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)

Calculate the wavelength and wave numbers of the first and second line

1.	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)
Answer» Solution :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`