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Question
What is the wavelength of light emitted when the electron in a hydrogen atom undergoes a transition from an energy level with n = 4 to an energy level with n = 2?
Solution
The ni = 4 to nf = 2 transition will give rise to a spectral line of the Balmer series. The energy involved in the transition is given by the relation,
`"E" = 2.18 xx 10^(-18) [1/"n"_"i"^2 - 1/"n"_"j"^2]`
Substituting the values in the given expression of E:
`"E" = 2.18 xx 10^(-18) [1/4^2 - 1/2^2]`
`= 2.18 xx 10^(18) [(1-4)/16]`
`=2.18xx10^(-18)xx(-3/16)`
E = – (4.0875 × 10–19 J)
The negative sign indicates the energy of emission.
Wavelength of light emitted (`lambda`) = `("hc")/"E"`
`("since" "E" = ("hc")/lambda)`
Substituting the values in the given expression of λ:
`lambda = ((6.626 xx 10^(-34))(3xx10^8))/(4.08875xx10^(-19))`
`lambda = 4.8631 xx 10^(-7) m`
`= 486.3 xx 10^(-9)` m
= 486 nm
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[RH = 1 × 105 cm−1, h = 6.6 × 10−34 Js, c = 3 × 108 ms−1]
