When an electron in hydrogen atom jumps from the third excited state to the ground state, how would the de Broglie wavelength associated with the electron change ? Justify your answer.

Explanation

(i) de Broglie wavelength associated with a moving charge particle a KE 'K' can be given as
`lambda = ( h )/ ( p ) = ( h )/( sqrt( 2mK)) `.....(1) `[ K = (1)/( 2) mv^(2) = ( p^(2))/( 2m)]`
(ii) The kinetic energy of the electron in any orbit of hydrogen atom can be given as
`K= - E =- ((13.6)/(n^(2)) eV) = ( 13.6)/(n^(2)) eV ` ....(2)
(iii) Let `k_(1)` and `K_(4)` be the KE of the elctron in ground state and third excited state, where `n_(1) = 1` shows ground state and `n_(2)=4` shows third excited state.
Using the concept of equation (1) & (2), we have
`(lambda_(1))/( lambda_(4)) = sqrt((K_(4))/( K_(1))) = sqrt((n_(1)^(2))/( n_(2)^(2)))`
`= ( lambda_(1))/( lambda_(4)) = sqrt((1^(1))/( 4^(2))) = (1)/(4)`
`rArr lambda = ( lambda_(4))/( 4)`
i.e., the wavelength in the ground in the ground state will decrease.