Advertisements
Advertisements
प्रश्न
Write the relationship of de Broglie wavelength λ associated with a particle of mass m in terms of its kinetic energy K.
उत्तर
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
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
State de Broglie hypothesis.
An electron and an alpha particle have the same kinetic energy. How are the de Broglie wavelengths associated with them related?
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.
How do we obtain characteristic x-ray spectra?
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.
