Star A has a surface temperature of 8000K and its thermal emission peaks around 360nm. Star B is twice as cold, with a temperature of 4000K. What wavelength does its curve peak?
The correct answer and explanation is:
To find the peak wavelength of thermal emission for a star, we use Wien’s Displacement Law, which states: λmax=bT\lambda_{\text{max}} = \frac{b}{T}
Where:
- λmax\lambda_{\text{max}} is the peak wavelength in meters,
- TT is the surface temperature in Kelvin (K),
- bb is Wien’s displacement constant, approximately 2.898 × 10⁻³ m·K.
Step 1: Apply Wien’s Law to Star B
Given:
- Temperature of Star B = 4000 K
- b=2.898×10−3 m\cdotpKb = 2.898 \times 10^{-3} \, \text{m·K}
λmax, B=2.898×10−34000=7.245×10−7 m\lambda_{\text{max, B}} = \frac{2.898 \times 10^{-3}}{4000} = 7.245 \times 10^{-7} \, \text{m}
Convert to nanometers (nm): 7.245×10−7 m=724.5 nm7.245 \times 10^{-7} \, \text{m} = 724.5 \, \text{nm}
✅ Final Answer: The peak wavelength of Star B is approximately 724.5 nanometers (nm).
Explanation (≈300 words):
Wien’s Displacement Law explains how the temperature of a blackbody (like a star) determines the wavelength at which it emits most of its radiation. This law shows an inverse relationship between temperature and peak wavelength: as the temperature increases, the peak wavelength shifts to shorter (bluer) wavelengths; as the temperature decreases, the peak shifts to longer (redder) wavelengths.
In this problem, Star A has a surface temperature of 8000 K and peaks at 360 nm, which is in the ultraviolet part of the spectrum. Star B is twice as cool, at 4000 K. Since the temperature is halved, the peak wavelength should double, which gives a good estimate even before calculating.
Using Wien’s Law: λmax=2.898×10−34000=724.5 nm\lambda_{\text{max}} = \frac{2.898 \times 10^{-3}}{4000} = 724.5 \, \text{nm}
This places Star B’s peak in the red/infrared part of the spectrum. That means Star B appears redder than Star A.
This concept is important in astronomy because it helps classify stars based on color and temperature. Hotter stars (blue or white) emit more energy at shorter wavelengths, while cooler stars (red or orange) emit more at longer wavelengths.
Understanding these temperature-wavelength relationships helps astronomers determine a star’s properties by just analyzing its spectrum.