$(a)\;1\qquad(b)\;2\qquad(c)\;3\qquad(d)\;4$

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$E_n=-13.6+12.1=1.5eV$

$\therefore n^2=\large\frac{E_1}{E_n}$

$\Rightarrow \large\frac{13.6}{1.5}$$=9$

$\Rightarrow n=3$

Thus transition occurs from $3^{rd}$ to $1^{st}$ orbit.

$\therefore$ spectral lines emitted $=\sum \Delta n$

$\Rightarrow \sum(3-1)$

$\Rightarrow \sum(2)$

$\Rightarrow 1+2=3$

Hence (c) is the correct answer.

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