Advertisements
Advertisements
Question
Calculate the de-Broglie wavelength of the electron orbitting in the n = 2 state of hydrogen atom.
Solution
As we know,
`mvr_2=(nh)/(2pi)`
The de-Broglie wavelength is given as,
`lambda=h/(mv)=h/(nh)xx(2pir_2)=pir_2`
As r2=0.529×n2×10−10 m
Therefore, λ=π×0.529×10−10×n2=3.14×0.529×10−10×22≃6.64×10−10 m
shaalaa.com
Is there an error in this question or solution?
APPEARS IN
RELATED QUESTIONS
The equivalent wavelength of a moving electron has the same value as that of a photon of energy 6 × 10–17 J. Calculate the momentum of the electron.
The kinetic energy of the electron orbiting in the first excited state of hydrogen atom is 3.4 eV. Determine the de Broglie wavelength associated with it.
Write the expression for the de Broglie wavelength associated with a charged particle having charge ‘q’ and mass ‘m’, when it is accelerated by a potential V.
