What is the maximum number of emission lines obtained when the excited electron of a H atom in n= 5 drops to the ground state?

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Solution

The correct option is D 10
Maximum number of emission lines from $n^{t h}$ energy state to ground state is given by:
$= \frac{n (n - 1)}{2} = \frac{5 (5 - 1)}{2} = \frac{20}{2} = 10$ lines.
option D

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