Advertisements
Advertisements
प्रश्न
A proton and an electron have the same kinetic energy. Which one has a greater de Broglie wavelength? Justify.
उत्तर
de-Broglie wavelength of the particle is λ = `"h"/"p" = "h"/sqrt(2"mK")`
i.e. `λ ∝ "h"/sqrt"m"`
As for me < mp,
So λe > λp
Hence wavelength of electron is greater than that of proton.
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
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.
Why we do not see the wave properties of a baseball?
Write the relationship of de Broglie wavelength λ associated with a particle of mass m in terms of its kinetic energy K.
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.
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.
