Advertisements
Advertisements
Question
What should be the velocity of the electron so that its momentum equals that of 4000 Å wavelength photon.
Advertisements
Solution
λ = 4000 × 10−10 m
λ = `"h"/"p" = "h"/("m"_"e""v"_"e")`
ve = `"h"/(λ"m")`
ve = `(6.626 xx 10^-34)/(4000 xx 10^-10 xx 9.11 xx 10^-31)`
= `(6.626 xx 10^-34)/(1.6565 xx 10^-27)`
ve = 0.1818 × 104
ve = 1818 ms−1
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.
A proton and an electron have the same kinetic energy. Which one has a greater de Broglie wavelength? Justify.
Write the relationship of de Broglie wavelength λ associated with a particle of mass m in terms of its kinetic energy K.
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.
Briefly explain the principle and working of electron microscope.
Describe briefly Davisson – Germer experiment which demonstrated the wave nature of electrons.
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?
