Calculate the wavelength and wave numbers of the first and second lines in tbe Balmer series of hydrogen spectrum. Given `R = 1.096xx10^(7) m^(-1)`

Explanation

For Balmer series `n_(1) = 2`, for first line=`n_(2)= 3`, for second line`= n_(2) = 4`
i. Wave number `barv` of the first line is given by
`bar(v_(3to2))=R[1/(n_(1)^(2))-1/(n_(2)^(2))]=1.096xx10^(7)xx(1/(2^(2))-1/(3^(2)))`
`=1.096xx10^(7)xx(1/4-1/9)=1.096xx10^(7)xx5/36=1.5236xx106m^(-1)`
`lamda=36/(1.0967xx10^(7)xx5)m=656.3xx10^(-9)m=656.3xx10^(-9)m=656.3nm`
Wavelength of `H_(alpha)` line = 656.3 nm
For `H_(beta)` line `n_(2//=4).R=10.97xx10^(6)(1/(2^(2))-1/(4^(2)))m^(-1)`
`=(10.97xx10^(6)xx3)/16m^(-1)=2.0568xx10^(6)m`
`lamda=16/(10.97xx10^(6)xx3)m`