The ground state energy of hydrogen atom is -13.6 eV.
(i) What is the kinetic energy of an electron in the second excited state ?
(ii) If the electron jumps to the ground state from the second excited state, calculate the wavelength of the spectral line emitted.

Explanation

(i) Ground state ( n =1) energy of hydrogen atom `E_(1) = - 13.6 eV`
`therefore ` Energy of electron in the second excited state `(n=3) E _(3) = (-13.6)/((3)^(2)) eV = - 1.51 eV`
`therefore ` Kinetic energy of electron in the second excited state `k_(3) = + 1.51 eV`
(ii) Energy of the photon emiited ` E_(1) + E_(2) = - 1.51 - (-13.6) = 12.09 eV = 12.09 xx eJ` .
`therefore ` Wavelenght of the spectral line emitted .
`lambda = (hc)/(E) = (6.63 xx 10^(-34) xx 3 xx 10^(8))/(120.9 xx 1.6 xx 10^(-19)) m = 1.025 xx 10^(-19) m = 102.5 nm`