Home
Class 11
CHEMISTRY
Calculate the wave number, wavelength an...

Calculate the wave number, wavelength and frequency first line of hydrogen spectrum or Calculate the maximum wave length of a line in the Lyman series.

Text Solution

Verified by Experts

First line is Lyman Series, where `n_(1) = 1, n_(2) = 2`. Maximum wave length corresponds to minimum frequency i.e., `n_(1) = 1, n_(2) = 2.`
Wave number `=barv=R[1/(n_(1)^(2))-1/(n_(2)^(2))]`.
R = Rydberg constant = `1.097xx10^(+7)m.n_(1)=1n_(2)=2`
`barv=1.097xx10^(+7)[1/(I^(2))-1/(2^(2))]`
Wave length `lamda=0.8227xx10^(7)=8.227xx10^(6)m^(-1)`
But `barv=1/(lamda)`
`thereforelamda=1/v=1/(8.27xx10^(6)m^(-1))=0.1215xx10^(-6)m=1.215xx10^(-17)m`
Frequency `lamda=C/(barv)=(3xx10^(8)m//s)/(1.215xx10^(-7)m)=2.469xx10^(-15) Hz` or `s^(1)`
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • STRUCTURE OF ATOM

    SUBHASH PUBLICATION|Exercise THREE MARKS QUESTIONS AND ANSWERS|16 Videos

  • STATES OF MATTER : GASES AND LIQUIDS

    SUBHASH PUBLICATION|Exercise NUMERICAL PROBLEMS AND ANSWERS|25 Videos

  • SUPER MODEL QUESTION PAPER (FOR ANSWER)

    SUBHASH PUBLICATION|Exercise PART-E|19 Videos

Similar Questions

Explore conceptually related problems

Calculate the wave number, wavelength and frequency of the first line in the Baliner series.

calculate the wave number , wavelength and frequency of the spectral line of hydrogen for the transition from n_(2)=4" to "n_(1)=2. (Given R=1.097xx10^(7)m^(-1))

Knowledge Check

  • The wavelength of the first line of Balmer series of H-atom of 6561 Å . The wavelength of the second line of the series is :

    A
    `13122Å`
    B
    `3280Å`
    C
    `4860Å`
    D
    `2180Å`

  • The wavelength of the first line of Balmer series of H-atom of 6561Å . The wavelength of the second line of the series is

    A
    `13122Å`
    B
    `3280Å`
    C
    `4860Å`
    D
    `2180Å`

    • Similar Questions

    Explore conceptually related problems

    The wavelength of the first line of Balmer series of hydrogen atom is lambda , the wavelength of the same line in doubly ionised lithium is

    How is wave number and wavelength of a wave related?

    The last member of Lymann series of Hydrogen atom is 912Å Calculate The wavelength of series limit of blamer series.

    The wavelength of first line in Balmer series is 6563 Å. Calculate the longest and the shortest wavelengths of the first spectral lines in the Lyman series.

    Calculate the value of Rydberg constant if the wavelength of the first member of Balmer series in the hydrogen spectrum is 6563 Å. Also find the wavelength of the first member of Lyman series in the same spectrum.

    Write the formula for the wave number of the spectral lines of Lyman series.

    Home
    Profile