The shortest wavelength in hydrogen spectrum of Lyman series when R_(H

1.	The shortest wavelength in hydrogen spectrum of Lyman series when R_(H) = 109678 cm^(-1) is
Answer» `1002.7 Å` `1215.67 Å` `1127.30 Å` `911.7 Å` Solution :`bar(v) = (1)/(lamda) = R_(H) ((1)/(n_(1)^(2)) - (1)/(n_(2)^(2)))` For LYMAN series, `n_(1) = 1` `lamda` is shortest of `1//lamda` is largest when `n_(1) = oo` Hence, `(1)/(lamda) = 109678 ((1)/(1^(2))) = 109678 CM^(-1)` or `lamda = (1)/(109678) cm = 911.7 xx 10^(-8) cm = 911.7 Å`

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