Advertisements
Advertisements
Question
An electron and an alpha particle have the same kinetic energy. How are the de Broglie wavelengths associated with them related?
Solution
λ = `"h"/"p"`
Kinetic energy of the particle K = `1/2 "mv"^2 = "P"^2/(2"m") = "h"/(2"m"λ^2)`
i.e. λ = `"h"/sqrt(2"mK")`; `λ ∝ 1/sqrt"m"`
`λ_"e"/λ_"α" = sqrt("m"_α/"m"_"e")`
APPEARS IN
RELATED QUESTIONS
Write the expression for the de Broglie wavelength associated with a charged particle of charge q and mass m, when it is accelerated through a potential V.
State de Broglie hypothesis.
Why we do not see the wave properties of a baseball?
What is Bremsstrahlung?
Derive an expression for de Broglie wavelength of electrons.
How do we obtain characteristic x-ray spectra?
What should be the velocity of the electron so that its momentum equals that of 4000 Å wavelength photon.
Calculate the de Broglie wavelength of a proton whose kinetic energy is equal to 81.9 × 10–15 J.
(Given: mass of proton is 1836 times that of electron).
A deuteron and an alpha particle are accelerated with the same potential. Which one of the two has
- greater value of de Broglie wavelength associated with it and
- less kinetic energy?
Explain.
An electron is accelerated through a potential difference of 81 V. What is the de Broglie wavelength associated with it? To which part of the electromagnetic spectrum does this wavelength correspond?
