Advertisements
Advertisements
Question
Write the relationship of de Broglie wavelength λ associated with a particle of mass m in terms of its kinetic energy K.
Advertisements
Solution
Kinetic energy of the particle, K = `1/2 "mv"^2 = "P"^2/(2"m")`
p = `sqrt(2"mK")`
de-Broglie wavelength of the particle λ = `"h"/"p" = "h"/sqrt(2"mK")`
APPEARS IN
RELATED QUESTIONS
State de Broglie hypothesis.
A proton and an electron have the same kinetic energy. Which one has a greater de Broglie wavelength? Justify.
An electron and an alpha particle have the same kinetic energy. How are the de Broglie wavelengths associated with them related?
Explain why photoelectric effect cannot be explained on the basis of wave nature of light.
Derive an expression for de Broglie wavelength of electrons.
Briefly explain the principle and working of electron microscope.
What should be the velocity of the electron so that its momentum equals that of 4000 Å wavelength photon.
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?
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.
