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?

Answer 1

de-Broglie wavelength,

`λ = h/(mv)`

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 `λ ∝ 1/m`

As mass of a proton, m_p > mass of an electron, m_e, the proton will have a smaller wavelength compared to the   electron.

(b) `λ = h/p  (∵ 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 = h/sqrt(2mE)` ,

where E = energy of the particle.
Let the energy of the proton and electron be E.
Wavelength of the proton,

`λ_p = h/sqrt(2m_pE)`      ....(1)

Wavelength of the electron,

`λ_e = h/sqrt(2m_eE)`      ....(2)

Dividing (2) by (1), we get : 

`λ_e/λ_p = sqrt(m_e)/sqrt(m_p)`

⇒ `λ_e/λ_p < 1`

⇒ `λ_e < λ_p`

It is clear that the proton will have smaller wavelength compared to the electron.