lambda=h/(mv)=(6.63xx10^(-34))/((1.67xx10^(-27))(2.77xx10^5))=1.43xx10^(-12)m#

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