Chapter 39: Problem 41
Two stars, both of which behave like ideal blackbodies, radiate the same total energy per second. The cooler one has a surface temperature \(T\) and a diameter 3.0 times that of the hotter star. (a) What is the temperature of the hotter star in terms of \(T\) ? (b) What is the ratio of the peak-intensity wavelength of the hot star to the peak-intensity wavelength of the cool star?
Short Answer
Step by step solution
Understand the Stefan-Boltzmann Law
Express the Power for Each Star
Equate the Powers
Solve for the Temperature of the Hotter Star
Understand Wien’s Displacement Law
Calculate the Ratio of Peak Wavelengths
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Stefan-Boltzmann Law
\[P = \sigma A T^4\]
where:
- \( P \) is the power emitted,
- \( \sigma \) is the Stefan-Boltzmann constant, approximately \( 5.67 \times 10^{-8} \text{ W/m}^2\text{K}^4 \),
- \( A \) is the surface area of the blackbody, and
- \( T \) is the temperature in Kelvin.
Wien's Displacement Law
\[\lambda_{max} = \frac{b}{T}\]
where:
- \( \lambda_{max} \) is the peak wavelength,
- \( b \) is Wien's displacement constant, approximately \( 2.898 \times 10^{-3} \text{ m K} \), and
- \( T \) is the temperature in Kelvin.
Peak-intensity wavelength ratio
For the hotter star:\[\lambda_1 = \frac{b}{T_1}\]
For the cooler star:\[\lambda_2 = \frac{b}{T}\]
This allows us to calculate the ratio of their peak wavelengths:
\[\frac{\lambda_1}{\lambda_2} = \frac{T}{T_1}\]
From our exercise, with the temperature of the hotter star being \( T \sqrt{3} \), the ratio simplifies to:\[\frac{1}{\sqrt{3}}\]
This tells us that the peak wavelength of the hotter star is shorter than that of the cooler star, precisely by the factor \( \sqrt{3} \). Such ratios are crucial in astrophysics, as they enable the understanding of the relative temperatures of celestial bodies through observational data.