Question

One of the lines in the emission spectrum of $L{i}^{2+}$ has the same wavelength as that of the second line of Balmer series in hydrogen spectrum, The electronic transition corresponding to this line is

• A. $n=4\to n=2$
• B. $n=8\to n=2$
• C. $n=8\to n=4$
• D. $n=12\to n=6$

Answer 1

The correct option is D $n=12\to n=6$

$\frac{1}{\lambda }=Z^{2}R(\frac{1}{n_1^2}-\frac{1}{n_2^2})$

For $Li,Z=3$
$\frac{1}{\lambda }=3^{2}R(\frac{1}{n_1^2}-\frac{1}{n_2^2})$

For $H,Z=1$, second line of Balmer series
$\frac{1}{\lambda }=R(\frac{1}{2^2}-\frac{1}{4^2})$

Equating:
$\frac{3}{16}=9(\frac{1}{n_1^2}-\frac{1}{n_2^2})$
$\frac{1}{48}=(\frac{1}{n_1^2}-\frac{1}{n_2^2})$

For integral values of $n_1$ and $n_2$,
$n_1=6$, and $n_2=12$