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

1 Answer
Dwight
Oct 17, 2017

The de Broglie wavelength would be 1.43×10−12m

Explanation:

I would approach the problem this way:

First, the de Broglie wavelength is given by λ=hp

which can be written as

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=12mv2

which becomes v=√2qVm

This gives v=√2⋅1.6×10−19×4001.67×10−27=2.77×105m/s

Back to the wavelength

λ=hmv=6.63×10−34(1.67×10−27)(2.77×105)=1.43×10−12m

This is quite a large wavelength when compared to the diameter of the proton at approx 10−15 m

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