Question

The spectral lines of atomic hydrogen wave number is equal to the difference between the wave numbers of the following two lines of the Balmer series: 486.1 and 410.2 nm. The wavelength of that line is  ______ × 10^-4 cm.

पर्याय

• 41.03

• 0.152

• 52.06

• 65.30

Answer 1

The spectral lines of atomic hydrogen wave number is equal to the difference between the wave numbers of the following two lines of the Balmer series: 486.1 and 410.2 nm. The wavelength of that line is  41.03 × 10^-4 cm.

Explanation:

Given: λ₁ = 486.1 × 10^-9 m

λ₂ = 410.2 × 10^-9 m

`barν = barν_2 - barν_1`

`barν = 1/lambda_2 - 1/lambda_1`

`barν = "R"_"H" [1/2^2 - 1/"n"_2^2] - "R"_"H"[1/2^2 - 1/"n"_1^2]`

`barν = "R"_"H" [1/"n"_1^2 - 1/"n"_2^2]`

For I case of Balmer series

`1/lambda_1 = "R"_"H" [1/2^2 - 1/"n"_1^2]`

`= 109678 [1/2^2 - 1/"n"_1^2]`

n₁ = 4

For II case

`1/lambda_1 = 1/(410.2 xx 10^-7)`

= `109678 [1/2^2 - 1/"n"_2^2]`

n₂ = 6

Transition 6 to 4

`barν = 1/lambda = 109678 (1/2^2 - 1/6^2) "cm"^-1`

∴ λ = 41.03 × 10^-4 cm