First Hour Exam |
1999 February 19 |
Question #1: c) It would move upward and to the right.
This one comes (quite literally) straight from class. Because the x-axis on the H-R Diagram represents temperature and temperature increases to the left, a decrease in temperature means that the star will move to the right. At the same time, it is becoming more luminous, and since the y-axis on the H-R Diagram represents luminosity, the star will move upward.
Another "Question to Ponder" straight out of class. Since the waves come from the same source (the light) and travel the same distance to each slit, they will arrive together, or in phase at the two slits. Then, since the distance from each slit to the point P is the same (because point P is on a line between them), the waves will arrive together at point P and constructively interfere since peaks will arrive together and valleys will arrive together. Constructive interference means that the light will be brighter, and therefore, you'll see a bright "fringe" of light.
One of the most recognizable constellations in the night sky. If you got this one wrong, take another look at the night sky.
The corona is a the diffuse outer part of the Sun's atmosphere, and for reasons that are still unclear, it's heated to very high temperatures of 1 to 2 million K. This isn't as hot as the Sun's core, but it is hotter than the other regions of the Sun listed in the question. This question was taken straight out of the reading.
The light from the hammer traverses the 100 meters almost instantaneously, because the speed of light is so high (3 x 108 m/s). Therefore, you'll see the hammer blow only an instant after it happens. However, the speed of sound is much slower, so it will take a bit of time for the sound waves from the hammer blow to travel the 100 meters to your ear.
How long? Well, distance = rate x time, so time = distance/rate, or 100 m / 340 m/s = 0.29 seconds.
The the magnitudes of the two stars differ by 3.0, and since one magnitude difference corresponds to a factor of 2.512 in flux, three magnitudes must correspond to a factor of 2.512 x 2.512 x 2.512 in flux, or a factor of 15.95.
Question #7: b) the 6000 K star
Tricky question (though not a trick question). Many of you realised that even though one star is hotter than another, it may not be more luminous, since the hot one could be smaller than the cooler one. That's correct reasoning, but it doesn't make use of all of the information available. Both stars are Main Sequence stars, and we know because of the shape of the Main Sequence in the H-R Diagram that hotter stars are also more luminous ones. Therefore, the hotter Main Sequence star really must be more luminous, and since both stars are located at the same distance, the hotter star will have the higher flux.
Question #8: c) 1.7 x 10-22 Joules
The key here is that I've asked what energy photon can be absorbed. To absorb a photon, the atom will have to move to a higher energy state, and there's only one available in this atom. The difference between the energy of the atom's current state, and the only stable state of higher energy is 1.7 x 10-22 Joules, so that's the energy of the only photon it can absorb.
Question #9: a) 1 x 10-6 m
This is a straightforward application of Wien's Law, which can be
found on the front of your exam:
Question #10: c) the M supergiant is a
million times more luminous
Luminosity is dependent on the intensity and the surface area. If the
two stars have the same surface temperatures, then they have the same
intensities, so the difference in luminosity comes down to the
difference in surface areas. The M supergiant has a radius that's 1000
times bigger, so it's surface area is 1000 x 1000 = 1,000,000 times
bigger, since surface area depends on the radius squared.
With a temperature of 3000 K, Wien's law say that the peak of the spectrum
will occur at a wavelength of