Advertisements
Advertisements
प्रश्न
Derive an expression for de Broglie wavelength of electrons.
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")`
Substituting the known values in the above equation, we get
λ = `(6.626 xx 10^-34)/(sqrt(2"V" xx 1.6 xx 10^-19 xx 9.11 xx 10^-31))`
λ = `(12.27 xx 10^-10)/sqrt("V")` meter (or) λ = `12.27/sqrt("V")` Å
For example, if an electron is accelerated through a potential difference of 100 V, then its de Broglie wavelength is 1.227 Å.
Since the kinetic energy of the electron, K = eV, then the de Broglie wavelength associated with electron can be also written as
λ = `"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
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.
A proton and an electron have the same kinetic energy. Which one has a greater de Broglie wavelength? Justify.
What is Bremsstrahlung?
Describe briefly Davisson – Germer experiment which demonstrated the wave nature of electrons.
How do we obtain characteristic x-ray spectra?
What should be the velocity of the electron so that its momentum equals that of 4000 Å wavelength photon.
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).
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.
