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One of the lines in the emission spectrum of $\mathrm{Li}^{2+}$ has the same wavelength as that of the $2^{\text {nd }}$ line of Balmer series in hydrogen spectrum. The electronic transition corresponding to this line is $\mathrm{n}=12 \rightarrow \mathrm{n}=\mathrm{x} .$ Find the value of $\mathrm{x}$
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The correct answer is:
6
For $2^{\text {nd line of Balmer series in hydrogen }}$ spectrum
$$
\frac{1}{\lambda}=\mathrm{R}(1)\left(\frac{1}{2^{2}}-\frac{1}{4^{2}}\right)=\frac{3}{16} \mathrm{R}
$$
$$
\text { For } \mathrm{Li}^{2+}\left[\frac{1}{\lambda}=\mathrm{R} \times 9\left(\frac{1}{\mathrm{x}^{2}}-\frac{1}{12^{2}}\right)=\frac{3 \mathrm{R}}{16}\right]
$$
which is satisfied by $n=12 \rightarrow n=6$
$$
\frac{1}{\lambda}=\mathrm{R}(1)\left(\frac{1}{2^{2}}-\frac{1}{4^{2}}\right)=\frac{3}{16} \mathrm{R}
$$
$$
\text { For } \mathrm{Li}^{2+}\left[\frac{1}{\lambda}=\mathrm{R} \times 9\left(\frac{1}{\mathrm{x}^{2}}-\frac{1}{12^{2}}\right)=\frac{3 \mathrm{R}}{16}\right]
$$
which is satisfied by $n=12 \rightarrow n=6$
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