The wavelength of first line of Balmer series is `6563Å`. The wavelength of first line of Lyman series will be

Options

A. `1215.4Å`
B. `2500Å`
C. `7500Å`
D. `600Å`

Answer

A

Explanation

`(lambda_("Lyman"))/(lambda_("Blamer"))=(((1)/2^(2)-(1)/(3^(2))))/(((1)/(1^(2))-(1)/(2^(2))))=(5)/(27)" " [:. (1)/(lambda)=R((1)/(n_(f)^(2))-(1)/(n_(i)^(2)))]`
`rArrlambda_("Lyman")=(5)/(27)xxlambda_("Balmer")=(5)/(27)xx6563=1215.4Å`