Question

Calculate the longest and shortest wavelength in the Balmer series of hydrogen atom. Given Rydberg constant =`1.0987xx10^7m^-1`.

Answer 1

The wavelength `(lambda)` of different spectral lines of Balmer series is given by
`1/lambda=R[1/(2^2)-1/(n_2^2)]`
Longest wavelength is of `H_(alpha)` line or 1st line of the series, for which `n_2=3`
`:. 1/lambda=1.097xx10^7[1/(2^2)-1/(3^2)]=1.097xx10^7xx5/36`
`lambda=36/(5xx1.097xx10^7)m=(36xx10^(10))/(5xx1.097xx10^7)A^@=6563A^@`
for shortest wavelength, `n_2=oo`
`:. 1/lambda=1.097xx10^7[1/(2^2)-1/(oo^(2))]=(1.097xx10^7)/4`
`lambda=4/(1.097xx10^7)m=(4xx10^(10)A^@)/(1.097xx10^7)=3646A^@`