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InterviewSolution
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The ground state energy of hydrogen atom is - 13.6 eV. If an electron makes a transition from an energy level - 0.85 eV to -3.4 eV, calculate the wavelength of the spectral line emitted. To which series of hydrogen spectrum does this wavelength belong? |
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Answer» Solution :The groundstateenergy of hydrogen `E_(1)`= 13.6 eV . `therefore ` Energy `E_(i) = -0.85e V` CORRESPONDS to a level `n_(i)`where ` - 0.85 = (-13.6)/(n_(i)^(2)) rArr n_(i) = 4` Againenergy `E_(f) = - 3.4` eV correspondsto a level `n_(f)`suchthat ` -3.4 = - (13.6)/(n_(f)^(2))`, which leads us totheresult `n_(f) = 2` As transition is takingplace from `n_(i) = 4`level to `n_(f) = 2` levelthewavelenghtbelongs to Balmer series of hydrogen SPECTRUM . The wavelenghtof spectrall lineis given by therelation. `hv = (HC)/(lambda) = E_(i) = E_(f) =- 0.85 eV - (-3.4 eV) = + 2.55 eV`. `therefore "" lambda = (hc)/(2.55ev) = (6.63 xx 10^(-34) xx 3 xx 10^(8))/((2.55 xx 1.60 xx 10^(-19))J)= 4.875 xx 10^(-7) m = 487.5` nm |
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