Atomic StructureHard

Question

To what series does the spectral line of atomic hydrogen belong if its wave number is equal to the di­fference between the wave numbers of the following two lines of the Balmer series: 486.1 and 410.2 nm?

Options

A.Lyman series
B.Balmer series
C.Paschen series
D.Brackett series

Solution

Both are visible radiations. For required series, we get only n1.

Now, $\frac{1}{486.1 \times 10^{- 9}} = 1.09 \times 10^{7} \times 1^{2}\left( \frac{1}{2^{2}} - \frac{1}{n_{1}^{2}} \right) \Rightarrow n_{1} = 4$

Hence, the series is Brackett series.

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