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In Bohr series of lines of hydrogen spectrum,third line from the red end corresponds to which one of the following inner orbit jumps of electron of Bohr orbit in atom of hydrogen

$\begin{array}{1 1}(a)\;4\to 1&(b)\;2\to 5\\(c)\;3\to 2&(d)\;5\to 2\end{array}$

1 Answer

The emission is in visible (VIBGYOR) region.Thus Balmer series is either (c) or (d).In (c) only one emission .The third line will be in the jump from 5 to 12.
Hence (d) is the correct answer.
answered Jan 13, 2014 by sreemathi.v

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