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

Answer 1

The correct option is D 1080 nm

According to the given data
$n_1+n_2=4$....(i)
$n_2-n_1=2$....(ii)
Adding equation (i) and (ii) we get
$2n_2=6$
$\therefore n_2=3$
From equation (i),
$n_1+3=4$
$\therefore n_1=1$
Therefore, the series is Lyman series.
The 3${}^{rd}$ line of this series will have a transition from
$n_2=4\to n_1=1$

We know,
$\frac{1}{\lambda }=R{Z}^2[\frac{1}{n_1^2}-\frac{1}{n_2^2}]$
$=1.097\times {10}^5\times 9[\frac{1}{1^2}-\frac{1}{4^2}]c{m}^{-1}$
$=9.873\times {10}^5\times \frac{15}{16}$
$=925593.75c{m}^{-1}$
$\Rightarrow \lambda =1.0803\times {10}^{-6}$ cm or $1080$ nm