Calculate the longest and shortest wavelengths of balmer series in hydrogen
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v` ={1/(n1)2-1/(n2)2} RH , where n1=2 , n2=3, RH=(1.097x10000000 1/m) answer is( 1.5236 x 1000000 1/m)
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Answer:
For the Balmer series, ni = 2. Thus, the expression of wavenumber is given by, Wave number is inversely proportional to wavelength of transition. Hence, for the longest wavelength transition, has to be the smallest.
Explanation:
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