In the hydrogen emission spectrum, which series contains the line with the smallest wavelength?

• A. Paschen series
• B. Brackett series
• C. Balmer series
• D. Lyman series

Answer: (D)

Verified step by step guidance

1. Understand that the hydrogen emission spectrum consists of several series of spectral lines, each corresponding to electron transitions ending at a specific energy level (n).
2. Recall that the wavelength of emitted light is related to the energy difference between the initial and final energy levels by the formula \(\lambda = \frac{hc}{\Delta E}\), where \(h\) is Planck's constant, \(c\) is the speed of light, and \(\Delta E\) is the energy difference.
3. Identify the series by their final energy levels: Lyman series ends at \(n=1\), Balmer at \(n=2\), Paschen at \(n=3\), and Brackett at \(n=4\).
4. Recognize that the largest energy difference (and thus the smallest wavelength) occurs when the electron falls to the lowest energy level, which is \(n=1\) in the Lyman series.
5. Conclude that the Lyman series contains the lines with the smallest wavelengths because transitions to \(n=1\) involve the greatest energy changes.