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

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Text Solution

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