Question

the wavelength of the second line of the balmer series in the hydrogen spectrum is 4861 Armstrong calculate the wavelength of the first line

Answer 1

have done the solution if you have any doubts then pls ask them in the comments section

Answer 2

The wavelength of the first line of the Balmer series is 6562.35 Å

For the first line of any series (For Balmer, n = 2), wavenumber (1/λ) is represented as:

1/λ = R [1/n² - 1/(n+1)²], R is the Rydberg constant.

Putting, n=2, we get:

1/λ = R [1/2² - 1/3²] = R[1/4 - 1/9] = 5R/36           ...(1)

The second line is represented as:

1/λ = R [1/n² - 1/(n+2)²], R is the Rydberg constant.

Putting, n=2, we get:

1/λ = R [1/2² - 1/4²] = R[1/4 - 1/16] = 3R/16          

Given, wavelength for this line is 4861 Å, So,

1/4861 = 3R/16

⇒ R = [16/(4861×3)]

Putting this value of R in eqn.(1), we get the wavelength of the first line as:

1/λ =  [(5×16)/(36×3×4861)]  

⇒ λ = [(5×16)/(36×3×4861)]⁻¹

       = 6562.35 Å