Calculate the wave number for the longest wavelength transition in the Balmer series of atomic hydrogen.
According to Balmer formula
ṽ=1/λ = RH[1/n12-1/n22]
For the Balmer series, ni = 2.
Thus, the expression of wavenumber(ṽ) is given by,
Wave number (ṽ) is inversely proportional to wavelength of transition. Hence, for the longest wavelength transition, ṽ has to be the smallest.
For ṽ to be minimum, nf should be minimum. For the Balmer series, a transition from ni = 2 to nf = 3 is allowed. Hence, taking nf = 3,we get:
ṽ= 1.5236 × 106 m–1
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Thanks
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This is correct ans
Wait then value is 1.097.isit right
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