$(a)\;5\qquad(b)\;3\qquad(c)\;6\qquad(d)\;1$

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$\Delta E=8.4375$

$R_h,E_1Li^{2+}=-R_h\times 9$

$R_h,E_4Li^{2+}=-\large\frac{R_h\times 9}{16}$

$E_1-E_4=\large\frac{9R_h}{1}-\frac{9R_h}{16}$

$\Rightarrow \large\frac{135R_h}{16}$

$\Rightarrow 8.4375R_h$

Thus de-excitation will lead from $4^{th}$ to $1^{st}$ orbit.

$\therefore $ No of spectral lines =$\sum \Delta n=\sum (4-1)=\sum 3$

$\Rightarrow 1+2+3=6$

Hence (c) is the correct answer.

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