Question

One of the lines in the emission spectrum of ${\text{Li}}^{2+}$ has the same wavelength as that of $2^{nd}$ line of Balmer series in hydrogen spectrum. The electronic transition corresponding to this line is :-

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

Answer 1

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

From Rydberg's formula,

$\frac{1}{\lambda }=R{Z}^2(\frac{1}{n_f^2}-\frac{1}{n_i^2})$

For $2^{nd}$ line of Balmer series in hydrogen spectrum,

$Z=1,n_f=2,n_i=4$

$\Rightarrow \frac{1}{\lambda }=R(1{)}^2(\frac{1}{2^2}-\frac{1}{4^2})=\frac{3R}{16}$

Now, for ${\text{Li}}^{2+}$, considering option $\left(D\right)$,
$Z=3,n_1=6,n_2=12$

$\Rightarrow \frac{1}{\lambda }=R(3{)}^2(\frac{1}{n_f^2}-\frac{1}{n_i^2})$

$\Rightarrow \frac{3R}{16}=9R(\frac{1}{n_f^2}-\frac{1}{n_i^2})$

$\Rightarrow \frac{1}{48}=(\frac{1}{n_f^2}-\frac{1}{n_i^2})$

Using hit and trial method, we get

$n_i=12;n_f=6$

Hence, option $\left(\text{D}\right)$ is correct.