Question:
The shortest wavelength in the Lyman series of hydrogen \([R_H = 1.097 \times 10^7~\text{m}^{-1}]\) is:
The shortest wavelength in the Lyman series of hydrogen \([R_H = 1.097 \times 10^7~\text{m}^{-1}]\) is:
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For Lyman series, shortest wavelength occurs for transition from \(n = \infty\) to \(n = 1\). Use \(\lambda_\text{min} = 1/R_H\).
Updated On: Jun 19, 2026
- 91.2 nm
- 364.6 nm
- 820.4 nm
- 2278.9 nm
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The Correct Option is A
Solution and Explanation
Step 1: Lyman series formula.
\[ \frac{1}{\lambda} = R_H \left(1 - \frac{1}{n^2}\right), \quad n > 1 \] Shortest wavelength occurs for \(n \to \infty\).
Step 2: Substitute limit.
\[ \frac{1}{\lambda_\text{min}} = R_H (1 - 0) = R_H \]
Step 3: Solve for \(\lambda_\text{min}\).
\[ \lambda_\text{min} = \frac{1}{R_H} = \frac{1}{1.097 \times 10^7} \approx 9.12 \times 10^{-8}~\text{m} = 91.2~\text{nm} \]
Step 4: Conclusion.
The shortest wavelength in the Lyman series is 91.2 nm.
\[ \frac{1}{\lambda} = R_H \left(1 - \frac{1}{n^2}\right), \quad n > 1 \] Shortest wavelength occurs for \(n \to \infty\).
Step 2: Substitute limit.
\[ \frac{1}{\lambda_\text{min}} = R_H (1 - 0) = R_H \]
Step 3: Solve for \(\lambda_\text{min}\).
\[ \lambda_\text{min} = \frac{1}{R_H} = \frac{1}{1.097 \times 10^7} \approx 9.12 \times 10^{-8}~\text{m} = 91.2~\text{nm} \]
Step 4: Conclusion.
The shortest wavelength in the Lyman series is 91.2 nm.
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