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Explain the series of spectral lines for H-atom whose fixed inner orbit numbers are 3 and 4. |
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Answer» Solution :According to Bohr's theory, electromagnetic radiation of a particular wavelength is emitted when there is a transition of the electron from a higher energy state to a lower energy state. Let the quantum number `n=n_(1)` REPRESENT a higher energy state and `n=n_(1)` represent a lower energy state `(n_(i)gtn_(f))`. For hydrogen, the wavelength `(lamda)` of the radiation arising due to transition from `n_(i)" to "n_(f)` is given by `(1)/(lamda)=R((1)/(n_(f)^(2)-(f)/(n_(i)^(2)))),` where R is the RYDBERG constant. The SPECTRAL SERIES arising due to the transitions to `n_(1)=3` from `n_(1)=4, 5,6,...,` etc. is called the Paschen series. The wavelengths in this series are given by `(1)/(lamda)=R((1)/(3^(2))-(1)/(n_(1)^(2)))(n_(1)=4,5,6,...,` etc.) The spectral series arising due to the transitions to `n_(1)=4" from "n_(1)=5,6,7,...,` etc. is called the Brackett series. The wavelengths in this series are given by `(1)/(lamda)=R((1)/(4^(2))-(1)/(n_(1)^(2)))(n_(1)=5,6,7,...,` etc.) |
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