The first four spectral in the Lyman series of a H-atom are λ= 1218 Å, 1028 Å, 974.3 Å and 951.4 Å. If instead of Hydrogen, we consider deuterium, calculate the shift in the wavelength of these lines.

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    Vishal Baghel | Contributor-Level 10

    3 months ago

    This is a Long Answer Type Questions as classified in NCERT Exemplar

    Explanation- as we know total energy in stationary orbit is

    En=- - m e 4 8 n 2 ε 0 2 h 2  where sign have usual meaning.

    According to bohr third postulate h ν = E f - E i

    ν = - m e 4 8 ε 0 2 h 3 ( 1 n f 1 - 1 n i 2 )

    λ 1

    Where  is the reduced mass

    Reduced mass for H=H=;me(1-me/M)

    D= D; me(1-me/2M)

     =me(1-me/2M)(1+me/2M)

    If for hydrogen deuterium, the wavelength

    λ D λ H = H D = (1+ m e 2 M )-1= (1- 1 2 × 1840 )

    λ D = λ H × 0.99973  so lines emitted are 1217.7A0,1027.7A0,974.04A0

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