InterviewSolution
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InterviewSolution
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A hydrogen atom in a having a binding of `0.85eV`makes transition to a state with excited energy `10.2eV`(a) identify the quantum number n of theupper and the lower energy state involved in the transition (b) Find the wavelength of the emitted radiation |
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Answer» a. Let `n_(1)` be the initial state of electron, Then. `E_(1) = - (13.6)/(n_(1)^(2)) e V` Here, `E_(1) = - 0.85 e V` `:. - 0.85 = - (13.6)/(n_(1)^(2))` or `n_(1) = 4` b. Let `n_(2)` be the final excitation state of electron. Since excitation energy is always measured with respect to the ground state, therefore `Delta E = 13.6 [1 - (1)/(n_(2)^(2))]` here `Delta E = 10.2 e V` `:. 10.2 = 13.6 [1 - (1)/(n_(2)^(2))]` or `n_(2) = 2` Thus the electron jumps from `n_(2) = 4` to `n_(2) = 2` c. The wavelength of the photon emitted for a transition between `n_(2) = 4` to `n_(2) = 2` is given by `(1)/(lambda) = R_(oo) [(1)/(n_(2)^(2)) - (1)/(n_(1)^(2))]` `(1)/(lambda) = 1.097 xx 10^(7) [(1)/(2^(2)) - (1)/(4^(2))]` `lambda = 4860 Å` |
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