The ground state energy of hydrogen atom is -13.6 eV. If an electron makes a transition from an energy level -1.51eV to -3.4 eV, calculate the wavelength of the spectral line emitted and name the series of hydrogen spectrum to which it belongs.

Explanation

Energy difference = Energy of emitted photon
`= E_(2) - E_(1)`
`= - 1.51 - ( - 3.4 ) = 1.89 eV`
`= 1.89 xx 1.6 xx 10^(-19) J`
`lambda = ( hc)/( E_(2) -E_(1))`
`= ( 6.6 xx 10^(-34) xx 3 xx 10^(8))/( 1.89 xx 1.6 xx 0^(-19)) = ( 19.8)/( (3.024) xx10^(-7))`
`= 6.548 xx 10^(-7) m = 6548Å`
This wavelength belongs to Balmer series of hydrogen spectrum.