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

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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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