Question:medium
The shortest wavelength of the line in hydrogen atomic spectrum of Lyman series when $R_H = 109678 \text{ cm}^{-1}$ is}
The shortest wavelength of the line in hydrogen atomic spectrum of Lyman series when $R_H = 109678 \text{ cm}^{-1}$ is}
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Shortest wavelength = highest energy = transition from infinity. For any series $n$, the series limit wavelength is simply $\lambda = n^2 / R_H$. For Lyman, it's $1/R_H$.
Updated On: May 6, 2026
- $1002.7 \text{ \AA}$
- $1215.67 \text{ \AA}$
- $1127.30 \text{ \AA}$
- $911.7 \text{ \AA}$
- $1234.7 \text{ \AA}$
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The Correct Option is D
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