Advertisements
Advertisements
प्रश्न
Answer the following question.
Obtain the expression for the ratio of the de-Broglie wavelengths associated with the electron orbiting in the second and third excited states of the hydrogen atom.
Advertisements
उत्तर
Second excited state (n = 3)
Third excited state (n = 4)
We know that,
`E = "hc"/λ`
`(13.6)/(n^2) = "hc"/λ`
`(13.6)/("n"^2"hc") = 1/λ`
`λ = ("n"^2"hc")/(13.6)`
Now,
`λ_2/λ_3 = ("n"_2^2"hc")/(13.6) xx (13.6)/("n"_3^2"hc")`
`λ_2/λ_3 = ("n"^2 2)/("n"^2 3) = (3)^2/(4)^2`
`λ_2/λ_3 = (9)/(16)`
`λ_2 : λ_3 = 9: 16`
APPEARS IN
संबंधित प्रश्न
A proton and an α-particle are accelerated through the same potential difference. Which one of the two has less kinetic energy? Justify your answer.
A proton and an α-particle have the same de-Broglie wavelength. Determine the ratio of their accelerating potentials
A proton and an alpha particle are accelerated through the same potential. Which one of the two has (i) greater value of de−Broglie wavelength associated with it and (ii) less kinetic energy. Give reasons to justify your answer.
An electron is accelerated through a potential difference of 64 volts. What is the de-Broglie wavelength associated with it? To which part of the electromagnetic spectrum does this value of wavelength correspond?
Light of wavelength 2000 Å falls on a metal surface of work function 4.2 eV.
If the same light falls on another surface of work function 6.5 eV, what will be the energy of emitted electrons?
Gas exerts pressure on the walls of the container because :
A particle is dropped from a height H. The de Broglie wavelength of the particle as a function of height is proportional to
A proton is accelerated through one volt the increase in its kinetic energy is approximately
The work function of a metal is 4.50 eV. Find the frequency of light to be used to eject electrons from the metal surface with a maximum kinetic energy of 6.06 × 10−19 J.
The de-Broglie wavelength (λ) associated with a moving electron having kinetic energy (E) is given by ______.
