Question

The atomic spectrum of $L{i}^{2+}$ ion arises due to the transition of an electron from $n_2$ to $n_1$. If $n_2+n_1=4$ and $n_2-n_1=2$. The wavenumber of $3^{rd}$ line of this series in $L{i}^{2+}$ ion is found to be $x\times 135c{m}^{-1}$. Find the value of x. $\left(RydbergConstant\right(R_H)\approx 109600c{m}^{-1})$

Answer 1

$n_1+n_2=4n_2-n_1=2$
$\Rightarrow 2n_2=6$
$\therefore n_2=3n_1=1$
For the third line of same series
$n_2^{\prime }=4\to n_1=1$
By using the relation,
$\stackrel{-}{\nu }=\frac{1}{\lambda }=R_H{Z}^2[\frac{1}{n_1^2}-\frac{1}{(n_2^{\prime }{)}^2}]$
$=109600\times 9[\frac{1}{1^2}-\frac{1}{4^2}]=109600\times 9\times \frac{15}{16}=6850\times 135c{m}^{-1}$