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What is the wavelength of light emitted when the electron of a hydrogen atom undergoes a transition from an energy level with `n= 4` to an energy level with `n = 2`? .What is the colour corresponding to this wavelength ? |
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Answer» The frequency (v) of radiation when an electron jumps from a higher orbit `v_(2)` to a lower orbit `v_(1)` is given by `v = 3.29 xx 10^(15)((1)/(n_(1)^(2)) - (1)/(n_(2)^(2)))` Therefore frequency v when an electron jumps from `n = 4 "to" n = 2` is given by `v = 3.29 xx 10^(15)((1)/((2)^(2)) -(1)/((4)^(2)))` `3.29 xx 10^(14) s^(-1)` we also know `v lambda = c` `or lambda = (c )/(v) = (3 xx 10^(6) ms^(-1))/(6.85 xx 10^(14) s^(-1))` `= 4.85 xx 10^(-7) m =- 485 nm` THe colour is blue |
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