Sirius B. The brightest star in the sky is Sirius, the Dog Star. It is actually a binary system of two stars, the smaller one (Sirius B) being a white dwarf. Spectral analysis of Sirius B indicates that its surface temperature is 24,000 K and that it radiates energy at a total rate of l.0 x 1025 W. Assume that it behaves like an ideal blackbody.
(a) What is the total radiated intensity of Sirius B?
(b) What is the peak-intensity wavelength? Is this wavelength visible to humans?
(c) What is the radius of Sirius B? Express your answer in kilometers and as a fraction of our sun’s radius.
(d) Which star radiates more total energy per second, the hot Sirius B or the (relatively) cool sun with a surface temperature of 5800 K? To find out, calculate the ratio of the total power radiated by our sun to the power radiated by Sirius B.
The Correct Answer and Explanation is:
To answer the questions about Sirius B, we’ll use the Stefan–Boltzmann law and Wien’s displacement law, assuming it is a perfect blackbody.
Given:
- Surface temperature, T=24,000 KT = 24{,}000 \, \text{K}
- Total radiated power (luminosity), L=1.0×1025 WL = 1.0 \times 10^{25} \, \text{W}
- Stefan–Boltzmann constant: σ=5.67×10−8 W/m2⋅K4\sigma = 5.67 \times 10^{-8} \, \text{W/m}^2\cdot\text{K}^4
- Sun’s surface temperature: T⊙=5800 KT_{\odot} = 5800 \, \text{K}
- Sun’s luminosity: L⊙=3.828×1026 WL_{\odot} = 3.828 \times 10^{26} \, \text{W}
- Sun’s radius: R⊙=6.96×105 kmR_{\odot} = 6.96 \times 10^5 \, \text{km}
(a) Total Radiated Intensity
Intensity II is the power per unit area, given by the Stefan–Boltzmann law: I=σT4=(5.67×10−8)⋅(24000)4=1.88×109 W/m2I = \sigma T^4 = (5.67 \times 10^{-8}) \cdot (24000)^4 = 1.88 \times 10^9 \, \text{W/m}^2
(b) Peak-Intensity Wavelength
Use Wien’s displacement law: λmax=bT,b=2.898×10−3 m\cdotpK\lambda_{\text{max}} = \frac{b}{T}, \quad b = 2.898 \times 10^{-3} \, \text{m·K} λmax=2.898×10−324000=1.21×10−7 m=121 nm\lambda_{\text{max}} = \frac{2.898 \times 10^{-3}}{24000} = 1.21 \times 10^{-7} \, \text{m} = 121 \, \text{nm}
This is in the ultraviolet range, not visible to the human eye (which sees 380–750 nm).
(c) Radius of Sirius B
Using the luminosity equation: L=4πR2σT4⇒R=L4πσT4L = 4\pi R^2 \sigma T^4 \Rightarrow R = \sqrt{\frac{L}{4\pi \sigma T^4}} R=1.0×10254π(5.67×10−8)(24000)4=1.0×10252.36×1011R = \sqrt{\frac{1.0 \times 10^{25}}{4\pi (5.67 \times 10^{-8}) (24000)^4}} = \sqrt{\frac{1.0 \times 10^{25}}{2.36 \times 10^{11}}} R≈4.24×1013≈6.51×106 m=6510 kmR \approx \sqrt{4.24 \times 10^{13}} \approx 6.51 \times 10^6 \, \text{m} = 6510 \, \text{km}
As a fraction of the sun’s radius: RR⊙=65106.96×105≈0.0094\frac{R}{R_{\odot}} = \frac{6510}{6.96 \times 10^5} \approx 0.0094
(d) Comparison of Total Radiated Power
L⊙LSirius B=3.828×10261.0×1025=38.3\frac{L_{\odot}}{L_{\text{Sirius B}}} = \frac{3.828 \times 10^{26}}{1.0 \times 10^{25}} = 38.3
So, the Sun radiates about 38 times more energy per second than Sirius B, despite its lower temperature.
Sirius B, the smaller component of the Sirius binary system, is a white dwarf star with extreme physical characteristics. Given a surface temperature of 24,000 K and assuming it behaves as an ideal blackbody, we use Stefan–Boltzmann’s law to determine its total radiated intensity (power per unit area). Plugging in the values yields an intensity of approximately 1.88×109 W/m21.88 \times 10^9 \, \text{W/m}^2, highlighting its incredible energy output per square meter.
To determine the wavelength at which Sirius B radiates most intensely, Wien’s displacement law is applied. The resulting peak wavelength is 121 nm, placing it in the ultraviolet region, which is invisible to the human eye. This explains why Sirius B, though intensely hot, does not significantly contribute to visible starlight on its own.
Next, by rearranging the luminosity formula for a blackbody and solving for radius, we find Sirius B’s radius to be about 6510 km. This is only about 0.94% the radius of our sun, consistent with its classification as a white dwarf—extremely dense and compact.
Finally, despite Sirius B’s high temperature, the sun radiates far more total energy per second—about 38 times more. This is due to the sun’s much larger surface area, which more than compensates for its cooler temperature. This contrast vividly demonstrates how surface area plays a crucial role in stellar energy output, and that high temperature alone doesn’t guarantee high luminosity.
