Problem #1: The surface of a star has a
temperature of 3800 K and emits as a blackbody.
a) At what wavelength does this star emit the most light?
b) What color would this star appear to be to the human eye?
Solution: We're given a temperature and asked at what wavelength the star emits the most light. Since this star (and all normal stars) emits like a blackbody, this question is equivalent to asking at what wavelength the blackbody spectrum peaks. Using the Wien Displacement Law,
where 0.003 K m is often referred to as the Wien Displacement constant (the "K" and "m" are units). We're given the star's temperature, so
Note that the Kelvin units on the top and bottom cancel out, leaving meters as the units for our answer. This makes a lot of sense, since our answer is supposed to be a wavelength.
The above answer is correct, but because it's expressed in meters, it's hard to have a feel as to what color the star's emission would be. We would have an easier time estimating the color if our answer were expressed in nanometers, since these are more normal units for optical wavelengths. Since there are 109 nm in a m,
A quick consultation with your book or class notes (or perhaps your memory from the spectroscopy lab) should lead you to conclude that light of this wavelength is red. In fact, it's so red it might be called infrared, since our eyes can only see light of wavelengths 400-700 nm or so. However, even though we can't see the light at the peak of the spectrum, we can still see this star, since blackbodies emit over a broad range in wavelength. Furthermore, since the peak occurs in the infrared portion of the spectrum, we should expect to see more red light than blue light, and so our consclusion that this star appears red still holds.
This star has a surface temperature quite a bit lower than that of the Sun (T = 5800 K), and so should have a peak wavelength longer than the Sun's peak (in the yellow), so it sounds like this answer passes a sanity check.
Problem #2: a) I have two spheres, one
with a radius of 0.2 m and the other with a radius of 0.5 m. I heat the
small one to a temperature of 2400 K, and the big one to a temperature of
of 2100 K.
a) What is the ratio of the intensities of the two spheres?
b) What is the ratio of the total power emitted
by these spheres?
Solution: There's an easy way and a hard way to do this problem. The hard way is to calculate the intensities and powers from the Stefan-Boltzmann Law and the surface area of a sphere, stuffing all of the relevant numbers in a calculator and coming up with numerical answers. However, since all we're interested in is the ratios of intensities and powers, we don't really need their actual values, and we can solve this problem with a lot less effort.
Let's set up the ratio of intensities directly. We know that for a heated object, the intensity (I) emitted by a patch of that object will be described by the Stefan-Boltzmann Law
where T is the temperature of the object and sigma is the Stefan-Boltzmann constant, which is 5.67 x 10-8 W/(m2 K4). So, if we want the ratio of the intensity of the big sphere to the intensity of the little sphere, we just use this expression twice
where the subscript B denotes quantities related to the "big" sphere and the subscript L denotes quantities related to the "little" sphere. The sigma's are constants and appear on the top and bottom, so they immediately cancel out, leaving
which says that the ratio of intensities is equal to the fourth power of the ratio of the temperatures. So we can now plug in the two temperatures and get
Does this make sense? I mean, "shouldn't the bigger object be brighter?" you might say. Ah, it sounds to me like you're confusing intensity with power. Remember that the intensity tells you how much each bit of the sphere is emitting, not the total emission from the sphere. The surface of little sphere will appear brighter than the surface of the big sphere because it's hotter. We don't yet know which of the objects generates more power.
While the intensity of a blackbody depends only on its temperature, the power emitted by each sphere also depends on the amount of surface area available for emission. As we discussed in class,
where P is the power emitted by an object, I is the intensity of the object, and S is the surface area of the object. Using the Stefan-Boltzmann relation for the intensity and the standard formula for the surface area of a sphere, we get
where R is the radius of the sphere. Now let's set up the ratio of powers just as we set up the ratios of intensities above:
where the B and L subscripts again refer to the "big" and "little" spheres. Once again the sigma's drop out, as do the 4's and pi's, so we end up with
From above, we learned that TB4/TL4 = 0.586 so the ratio expression reduces further to
Now we can resort to numbers, and putting in the values given in the beginning of the problem,
Note that since this is a ratio of quantities, there are no units. The big sphere emits 3.66 times the power that the little sphere emits.
Problem #3: Calculate the total power emitted at all wavelengths by a star whose surface temperature is 7300 K, and whose radius is 2.5 solar radii. (You may assume that the star radiates as a blackbody.)
Solution: This problem is essentially the same as the last one, only this time, we'll use real numbers and therefore we'll have to be careful about units. The expression for power from the last problem works just fine here since we're considering a spherical blackbody emitter, so
We're given the temperature and told that the radius of the star is 2.5 solar radii. Well, the radius of the Sun is 6.96 x 108 m (from the back of your book), so 2.5 solar radii must be two and a half of these, or
OK, now let's solve for the total power emitted using these quantities,
While it's hard to keep track of all of the units, you should be able to see that everything cancels out except the watts. That's good, since we are trying to calculate a power here, and power is measured in watts. Before we finish here, let's just perform a quick sanity check on our answer. The luminosity (which is another name for power radiated) of the Sun is 3.8 x 1026 so our result is in the same ballpark. In addition, since the star is both hotter and bigger than the Sun, we should expect that its luminosity be greater than that of the Sun. It is, so this answer survives a sanity check.