Calculate the wave number for thelongest wavelength transition in the Balmer series of atomic hydrogen.

Calculate the wave number for thelongest wavelength transition in the

1.	Calculate the wave number for thelongest wavelength transition in the Balmer series of atomic hydrogen.
Answer» Solution :For BALMER series, `bar(v) = 109, 677 ((1)/(2^(2)) - (1)/(n^(2))) cm` `lamda` will be maximum or `bar(v)` is minimum `(bar(v) = (1)/(lamda)) " for " n = 3`. The value will be `bar(v) = 109, 677 cm^(-1) ((1)/(2^(2)) - (1)/(3^(2))) = 109, 677 XX (5)/(36) cm^(-1) = 15232.9 cm^(-1) = 1.523 xx 10^(6) m^(-1)`