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. `(20lambda)/27`
B. `(3lambda)/16`
C. `(5lambda)/36`
D. `(3lambda)/4`

Answer

A

Explanation

`1/lambda_(1)= R(1/4-1/9) Rightarrow lambda_(1)=(4xx9)/(5R)`
Similarly `1/lambda_(2)=R(1/4-1/4^(2))`
`Rightarrowlambda_(2)=16/(3R)=16/3xx(5lambda)/(4xx9)=20/(27lambda)`