Question

Hydrogen spectrum consists of discrete bright lines in a dark background and it is specifically known as hydrogen emission spectrum. There is one more type of hydrogen spectrum that exists where we get dark lines on the bright background, it is known as absorption spectrum. Balmer found an empirical formula by the observation of a small part of this spectrum and it is represented by

`1/lambda = "R"(1/2^2 - 1/"n"^2)`, where n = 3, 4, 5,....

For Lyman series, the emission is from first state to n^th state, for Paschen series, it is from third state to n^th state, for Brackett series, it is from fourth state to n^th state and for P fund series, it is from fifth state to n^th state.

Which of the following lines of the H-atom spectrum belongs to the Balmer series?

Options

• 1025 Å

• 1218 Å

• 4861 Å

• 18751 Å

Answer 1

4861 Å

Explanation:

The shortest Balmer line has energy

= |(3.4 - 1.51)|eV = 1.89 eV

and the highest energy = |(0 - 3.4)| = 3.4 eV

The corresponding wavelengths are

`(12400 "eV"Å)/(1.89 "eV") = 6561 Å` and 

`(12400 "eV"Å)/(3.4 "eV") = 3647 Å`

Only 4861 Å is between the first and last line of the Balmer series.