Calculate the wavenumber for the longest wavelength transition in the Balmer series of atomic hydrogen?

Answers (1)
S Sumit

In Balmer series, n1 = 2. Hence, á¹½ = R(1/2- 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 = (1.097×10m-1) (1/2- 1/32) = 1.097×107×5/36 m-1 = 1.523×10m-1

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