Calculate the wavelength of the two spectral lines with the longest wavelengths (called the first two lines ) in the visible region of the atomic spectrum of hydrogen,
Strategy: Use Balmer's formula, Eq. to calculate the wavelength of visible lines in the atmoic emission spectrum of hydrogen. To calculate the wavelengths of the first two lines, used the two smallest allowed integers, `n = 3`and `n =4`, in the Balmer formula because `lambda` is inversely related to `n`.

Explanation

First longest wavelength
` bar(v) = (1)/(lambda) = R ((1)/(2^(2))-(1)/(n^(2)))`
`= 1.097 xx 10^(7)m^(-1) ((1)/(2^(2))-(1)/(3^(2)))`
`= 1.097 xx 10^(7)m^(-1) ((5)/(36))`
`= 0.1524 xx 10^(7)m^(-1)`
`lambda = 6.562 xx 10^(-7) m = 656.2 nm`
Second longest wavelength.
`bar(v) = (1)/(lambda) = R ((1)/(2^(2))-(1)/(n^(2)))`
`= 1.097 xx10^(7) m^(-1) ((1)/(2^(2))-(1)/(4^(2)))`
`= 1.097 xx 10^(7)m^(-1)((3)/(16)) = 0.2057xx10^(7) m^(-1)`
`lambda = (1)/(0.2057xx10^(7))m`
`= 4.861 xx 10^(-7)m = 486.1 nm`