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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. |
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Answer» Solution :`(1)/(lambda)=R[(1)/(n_(F)^(2))-(1)/(n_(i)^(2))]` For First LINE of Balmer series `n_(f)=2 and n_(i)=3` `(1)/(lambda_(B_(2)))=R[(1)/(4)-(1)/(9)]=(5R)/(36)` For Longest wavelength of Lyman series `(1)/(lambda_(L_(1)))=R[1-(1)/(4)](3R)/(4)` to find `lambda_(L_(1))=1215Å` For shortest wavelength of Lyman series `n_(f)=1 and n_(i)=OO` `lambda_(L_(oo))=911.54Å` |
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