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

Calculate the wave number for the longest wavelength transition in the

1.	Calculate the wave number for the longest wavelength transition in the Balmer series of atomic hydrogen.
Answer» Solution :For Balmer SERIES, `n_(1)=2`. Hence, `BAR(v)=R(1/2^(2)-1/n_(2)^(2))` `bar(v)=1/lambda`. For `lambda` to be longest (maximum), `bar(v)` should be minimum. This can be so when `n_(2)` is minimum, i.e., `n_(2)=3`. Hence, `bar(v)=(1.097xx10^(7) m^(-1)) (1/2^(2)-1/3^(2))=1.097xx10^(7)xx5/36 m^(-1) =1.523xx10^(6) m^(-1)`