The first line of Balmer series has wvaelength `6563 Å`. What will be the wavelength of the ifrst member of Lyman series?

Options

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

Answer

A

Explanation

`(1)/(lambda_("Balmer")) = R [(1)/(2^(2)) - (1)/(3^(2))] = (5R)/(36), (1)/(lambda_("Lyman")) = R [(1)/(1^(2)) - (1)/(2^(2))] = (3R)/(4)`
`:. Lambda_("Lyman") = lambda_("Balmer") xx (5)/(27) = 1215.4 Å`