Download Astronomy Study Guide - Light Chapter 4, 6, 7 - Answers

Astronomy Study Guide - Light
Chapter 4, 6, 7 - Answers
1) A hot air balloon rises: as we heat the gas in a balloon, the internal pressure increases and the balloon expands.
Therefore the density of the air inside decreases and when the average density of the entire balloon (balloon material
plus basket plus air inside) becomes less than the density of air outside, the balloon rises, gaining gravitational
energy.
2) The brakes on a car: applying the brakes on a car slows it down through friction of the brake pads with the brake
drums. The car slows down, losing kinetic energy and the pads warm up, gaining thermal energy (try touching your
wheels - but be careful because they can become very hot - after using your brakes for a long time, e.g. going down
a steep mountain road).
3) The number of protons in an atom.
4) Atomic weight is the number of protons plus neutrons in an atom.
5) The ice melts into liquid, and then evaporates into gas. At higher temperatures, the water molecules dissociate
atoms. At very high temperatures, the atoms are ionized.
6) The atomic number of fluorine is equal to the number of protons, 9. The atomic weight is equal to the number of
protons plus neutrons, 19. If we added a proton, it would no longer be fluorine. If we added a neutron instead, it
would just be another isotope of fluorine, with atomic number 9 but atomic weight 20.
7) The atomic number would still be 8 because the number of protons wouldn't change, but the atomic weight would
increase to 18.
8) The most common isotope of gold contains 79 protons and 118 neutrons. If it is neutral, it also contains 79
electrons. If the gold is triply ionized instead, it is missing 3 electrons and so has only 76 electrons.
9) U-238 has 146 neutrons, and U-235 has 3 fewer, or 143 neutrons.
10) The electrons can have only specific energies and not amounts of energy in between.
11) Every chemical element has a unique set of atomic energy levels and therefore a unique set of spectral lines.
Thus, by identifying spectral lines, we can identify the elements that produced them.
12) B
13) A
14) E
15) D
16) Different ionization states of the same element have different sets of spectral lines. Thus, we can identify the
ionization state, which tells us about the temperature because higher temperatures are required to reach higher
ionization states.
17) 1. Hotter objects emit more radiation per unit surface area.
2. Hotter objects emit photons with higher average energy.
18) From the first rule of thermal radiation, we know that tripling the temperature of an object increases the amount
of thermal radiation it emits per unit area by a factor of
34 = 81. Thus, increasing the surface temperature of the Sun from 6,000 K to 18,000 K would increase its thermal
radiation by a factor of 81. The higher temperature of the Sun would shift the peak of its thermal radiation spectrum
from its current place in the visible light region into the ultraviolet. The hotter Sun would emit more energy at all
wavelengths, with the greatest output coming in the ultraviolet.
19) By comparing the wavelength of the spectral lines in the object's spectrum to the rest wavelengths of the same
lines, we measure the Doppler shift. This tells us the object's radial motion: A shift toward shorter wavelength
means the object is moving toward us, and a shift to longer wavelength means it is moving away from us. We cannot
learn anything about the object's tangential motion from its spectral lines because this does not affect the line
positions.
20) Lines are wider for faster-rotating objects because parts of the object are rotating toward us and parts are rotating
away from us.
21) The separation of the headlights is s = 1 m, and their distance is d = 2 km. Thus, their angular separation is:
a=
1m
¥ 360! = 0.028!
2p ¥ 2,000m
This is a slightly wider separation than the angular resolution of your eyes, so you can resolve the two headlights.
22) We are given
a = 0.5° and d = 380,000. Thus, we must first solve the angular separation equation for s:
a=
s
2pd
¥ 360! fi s =
¥a
2pd
360!
Now we substitute the numbers.
s=
2p ¥ 380,000km
¥ 0.5! = 3,316km
360!
The Moon's diameter is about 3,300 km.
23) Diagrams should look similar to Figures S2.7a and S2.8a in the text.
24) Spectral resolution is a measure of the amount of detail that can be seen in a spectrum (i.e. how well it can
separate out two spectral features from each other). It depends on how widely the light from a telescope is spread out
but the trade off is that the more the light is spread out, the dimmer it becomes until it becomes undetectable. Since a
large telescope gathers more light than a smaller telescope, it can spread the light out further and achieve a higher
resolution.
25) Answers will vary somewhat with location, but in general campuses are bright environments that are poor
observing sites from a light pollution standpoint. Other factors students may mention might include light from
surrounding or nearby cities, excessive cloudiness or rain, windy area with a lot of turbulence, and low altitude with
lots of atmosphere above.
26) It would be worse than 0.01 arcsecond because of the distortion caused by atmospheric turbulence.
27) It would be worse than 0.01 arcsecond because radio waves have much longer wavelength than visible light, and
hence poorer angular resolution for the same size telescope.
28) It would be better than 0.01 arcsecond because a larger telescope means a better diffraction-limited resolution
for the same wavelength of light, and the location in space eliminates problems of atmospheric distortion.
29) A
30) C
31) D
32) C
33) B