Home
Class 11
CHEMISTRY
Calculate the wave number of the spectra...

Calculate the wave number of the spectral line of shortest wave length appearing in Given `R=1.09xx10^(7)m^(-1)`

Text Solution

Verified by Experts

`barv=R(1/(n_(1)^(2))-1/(n_(2)^(2)))m^(-1)`
`barv=1.09xx10^(7)m^(-1)(1/(2^(2))-1/0)`
`barv=1.09xx10^(7)(1/4)`
`barv=0.2725xx10^(7) barv=2.725xx10^(6)m^(-1)`
Promotional Banner

Topper's Solved these Questions

  • PUE DEPT. MODEL QUESTION PAPER

    SUBHASH PUBLICATION|Exercise PART-D|46 Videos

  • PUE DEPT. MODEL QUESTION PAPER

    SUBHASH PUBLICATION|Exercise PART-E|13 Videos

  • PUE DEPT. MODEL QUESTION PAPER

    SUBHASH PUBLICATION|Exercise PART -B|11 Videos

  • P-BLOCK ELEMENTS

    SUBHASH PUBLICATION|Exercise Three mark questions and answers|24 Videos

  • REDOX REACTIONS

    SUBHASH PUBLICATION|Exercise Three Marks Questions|29 Videos

Similar Questions

Explore conceptually related problems

Calculate the wave number of the spectral line of shortest wavelength appearing in the Balmer series of H-spectrum. (R=1.09xx10^7m^(-1))

What is meant by the wave number of spectral line ?

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

write the formula for the wave number of a spectral line of Balmer series.

Calculate the shortest and longest wavelength of Balmer series of hydrogen atom. Given R=1.097 xx 10^(7)m^(-1) .

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))

The wave number of the series limiting line for the Lyman series for hydrogen atom is (R =109678cm^(-1) )

calculate the longest wavelength in Balmer series and the series limit . (Given R=1.097xx10^(7)m^(-1) )