Advertisements
Advertisements
प्रश्न
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.
उत्तर
An electron of mass m is accelerated through a potential difference of V volt. The kinetic energy acquired by the electron is given by `1/2` mv2 = eV
Therefore, the speed v of the electron is v = `sqrt((2"eV")/"m")`
Hence, the de Broglie wavelength of the electron is λ = `"h"/"mv" = "h"/sqrt(2"emV"`
APPEARS IN
संबंधित प्रश्न
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
An electron and an alpha particle have the same kinetic energy. How are the de Broglie wavelengths associated with them related?
What is Bremsstrahlung?
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.