Question:
The energy of spectral line of lowest frequency in Lyman series of $Li^{2+}$ spectrum is x J. The energy of second spectral line in Balmer series of $He^{+}$ spectrum is y J. The ratio of x and y is
The energy of spectral line of lowest frequency in Lyman series of $Li^{2+}$ spectrum is x J. The energy of second spectral line in Balmer series of $He^{+}$ spectrum is y J. The ratio of x and y is
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Lowest frequency means lowest energy, which corresponds to an adjacent level transition ($\Delta n = 1$). Always square the atomic number ($Z^2$) first!
Updated On: Jun 3, 2026
- 3:1
- 1:3
- 1:9
- 9:1
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The Correct Option is D
Solution and Explanation
Step 1: Concept
The energy of a photon emitted during an electronic transition is given by $E = R_{H} \cdot h \cdot c \cdot Z^2 \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$, which means $E \propto Z^2 \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$.
Step 2: Meaning
For the lowest frequency line, the energy must be at its minimum value. In the Lyman series ($n_1 = 1$), the lowest energy transition is from $n_2 = 2$. The second line in the Balmer series ($n_1 = 2$) results from a transition from $n_2 = 4$.
Step 3: Analysis
For the $Li^{2+}$ ion ($Z = 3$), the energy $x$ of the lowest frequency Lyman line is: $$x \propto 3^2 \left(\frac{1}{1^2} - \frac{1}{2^2}\right) = 9 \cdot \left(1 - \frac{1}{4}\right) = 9 \cdot \frac{3}{4} = \frac{27}{4}$$ For the $He^{+}$ ion ($Z = 2$), the energy $y$ of the second Balmer line is: $$y \propto 2^2 \left(\frac{1}{2^2} - \frac{1}{4^2}\right) = 4 \cdot \left(\frac{1}{4} - \frac{1}{16}\right) = 4 \cdot \frac{3}{16} = \frac{3}{4}$$ Taking the ratio of $x$ and $y$: $$\frac{x}{y} = \frac{\frac{27}{4}}{\frac{3}{4}} = \frac{27}{3} = \frac{9}{1}$$
Step 4: Conclusion
The required ratio of $x$ and $y$ is 9:1.
Final Answer: (D)
The energy of a photon emitted during an electronic transition is given by $E = R_{H} \cdot h \cdot c \cdot Z^2 \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$, which means $E \propto Z^2 \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$.
Step 2: Meaning
For the lowest frequency line, the energy must be at its minimum value. In the Lyman series ($n_1 = 1$), the lowest energy transition is from $n_2 = 2$. The second line in the Balmer series ($n_1 = 2$) results from a transition from $n_2 = 4$.
Step 3: Analysis
For the $Li^{2+}$ ion ($Z = 3$), the energy $x$ of the lowest frequency Lyman line is: $$x \propto 3^2 \left(\frac{1}{1^2} - \frac{1}{2^2}\right) = 9 \cdot \left(1 - \frac{1}{4}\right) = 9 \cdot \frac{3}{4} = \frac{27}{4}$$ For the $He^{+}$ ion ($Z = 2$), the energy $y$ of the second Balmer line is: $$y \propto 2^2 \left(\frac{1}{2^2} - \frac{1}{4^2}\right) = 4 \cdot \left(\frac{1}{4} - \frac{1}{16}\right) = 4 \cdot \frac{3}{16} = \frac{3}{4}$$ Taking the ratio of $x$ and $y$: $$\frac{x}{y} = \frac{\frac{27}{4}}{\frac{3}{4}} = \frac{27}{3} = \frac{9}{1}$$
Step 4: Conclusion
The required ratio of $x$ and $y$ is 9:1.
Final Answer: (D)
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