Advertisements
Advertisements
Question
A proton and an electron have the same kinetic energy. Which one has a greater de Broglie wavelength? Justify.
Advertisements
Solution
de-Broglie wavelength of the particle is λ = `"h"/"p" = "h"/sqrt(2"mK")`
i.e. `λ ∝ "h"/sqrt"m"`
As for me < mp,
So λe > λp
Hence wavelength of electron is greater than that of proton.
APPEARS IN
RELATED QUESTIONS
In an electron microscope, the electrons are accelerated by a voltage of 14 kV. If the voltage is changed to 224 kV, then the de Broglie wavelength associated with the electrons would
Why we do not see the wave properties of a baseball?
Write the relationship of de Broglie wavelength λ associated with a particle of mass m in terms of its kinetic energy K.
Derive an expression for de Broglie wavelength of electrons.
Briefly explain the principle and working of electron microscope.
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).
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?
The ratio between the de Broglie wavelength associated with proton accelerated through a potential of 512 V and that of alpha particle accelerated through a potential of X volts is found to be one. Find the value of X.
