If the wavelength of the first member of Balmer series in hydrogen spectrum is 6563 Å, calculate the wavelength of the first member of Lymen series in the same spectrum.

Explanation

From Rydberg.s formula
`1/lambda_"if"=R[1/n_f^2-1/n_i^2]`
where `n_i` and `n_f` are initial and final energy states of transition.
`lambda_"if"` is the wavelength of the emitted spectral lines.
For the first member in Balmer series , `n_f`=2 and `n_i` =3
`therefore 1/lambda_32=R[1/2^2-1/3^2]=(5R)/36`
`therefore lambda_32=36/(5R)`
where R=Rydberg.s constant
`rArr` 6,563 Å = `36/(5R)`
`rArr R=36/(5xx6,563xx10^(-10))`...(i)
Now for the first member of Lyman series,
`n_f` =1 and `n_i`=2
`therefore 1/lambda_21=R[1/1^2-1/2^2]=(3R)/4`
`rArr lambda_21=4/(3R)`
Using (i), `lambda_21=(4xx5xx6,563xx10^(-10))/(3xx36)` =1,215.37 Å