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

1 Answer
Oct 17, 2017

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)))=2.77xx10^5m/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