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InterviewSolution
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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. |
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Answer» Solution :(i) Groundstate ( n =1) energyof HYDROGENATOM`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) Energyof the photon emiited ` E_(1) + E_(2) = - 1.51 - (-13.6) = 12.09 eV = 12.09 xx eJ` . `therefore ` Wavelenghtof thespectral 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` |
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