Calculate the de-Broglie wavelength associated with the electron revolving in the first excited state of hydrogen atom. The ground state energy of the hydrogen atom is - 13.6 eV.

Explanation

As per question ground state energy of hydrogen atom `E_(1) =-13.6 eV`
`:.`Energy in 1st excited state (i.e., for n = 2 state) will be `E_(2)= (E_(1))/(2)^(2)= -(13.6)/(4) = -3.4 eV`
`:.` K.E. of electron on 1st excited state `K_(2) = -E_(2) =+3.4 eV`
`:.` de Broglie wavelength in 1st excited state `lambda_(2)` = `(1.227)/(sqrt(K("ineV")))nm= (1.227)/(sqrt(3.4))=0.647`nm