The only e^- in the H-atom resides under ordinary conditions on the first orbit when energy is supplied, the e^- moves to higher energy shells depending upon the amount of energy absorbed. When an e emits energy i.e., the e^- returns to the lowest energy state, from this Lyman, Balmer, Paschen, Bracket, Pfund series are there, so different spectral lines in the spectra of atoms correspond to different transitions of e^- s from higher to lower energy levels: If the shortest wavelength of H atom in Lyinan series x, then longest wavelength in Balmer series of He^+ is

The only e^- in the H-atom resides under ordinary conditions on the fi

1.	The only e^- in the H-atom resides under ordinary conditions on the first orbit when energy is supplied, the e^- moves to higher energy shells depending upon the amount of energy absorbed. When an e emits energy i.e., the e^- returns to the lowest energy state, from this Lyman, Balmer, Paschen, Bracket, Pfund series are there, so different spectral lines in the spectra of atoms correspond to different transitions of e^- s from higher to lower energy levels: If the shortest wavelength of H atom in Lyinan series x, then longest wavelength in Balmer series of He^+ is
Answer» `(36x)/(5)` `(X)/(4)` `(9x)/(5)` `(5x)/(9)` Solution :Shortest WAVELENGTH `implies1/(R_H) =x` For Balmer SERIES of `He^(+2)` LONGEST `implies1/lambda = R_H [1/(2^2) - 1/(3^2)] xx 4 implieslambda implies(9x)/(5)`