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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