The wavelength of the first line in blamer series in the hydrogen spectrum is `lambda`. What is the wavelength of the second line:

Options

A. 20/27 `lamda`
B. 3/16 `lamda`
C. 5/36 `lamda`
D. 3/4 `lamda`

Answer

A

Explanation

Wavelength in Balmer series of hydrogen spectrum is given by relation,
`1/lamda=R[1/(2^(2))-1/(n^(2))]` where, n = 3, 4, 5
So, wavelength of first line,
`1/(lamda_(1))=R[1/4-1/(3^(2))]`
`=(5R)/36`
`!R=36/(5lamda)` `(becauselamda_(1)=lamda,`given`)` ....(i)
Similarly wavelength of second line,
`1/(lamda_(2))=R[1/4-1/(4^(2))]=(3R)/(16)`
`!lamda_(2)=16/(3R)` ....(ii)
From Eqs. (i) and (ii), we get
`lamda_(2)=20/27lamda`
Hence, the wavelength of the second line in the same series is `20/27lamda`