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Calculate the wave number for the longest wavelength transition in the Balmer series of atomic hydrogen.

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Wave number \(\bar{V}\) = 109677 \(\big(\frac{1}{n_1^2}-\frac{1}{n_2^2}\big)\)

n1 = 2

n2 = 3 For Balmer series

\(\therefore\) \(\bar{V}\) = 109677 \(\Big(\frac{1}{(2)^2}-\frac{1}{(3)^2}\Big)\)

= 109677 \(\Big(\frac{1}{4}-\frac{1}{9}\Big)\)

= 109677 \(\Big(\frac{9-4}{36}\Big)\)

= 109677 x \(\frac{5}{36}\)

= 1.523 × 104 cm−1

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