Question

One of the lines in the emission spectrum of $L{i}^{2+}$ has the same wavelength as that of the $2^{nd}$ line of Balmer series in hydrogen spectrum. The electronic transition corresponding to this line is $n=12\to n=x$. Find the value of $x$.

• A. $8$
• B. $6$
• C. $7$
• D. $5$

Answer 1

Answer: (B)

$6$

For $2^{nd}$ line of Balmer series in hydrogen spectrum
$\frac{1}{\lambda }=R\left(1\right)(\frac{1}{2^2}-\frac{1}{4^2})=\frac{3}{16}R$
For $L{i}^{2+}[\frac{1}{\lambda }=R\times 9(\frac{1}{x^2}-\frac{1}{{12}^2})=\frac{3R}{16}]$
Which is satisfied by $n=12\to n=6$.