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Calculate the wavelength for the emissio...

Calculate the wavelength for the emission transition if it starts from the orbit having radius `1.3225 nm` ends at `211.6 p m`. Name the series to which this transition belongs and the region of the spectrum.

Text Solution

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Radius of nth orbit of H-like particles `= (0.529 n^(2))/(Z) Å = (52.9 n^(2))/(Z)` pm
`r_(1) = 1.3225 nm = 1322.5 " pm" = (52.9 n_(1)^(2))/(Z)`
`r_(2) = 211.6 " pm" = (52.9 n_(2)^(2))/(Z) :. (r_(1))/(R_(2)) = (1322.5)/(211.6) = (n_(1)^(2))/(n_(2)^(2)) or (n_(1)^(2))/(n_(2)^(2)) = 6.25 or (n_(1))/(n_(2)) = 2.5`
`:.` If `n_(1) = 2, n_(1) = 5`. Thus, the transition is from 5th orbit to 2nd orbit. it belongs to Balmer series.
`bar(v) = 1.097 xx 10^(7) m^(-1) ((1)/(2^(2)) - (1)/(5^(2))) = 1.097 xx (21)/(100) xx 10^(7) m^(-1)`
or `lamda = (1)/(v) = (100)/(1.097 xx 21 xx 10^(7)) m = 434 xx 10^(-9) m = 434 nm`
It lies in the visible region
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(a) In hydrogen atom an electron undergoes transitions from 2nd excited state to the 1st excited state and then to the ground state. Identify the spectral series to which transition belong . (b) Find out the ratio of the wavelenght of the emitted radiations in the two cases.

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Knowledge Check

  • By calculating the wavelength for the emission transition, if it starts from the orbit having radius 1.3225 nm and ends at 211.6 pm, name the series to which this transition belongs and the region of the spectrum.

    A
    Balmer series, visible region.
    B
    Lyman series, UV-region
    C
    Paschen series, IR-region.
    D
    Brackett series, IR-region.

  • What would be the wavelength and name of series respectively for the emission transition for H-atom if it starts from the orbit having radius 1.3225 nm and ends at 211.6 pm?

    A
    434 nm, Balmer
    B
    434 pm, Paschen
    C
    545 pm, Pfund
    D
    600 nm, Lyman

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