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In hydrogen atom an electronic transition starts from the orbit having radius 1.3225 nm and ends at an orbit with 211.6 pm . The region of spectrum is

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Detailed Solution

$r_{n} = 0.0529 nm \times n^{2} = \frac{1.3225}{0.0529} = 25 \Leftrightarrow n = 5 r_{n} = 52.9 pm \times n^{2} = \frac{211.6}{52.9} = 4 \Leftrightarrow n = 2 Transition is from n = 5 to n = 2 Balmer series, visible region$

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