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The electron in $Li^{2+}$ ions are excited from ground state by absorbing $8.4375R_h$ energy/electron.How much emission lines are expected during de-excitation of electrons to ground state ?

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

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1 Answer

$\Delta E=8.4375$
$R_h,E_1Li^{2+}=-R_h\times 9$
$R_h,E_4Li^{2+}=-\large\frac{R_h\times 9}{16}$
$E_1-E_4=\large\frac{9R_h}{1}-\frac{9R_h}{16}$
$\Rightarrow \large\frac{135R_h}{16}$
$\Rightarrow 8.4375R_h$
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.
answered Jan 17, 2014 by sreemathi.v
 

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