Question

The shortest wavelength of the line in hydrogen atomic spectrum of Lyman series when $R_H = 109678 \text{ cm}^{-1}$ is}

• A. $1002.7 \text{ \AA}$
• B. $1215.67 \text{ \AA}$
• C. $1127.30 \text{ \AA}$
• D. $911.7 \text{ \AA}$
• E. $1234.7 \text{ \AA}$

Hint:

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 $1/R_H$.

Answer 1

The Correct Option is D

Concept: The Rydberg formula predicts the wavelengths of spectral lines in many chemical elements. For hydrogen, the wavenumber ($\bar{\nu}$) is given by: \[ \bar{\nu} = \frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
• Lyman Series: The series where transitions end at the ground state ($n_1 = 1$).
• Shortest Wavelength (Series Limit): This occurs when the transition is from the highest possible energy level ($n_2 = \infty$).

Step 1: Apply the condition for the shortest wavelength. For shortest wavelength in Lyman series, $n_1 = 1$ and $n_2 = \infty$. \[ \frac{1}{\lambda} = R_H \left( \frac{1}{1^2} - \frac{1}{\infty^2} \right) = R_H (1 - 0) = R_H \] \[ \lambda = \frac{1}{R_H} \]

Step 2: Calculate the numerical value. Given $R_H = 109678 \text{ cm}^{-1}$. \[ \lambda = \frac{1}{109678} \text{ cm} \approx 9.117 \times 10^{-6} \text{ cm} \] Convert to Angstroms ($1 \text{ cm} = 10^8 \text{ \AA}$): \[ \lambda = 9.117 \times 10^{-6} \times 10^8 = 911.7 \text{ \AA} \]