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

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

1.	If lambda = C_2 [(n^2)/(n^2 - n^2)] for Balmer series , what is the value of C_2?
Answer» `2/(R_H)` `2R_H` `4R_H` `4/(R_H)` SOLUTION :`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) impliesC_2 = 4//R_H`