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Question
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.
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Solution
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.
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