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Explain by drawing a line spectra diagram of the transition between energy levels in an atom. |
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Answer» Solution :The formula for the wavelength of light emitted from a transition between different ATOMIC states. `(1)/(lambda_(if))~~R[(1)/(n_(f)^(2))-(1)/(n_(i)^(2))] and n_(f) lt n_(i)` If `n_(f)=1 and n_(i) =2,3,4...`then the wave number of Lyman series line `alpha, beta, gamma, delta`..... resprectively obtained. If `n_(f)=2 and n_(i)= 3,4,5`... then wave number of Balmer series line `alpha, beta, gamma, delta,`.... respectively obtained. If `n_(f)=3 and n_(i)= 4,5,6`... then wave number of Balmer series line `alpha, beta, gamma, delta, `..... respectively obtain If `n_(f)=5 and n_(i)= 6,7,8`... then wave number of pfund series lineed. `alpha, beta, gamma, delta`.,.... respectively obtained. The spectral line from all the different TRANSITIONS of the electron in hydrogen atom is SHOWN in figure below.  To remember the short of this series : LAY BAPA BE FUND means LA = For Lyman `n_(f) = 1` BA = For Balmer `n_(f) = 2` PA = For Paschen `n_(f) =3` BE = For Brackett `n_(f) = 4` FUND = For Pfund `n_(f)=5` |
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