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The last member of Lymann series of Hydr...

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

Text Solution

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Rydberg formula for the wavelengths of spectral lines in hydrogen spectrum is
`(1)/(lamda)=R((1)/(n_(f)^(2))-(1)/(n_(i)^2))`
`:.R=(1)/(913.4Å)`
`:.` The short wavelength limit `lamda_(B)` for the Lyman series would be
`(1)/(lamda_(B))=R((1)/(oo^(2))-(1)/(2^(2)))=(R)/(4)`
`:.lamda_(B)=(4)/(R)=4xx913.4Å`
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Knowledge Check

  • The wavelength of the first spectral line in the Balmer series of hydrogen atom is 6561 Å. The wavelength of the second spectral line in the Balmer series of singly ionized helium atom is

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