Question

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

Answer 1

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

Explanation:

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