Question

Which electronic transition in $L{i}^{2+}$ ion would emit radiation of same wavelength as the wavelength of second

Balmer line of H - atom?

• A. $12\to 9$
• B. $12\to 6$
• C. $4\to 3$
• D. $6\to 3$

Answer 1

The correct option is B $12\to 6$

$R_H[\frac{1}{2^2}-\frac{1}{4^2}]=R_H[\frac{9}{n_1^2}-\frac{9}{n_2^2}]$ Putting $n_1=6,n_2=12$ makes both side equal.