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

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`1/lamda=R[1/n_(2)^(1)-1/n_(2)^(2)]` Where `n_(1)=2and n_(2)=3,4,5.......oo`
(i) Longest wavelength in Balmer series corresponds to the of `H_(2)` line for which `n_(2)=3`
`1/lamda=1.097xx10^(7)[1/2^(2)-1/3^(2)]=1.097xx10^(7)xx5/36`
`:. lamda=(36)/(5xx1.097xx10^(7))=6563Å`
(ii) For series limit `n_(2)=oo` Hence,
`1/lamda_(oo)=1.097xx10^(-1)[1/2^(2)-1/oo^(2)]=(1.097xx10^(7))/(4)`
`:.lamda_(oo)=4/(1.097xx10^(7))=3646Å`
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