Difficulty: Easy
Correct Answer: 0.55 μm
Explanation:
Introduction / Context:
Understanding where the solar spectrum peaks is fundamental in remote sensing, photovoltaic design, and human vision considerations. The Sun approximates a blackbody emitter near 5770 K, so its spectral irradiance peaks in the visible range near green light.
Given Data / Assumptions:
Concept / Approach:
By Wien’s displacement principle, the peak wavelength is λ_max ≈ 2898 μm·K / T. For T ≈ 5770 K, λ_max ≈ 0.50 μm. Accounting for actual solar spectrum and common textbook rounding, many references cite ≈ 0.55 μm (green). Remote-sensing bands and camera sensors are aligned with this visible peak for maximal signal.
Step-by-Step Solution:
Estimate λ_max using λ_max = 2898 / 5770 ≈ 0.50 μm.Recognize practical peak in extraterrestrial/atmospheric curves appears around 0.50–0.55 μm.Select the standard rounded value 0.55 μm as the best choice.Reject alternatives at 0.40 μm (violet) and 0.70 μm (red) as off-peak.
Verification / Alternative check:
Solar irradiance standards (AM0/AM1.5) show maximum visible-region energy around green wavelengths, consistent with the 0.5–0.55 μm range.
Why Other Options Are Wrong:
0.40 μm: Too short (near violet); intensity is lower than at green.0.70 μm: Too long (red); descending limb of solar curve.1.00 μm: Near-infrared; below the visible peak.None of these: Incorrect since 0.55 μm is appropriate.
Common Pitfalls:
Confusing human eye sensitivity peak with physical solar peak; fortunately both are close in the green region, aiding daylight vision efficiency.
Final Answer:
0.55 μm.
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