Question

The atomic spectrum of $L{i}^{2+}$ ion arises due to the transition of an electron from $n_2$ to $n_1$. If $n_1+n_2=4$ and $(n_2-n_1)=2$, find the wavelength of $3^{rd}$ line of this series in $L{i}^{2+}$ ion.

• A. $1.08nm$
• B. $10.8nm$
• C. $108nm$
• D. $1080nm$

Answer 1

The correct option is B $10.8nm$

$n_1+n_2=4$
$n_2-n_1=2$

Solving both we get, $n_2=3$ and $n_1=1$ (Lyman series)

For the $3^{rd}$ line of the same series;
$n_4=4\to n_1=1$

$\frac{1}{\lambda }=R{Z}^2[\frac{1}{n_1^2}-\frac{1}{n_2^2}]=R\times 9[\frac{1}{1^2}-\frac{1}{4^2}]=92559375/m$

$\lambda =10.80nm$