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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?
Solution
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, mp > mass of an electron, me, 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.
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