What is the ratio of de-Broglies wavelengths of electron and proton moving with same velocities?

In order to continue enjoying our site, we ask that you confirm your identity as a human. Thank you very much for your cooperation.

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

No worries! We‘ve got your back. Try BYJU‘S free classes today!

No worries! We‘ve got your back. Try BYJU‘S free classes today!

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Suggest Corrections

0

Find the ratio of the de Broglie wavelengths of an electron and a proton when both are moving with the (a) same speed, (b) the same kinetic energy, and (c) the same momentum. State which of the two will have a longer wavelength in each case.

Data: mp = 1836 me

(a) The de Broglie wavelength, λ = `"h"/"p" = "h"/"mv"`

`lambda_"e"/lambda_"p" = ("m"_"p"/"m"_"e")("v"_"p"/"v"_"e")` = 1836 as vp = ve

Thus, λe < λp.

(b) λ = `"h"/"p" = "h"/sqrt"2mK"`, where K denotes the kinetic energy `(1/2 "mv"^2)`

∴ `lambda_"e"/lambda_"p" = sqrt(("m"_"p" "K"_"p")/("m"_"e""K"_"e")) = sqrt("m"_"p"/"m"_"e") = sqrt1836 = 42.85`

as Kp = Ke

Thus, λe > λp.

(c) λ = `"h"/"p"` 

∴ `lambda_"e"/lambda_"p" = "p"_"p"/"p"_"e" = 1` as pp = pe.

Concept: De Broglie Hypothesis

  Is there an error in this question or solution?