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.019 1. When Johann Balmer found his famous series for hydrogen in wavelengths in the visible and near ultraviolet regions fr
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1) longest possible wavelength associated transition.it is having lowest energy in the series.

so that, in balmer series lowest possible energy transition is between 2 and 3 levels

wavenumber = 1/wavelength = 109677(1/n1^2-1/n2^2) cm-1

     n1 = 2 , n2 = 3

   1/wavelength = 109677(1/2^2-1/3^2)

            1/l = 15233 cm-1

        l = wavelength = 1/15233 = 6.56*10^-5 cm

   wavelength = 6.56*10^-5*10^7 = 656 nm

2) shortest wavelength associated transition.it is having highest energy in the series.

so that, in balmer series highest possible energy transition is between 2 and infinity levels

wavenumber = 1/wavelength = 109677(1/n1^2-1/n2^2) cm-1

     n1 = 2 , n2 = infinity

   1/wavelength = 109677(1/2^2-1/(infinity)^2)

            1/l = 27419.25 cm-1

        l = wavelength = 1/27419.25 = 3.65*10^-5 cm

   wavelength = 3.65*10^-5*10^7 = 365 nm

Balmer series lines are present in visible region.so that its wavelengths are in the range of 365 to 656 nm.

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