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Step-by-Step Solution
Step 1: Recall the Rydberg Formula for Hydrogen-like Ions
For a hydrogen-like ion, the wavelength $ \lambda $ corresponding to an electronic transition from
$ n_2 $ to $ n_1 $ (with $ n_2 > n_1 $) is given by:
$ \frac{1}{\lambda} = R \, Z^2 \Bigl(\frac{1}{n_1^2} - \frac{1}{n_2^2}\Bigr), $
where:
• $ R $ is the Rydberg constant,
• $ Z $ is the atomic number of the ion,
• $ n_1 $ and $ n_2 $ are principal quantum numbers (integers).
Step 2: Identify the Second Balmer Line in Hydrogen
In the hydrogen atom ($ Z = 1 $), the Balmer series corresponds to transitions ending at $ n_1 = 2 $.
The second line of this series is given by:
$ n_2 = 4 \quad \longrightarrow \quad n_1 = 2.
The wavelength $ \lambda_{\text{Balmer}} $ for this transition is:
$ \frac{1}{\lambda_{\text{Balmer}}}
= R \times 1^2 \Bigl(\frac{1}{2^2} - \frac{1}{4^2}\Bigr)
= R \Bigl(\frac{1}{4} - \frac{1}{16}\Bigr)
= R \Bigl(\frac{4 - 1}{16}\Bigr)
= \frac{3R}{16}.
Step 3: Match This Wavelength with Li2+ Emission
For Li2+ ($ Z = 3 $), the wavelength $ \lambda_{\text{Li}^{2+}} $ of any line is given by:
$ \frac{1}{\lambda_{\text{Li}^{2+}}}
= R \times 3^2 \Bigl(\frac{1}{n_1^2} - \frac{1}{n_2^2}\Bigr)
= 9R \Bigl(\frac{1}{n_1^2} - \frac{1}{n_2^2}\Bigr).
We want this to be equal to $ \frac{3R}{16} $ (the value for the second Balmer line), so we set:
$ 9 \Bigl(\frac{1}{n_1^2} - \frac{1}{n_2^2}\Bigr) = \frac{3}{16}, $
which ultimately leads to the transition:
$ n_2 = 12 \quad \longrightarrow \quad n_1 = 6. $
Step 4: Final Conclusion
Hence, the emission line in Li2+ that has the same wavelength as the second Balmer line
in hydrogen corresponds to:
$ n = 12 \;\longrightarrow\; n = 6. $
