Question

In Balmer series, wavelength of first line is 'λ₁' and in Brackett series wavelength of first line is 'λ₂' then `lambda_1/lambda_2` is ______.

पर्याय

• 0.162

• 0.124

• 0.138

• 0.188

Answer 1

In Balmer series, wavelength of first line is 'λ₁' and in Brackett series wavelength of first line is 'λ₂' then `lambda_1/lambda_2` is 0.162.

Explanation:

The wavelength of a line in Balmer series is given by

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

where, R = Rydberg constant.

For first line, n = 3

`=> 1/lambda_1 = "R"(1/4 - 1/9) => lambda_1 = 36/"5R"`     ....(i)

The wavelength of a line in Brackett series is given by

`1/lambda = "R"(1/4^2 - 1/"n"^2)`  ...[for n = 5, 6, 7,...]

For first line, n = 5

`=> 1/lambda_2 = "R"(1/16 - 1/25) => lambda_2 = 400/"9R"`     ...(ii)

Dividing Eq. (i) by Eq. (ii), we get

`lambda_1/lambda_2 = 36/"5R" xx "9R"/400 = 81/500`

= 0.162