For the first member of Balmer series of hydrogen spectrum, the wavelength is `lamda`. What is the wavelength of the second member?

Options

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

Answer

A

Explanation

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