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.
Advertisements
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
State de Broglie hypothesis.
Why we do not see the wave properties of a baseball?
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.
Briefly explain the principle and working of electron microscope.
Describe briefly Davisson – Germer experiment which demonstrated the wave nature of electrons.
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.
