If `lambda = C_2 [(n^2)/(n^2 - n^2)]` for Balmer series , what is the value of `C_2 ` ?

Options

A. `2/(R_H)`
B. `2R_H`
C. `4R_H`
D. `4/(R_H)`

Answer

D

Explanation

`1/(lambda) = 1/(C_2) [(n^2 - 2^2)/(n^2)] = 1/(C_2 ) [1-2/(n_2^2)]`
`=(2^2)/(C_2) [1/(2^2) -1/(n_2^2)] , R_H = (2^2)/(C_2) implies C_2 = 4//R_H`