Problem#1
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 (b) What is the ratio of the peak-intensity wavelength of the hot star to the peak-intensity wavelength of the cool star?
Answer:
Since the stars radiate as blackbodies, they obey the Stefan-Boltzmann law and Wien’s displacement law.
The Stefan-Boltzmann law says that the intensity of the radiation is I = σT4, so the total radiated power is P = eAσT4 . Wien’s displacement law tells us that the peak-intensity wavelength is λm = (constant)/T.
(a) The hot and cool stars radiate the same total power, so the Stefan-Boltzmann law gives
AhσTh4 = AcσTc4
4πRh2Th4 = 4π(3Rh)2Tc4
Th4 = 9T4
Th = T√3 = 1.7T, rounded to two significant digits.
(b) Using Wien’s law, we take the ratio of the wavelengths, giving
λm, (hoot)/λm, (cool) = Tc/Th = T/(T√3) = 0.58
Although the hot star has only1/9 the surface area of the cool star, its absolute temperature has to be only 1.7 times as great to radiate the same amount of energy.
Problem#2
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 1.0 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.
Answer:
Since the stars radiate as blackbodies, they obey the Stefan-Boltzmann law.
The Stefan-Boltzmann law says that the intensity of the radiation is I = σT4, so the total radiated power is
(a) I = σT4 = (5.67 X 10-8 W/m2.K4)(24000 K)4 = 1.9 x 1010 W/m2
(b) Wien’s law gives λm = (0.00290 m.K)/(24000 K) = 1.2 x 10-7 m = 20 nm
This is not visible since the wavelength is less than 400 nm
(c) P = AI à 4πR2 = P/I = (1.00 x 1025 W)/(1.9 x 1010 W/m2)
Which gives Rsirius = 6.51 x 106 m = 6510 km
(d) Using the Stefan-Boltzmann law, we have
Psun/Psirius = σAsunTsun4/σAsiriusTsirius4
Psun/Psirius = {σ(4π)Rsun2Tsun4}/{σ(4π)Rsun2Tsirius4}
Psun/Psirius = Rsun2Tsun4/Rsun2Tsirius4
Psun/Psirius = (Rsun/0.00935Rsun)2(5800 K/24000K)4 = 39
Even though the absolute surface temperature of Sirius B is about 4 times that of our sun, it radiates only 1/39 times as much energy per second as our sun because it is so small.
Problem#3
Blue Supergiants. A typical blue supergiant star (the type that explodes and leaves behind a black hole) has a surface temperature of 30,000 K and a visual luminosity 100,000 times that of our sun. Our sun radiates at the rate of 3.86 x 1026 W. (Visual luminosity is the total power radiated at visible wavelengths.) (a) Assuming that this star behaves like an ideal blackbody, what is the principal wavelength it radiates? Is this light visible? Use your answer to explain why these stars are blue. (b) If we assume that the power radiated by the star is also 100,000 times that of our sun, what is the radius of this star? Compare its size to that of our sun, which has a radius of 6.96 x 105 km. (c) Is it really correct to say that the visual luminosity is proportional to the total power radiated? Explain.
Answer:
Apply the Wien displacement law to relate λm and T. Apply the Stefan-Boltzmann law to relate the power output of the star to its surface area and therefore to its radius.
For a sphere A = 4πr2. Since we assume a blackbody, e = 1
(a) from Wien’s law:
λm = (2.90 x 10-3 m.K)/T = (2.90 x 10-3 m.K)/30000K = 9.7 x 10-8 m = 97 nm
This peak is in the ultraviolet region, which is not visible. The star is blue because the largest part of the visible light radiated is in the blue/violet part of the visible spectrum.
(b) P = AσT4 (Stefan-Boltzmann law)
(100,000)(3.86 x 1026) = (5.67 x 10-8)(4πR2)(30,000)4
R = 8.2 x 109 m
Rstar/Rsun = (8.2 x 109 m)/(6.96 x 108 m) = 12
(c) The visual luminosity is proportional to the power radiated at visible wavelengths. Much of the power is radiated nonvisible wavelengths, which does not contribute to the visible luminosity.
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