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Get Answers to all your Questions
Calculate the wavenumber for the longest wavelength transition in the Balmer series of atomic hydrogen?
Answers (1)
In Balmer series, n1 = 2. Hence, á¹½ = R(1/22 - 1/n22)
á¹½ = 1/λ i.e it is inversely proportional
For λ to be maximum, á¹½ should be minimum.
This is possible only when n2 is minimum i.e. n2 = 3. Hence, á¹½ = (1.097×107 m-1) (1/22 - 1/32) = 1.097×107×5/36 m-1 = 1.523×106 m-1
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