1.

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 ?

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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