Question

The first line of Balmer series has wavelength 6563 A. What will be the wavelength of the first member of Lyman series?

पर्याय

• 1215.4 Å

• 2500 Å

• 7500 Å

• 600 Å

Answer 1

1215.4 Å

Explanation:

`1/lambda_"Balmer" = "R"_"H" [1/2^2 - 1/3^2] = (5"R"_"H")/36`

`therefore 1/(lambda_"L""Balmer") = 3"R"_"H"/4`

`therefore 1/(lambda_"L""Balmer") = lambda_"Balmer" xx 5/27 = 1215 .4  Å`