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Calculate the wave number for the longes...

Calculate the wave number for the longest wavelength transition in the Balmer series fo atomic hydrogen . `( R_(H) = 109677 cm^(-1)).`

Text Solution

Verified by Experts

For Balmer series ` n_1 =2`
If this line possesses longest wavelength
( i.e., lowest energy ) then ` n_2 =3`
` overline (v) = 1/ (lambda) = 109677 [1/2^2 - 1/3^2]`
` = 1. 523 xx 10^4 cm^(-1) = 1. 523 xx 10^6 m^(-1)`.
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Knowledge Check

  • The wave number for the longest wavelength transition in the Balmer series of atomic hydrogen is

    A
    `15.2 xx 10^(6)m^(-1)`
    B
    `13.6 xx 10^(6)m^(-1)`
    C
    `1.5 xx 10^(6)m^(-1)`
    D
    `1.3 xx 10^(6)m^(-1)`

  • The wavelength of the third line of the Balmer series for a hydrogen atom is -

    A
    `(21)/(100R_(H))`
    B
    `(100)/(21 R_(H))`
    C
    `(21R_(H))/(100)`
    D
    `(100R_(H))/(21)`

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