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The first emission line of Balmer series in H spectrum has the wave number equal to :A. `(9R_H)/(400) cm^(-1)`B. `(9R_H)/(144)cm^(-1)`C. `(3R_H)/4cm^(-1)`D. `(5R_H)/(36) cm^(-1)` |
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Answer» Correct Answer - D `overlinev = 1/(lambda)= R_H [ 1/2^2 - 1/3^2], n_1 = 2` for Balmer series , ` n_2 = 3` for `H_ (alpha)` line . |
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