Calculate the wave number for the longest wavelength transition in the Balmer series fo atomic hydrogen . `( R_(H) = 109677 cm^(-1)).`

Explanation

According to Balmer formula, `bar(v)=(1)/(lambda)=R_(H)[(1)/(n_(1)^(2))-(1)/(n_(2)^(2))]`
In order that the wavelength `(lambda)` may be the maximum, wave number `(bar(v))` must be the least. This is possible in case `n_(2)-n_(1)` is minimum. Now, for Balmer series, `n_(1)=2` and `n_(2)` must be 3. Substituting these values in the Balmer formula, `bar(v)=(1.097 xx 10^(7)m^(-1))((1)/(2^(2))-(1)/(3^(2)))=1.097 xx 10^(7)m^(-1)((5)/(36))=1.523 xx 10^(6) m^(-1)`