For CaCl2, the dominant flame color is ______ and the wavelength of that color is ________.
对于 CaCl2，主导火焰颜色为______，该颜色的波长为________。

Question

BlackTom AI · Accepted Answer

Calcium ions in a flame emit primarily in the orange–yellow part of the visible spectrum, with a dominant emission peak near 597 nm.

Relevant concepts: Flame tests rely on electronic excitation of metal cations. Thermal energy promotes valence electrons to higher levels; when they relax back down they emit photons at characteristic wavelengths. Each metal ion has a set of spectral lines; the perceived flame color is the strongest visible emission(s) blended by the eye.

Step-by-step reasoning:
1. When CaCl2 is placed in a flame, Ca2+ ions are the chromophore. Heating excites electrons in Ca2+ and permitted electronic transitions produce emission lines in the visible.
2. Spectroscopic measurements of calcium-containing flames show the strongest visible emission in the yellow–orange region, centered around 597 nm. That wavelength corresponds to a photon frequency of:
   nu = c / lambda = (3.00 × 10^8 m/s) / (597 × 10^-9 m) ≈ 5.02 × 10^14 Hz.
3. The eye perceives radiation near 597 nm as orange–yellow, consistent with standard descriptions of calcium flame color (often described as brick-red to orange-yellow, depending on conditions).

Why this option is correct: The option "orange-yellow, 597 nm" directly matches the dominant visible emission observed for calcium salts like CaCl2. The wavelength 597 nm falls squarely in the orange–yellow portion of the spectrum and corresponds to the strongest visible line for Ca under typical flame-test conditions.

Why the other options are incorrect (briefly):
- ~589–587 nm (yellow) is characteristic of sodium (Na), not Ca.
- Violet/blue lines (423–474 nm) or green/blue (492–535 nm) correspond to elements such as K (violet component) or Cu/Ba (green/blue), not calcium.
- Deep red (≈701 nm) or red–orange (~609–622 nm) are more characteristic of other alkaline earths or different ions (e.g., Sr or Li produce reddish colors).
  
Conclusion: The correct, experimentally supported identification is orange-yellow, 597 nm, which matches the provided answer.

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For CaCl2, the dominant flame color is ______ and the wavelength of that color is ________. 对于 CaCl2，主导火焰颜色为______，该颜色的波长为________。多项选择题