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Question

Consider the de-Broglie wavelength of an electron and a proton. Which wavelength is smaller if the two particles have (a) the same speed (b) the same momentum (c) the same energy?

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Solution

de-Broglie wavelength,
λ=hmv,

where h = Planck's constant
m = mass of the particle
v = velocity of the particle
(a) It is given that the speed of an electron and proton are equal.
It is clear from the above equation that
λ1m
As mass of a proton, mp > mass of an electron, me, the proton will have a smaller wavelength compared to the electron.

(b) λ=hp p=mv
So, when the proton and the electron have same momentum, they will have the same wavelength.
(c) de-Broglie wavelength λ is also given by
λ=h2mE,
where E = energy of the particle.
Let the energy of the proton and electron be E.
Wavelength of the proton,
λp=h2mpE ...1
Wavelength of the electron,
λe=h2meE ...2
Dividing (2) by (1), we get:
λeλp=mempλeλp<1λe<λp
It is clear that the proton will have smaller wavelength compared to the electron.

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