Advertisements
Advertisements
प्रश्न
What should be the velocity of the electron so that its momentum equals that of 4000 Å wavelength photon.
उत्तर
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
संबंधित प्रश्न
State de Broglie hypothesis.
Why we do not see the wave properties of a baseball?
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?
What is Bremsstrahlung?
Derive an expression for de Broglie wavelength of electrons.
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.
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.
