Physics

Physics

Problem

An isolated star radiates total power P=3.8×1026 WP = 3.8\times10^{26}\,\text{W} and has radius R=7.0×108 mR = 7.0\times10^{8}\,\text{m}. Constants: σ=5.67×10−8 W m−2 K−4\sigma = 5.67\times10^{-8}\,\text{W}\,\text{m}^{-2}\,\text{K}^{-4}, b=2.90×10−3 m Kb = 2.90\times10^{-3}\,\text{m}\,\text{K}. What is the star’s peak-emission wavelength? a) 3.5×10−7 m3.5\times10^{-7}\,\text{m} b) 1.6×10−6 m1.6\times10^{-6}\,\text{m} c) 5.0×10−7 m5.0\times10^{-7}\,\text{m} d) 7.1×10−7 m7.1\times10^{-7}\,\text{m}
Intuitive explanation

The star radiates power according to the Stefan–Boltzmann law, which links total power to surface temperature—once you find TT, Wien's displacement law tells you the wavelength where the star shines brightest. The key is recognizing that all the radiated power spreads uniformly over the star's spherical surface, so you can solve for temperature first, then use the wavelength-temperature relationship.

Intuitive explanation

The star radiates power according to the Stefan–Boltzmann law, which links total power to surface temperature—once you find TT, Wien's displacement law tells you the wavelength where the star shines brightest. The key is recognizing that all the radiated power spreads uniformly over the star's spherical surface, so you can solve for temperature first, then use the wavelength-temperature relationship.

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