Astronomy Midterm #2 study guide
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1.)
The sketch below shows Isaac Newton’s drawing of his experiment that used a glass prism to spread
the Sun’s light out into a rainbow of colors.
This act of separating the different colors of light through being refracted by different amounts is known as
(a) reflection.
(b) precession.
(c) apparent brightness.
(d) rarification.
(e) dispersion.
2. In the early days of our Universe (i.e.,∼14 billion years ago), the only elements that scientists b
existed were
(a) oxygen, hydrogen, and carbon.
(b) oxygen, hydrogen, and iron.
(c) hydrogen, helium, and lithium.
(d) hydrogen, helium, and iron.
(e) cadmium and zinc.
3. Through which one of the following measurements will an astronomer be able to most accurately determine
the distance to a very nearby star?
(a) By carefully measuring the wavelength at which the star’s continuous spectrum peaks.
(b) By very accurately measuring the position of the nearby star in the sky relative to more distant
stars over the course of six months.
(c) By taking the star’s spectrum and carefully measuring the Doppler shifts of its spectral lines.
(d) By measuring the star’s proper motion.
(e) None of the above — at this time, astronomers have no reliable techniques to measure distances
to even the closest stars.
4.) The diagrams below represent different stages of evolution in the lives of
stars, where the chemical symbols indicate the primary constituent of each
layer of the star (that is, H = hydrogen, He =helium, C = carbon, O =
oxygen, Ne = neon, Si = silicon, and Fe = iron), and arrows (→) indicate
layers in which one element is being actively fused into another element. Of
the choices given below (labeled ‘a’ through ‘e’), which one do astronomers
believe most accurately represents the current structure of our Sun?
Answer
is A
5.)
Light bulb ‘A’ is a 50-watt light bulb. Light bulb ‘B’ is a 100-watt light bulb. Both light bulbs are turned on,
and are located at a distance of 10 meters from an observer. Which one of the following statements is TRUE?
(a) Light bulbs ‘A’ and ‘B’ will have identical apparent brightnesses at the location of the observer.
(b) Light bulb ‘A’ has 1/8 the luminosity of light bulb ‘B’.
(c) Light bulb ‘B’ will have four times the apparent brightness of Light bulb ‘A’ at the location of the observer.
(d) Light bulb ‘B’ will have twice the apparent brightness of Light bulb ‘A’ at the location of the observer.
(e) Light bulbs ‘A’ and ‘B’ have identical luminosities.
6. T or
F
. Astronomers currently believe that elements heavier than iron (i.e., with more protons in
their nuclei) can be produced in the cores of low-mass stars during the final years of their lives.
7.)
As defined by astronomers, a standard candle is:
(a) An object of known surface temperature.
(b) An object of known luminosity.
(c) An object of known mass.
(d) An object of known apparent brightness.
(e) An object of known density.
8. Samantha is generating waves on the surface of a pond by tapping her finger into the water two times
per second. This creates a series of waves in which the individual wave crests are 30 centimeters
apart. With what velocity are the waves traveling through the water?
(a) 2 centimeters per second.
(b) 15 centimeters per second.
(c) 30 centimeters per second.
(d) 60 centimeters per second.
(e) It is impossible to determine the velocity of the waves from the given information.
9. Which of the following is a correctly ordered listing of some of the regions of the electromagnetic
spectrum, progressing from shorter to longer wavelengths?
(a) radio waves, gamma rays, X-rays, ultraviolet rays, visible light, infrared rays.
(b) ultraviolet rays, infrared rays, visible light, gamma rays, X-rays, radio waves.
(c) gamma rays, X-rays, ultraviolet rays, visible light, infrared rays, radio waves.
(d) X-rays, ultraviolet rays, visible light, infrared rays, radio waves, gamma rays.
(e) gamma rays, X-rays, ultraviolet rays, infrared rays, visible light, radio waves.
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10. Which of the following evolutionary stages do we not expect our own Sun to go through?
(a) Spending billions of years as a main-sequence star.
(b) Expanding to become red giant.
(c) Giving off a planetary nebula.
(d) Exploding as a Type II supernova.
(e) Fusing helium into carbon and oxygen in its core.
11. Which one of the following is NOT a “rule” that electrons in atoms must obey, according to the
quantum model of the atom?
(a) Electrons can only exist in “allowed” orbits around an atom’s nucleus.
(b) Electrons can “jump” between orbits only by absorbing or emitting photons.
(c) Only two electrons are allowed in each orbital.
(d) Electrons prefer to be in the smallest possible orbits, the ones with the lowest energies that are
available to them.
(e) The energies of the allowed electron orbits are the same for all elements.
12. T or
F
. According to theoretical calculations, if our Sun had been born with only half the amount
of mass that it actually does have, it would never have been able to generate power through nuclear
fusion, and thus would not have become a star.
Questions 13 – 15 refer to the figure above, which shows a portion of the Sun’s spectrum, plotted with
relative brightness on the Y-axis and wavelength (in Angstrom units, A) on the X-axis. Note,
̊
in particular, the spectral “feature” that is indicated by the arrow; it occurs precisely at a wavelength
Of λ= 6563 A.
̊
13. By determining the wavelength at which maximum brightness occurs in the spectrum of the Sun (i.e.,
the broad “peak” of the spectrum shown in the figure, located around λ ≈ 4700
̊A) astronomers a
determine which one of the following properties of the Sun?
(a) The total mass of the Sun.
(b) The temperature of the Sun’s photosphere.
(c) The rate at which nuclear fusion is occurring in the Sun’s core.
(d) The density of the material in the Sun’s atmosphere.
(e) The velocity with which the Sun is moving through space, relative to the Earth.
14. The spectral “feature” that is indicated by the arrow in the figure is most generally referred to as
(a) an absorption line.
(b) an emission line.
(c) a continuous line.
(d) a Doppler line.
(e) a dense line.
15. The measurement that the spectral “feature” that is indicated by the arrow in the figure occurs right at a wav
Λ = 6563 A allows astronomers to conclude which of the following?
̊
(a) That the Sun’s atmosphere contains the element hydrogen.
(b) That the Sun must have been moving very rapidly towards the earth when this spectrum was taken.
(c) That the Sun’s core must be fusing helium into carbon and oxygen.
(d) That the Sun’s core must have a temperature of about 15,000,000 K.
(e) Both choices (a) and (b) can be concluded from this measurement.
16.) What supports a white dwarf star against gravitational collapse?
(a) Gas pressure resulting from the ongoing fusion of hydrogen to produce helium in its core.
(b) Neutron degeneracy pressure.
(c) Radiation pressure resulting from the recently started fusion of helium to produce carbon and oxygen in its c
(d) Electron degeneracy pressure.
(e) Both (a) and (d) are responsible for supporting a white dwarf star against gravitational collapse.
17. The scientist whose PhD thesis in 1925 laid the foundations for our current understanding of the precise
chemical composition of the Sun was
(a) Isaac Newton.
(b) Neils Bohr.
(c) Cecilia Payne-Gaposchkin.
(d) Albert Einstein.
(e) Raymond Davis.
18. The very short-range but powerful attractive force that binds protons and neutrons together is known as the
(a) electric force.
(b) magnetic force.
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(c) strong nuclear force.
(d) gravitational force.
(e) weak nuclear force.
19. Which one of the following things was first discovered to exist in nature before it had even been proposed
to exist by scientists?
(a) A neutron star.
(b) A white dwarf star.
(c) A neutrino.
(d) The planet Neptune.
(e) None of the above is correct, since all of these things were proposed to exist before they were actually found
20. Which one of the following is NOT a property of neutrinos?
(a) Neutrinos rarely interact with matter.
(b) Neutrinos come in 3 types.
(c) Neutrinos can transform from one type into another.
(d) Neutrinos travel at nearly the speed of light.
(e) Neutrinos are extremely massive particles, estimated (although not known precisely) to have a
mass greater than 100 times the mass of a proton.
21.
T
or F. Every star, regardless of its initial mass, spends the majority of its life as a “main-sequence” star.
22. T or
F
. The proportion of “heavy” elements (that is, elements with more protons in their nuclei than the eleme
hydrogen) in our Universe is predicted to be decreasing with time.
23.) When does fusion end in the core of a “high-mass” star?
(a) After production of the elements carbon and oxygen.
(b) At the moment there is no more hydrogen available.
(c) After production of the element uranium, the “heaviest” naturally occurring element.
(d) After production of the element iron.
(e) None of the above is correct, since “high-mass” stars are powered by nuclear fission, not fusion.
24. Which one of the following statements about planetary nebulae is FALSE?
(a) With few exceptions, the pictures scientists have taken of all planetary nebulae look very similar to
one another.
(b) The spectrum of a planetary nebula is an emission spectrum.
(c) The material that makes up a planetary nebula comes from mass-loss of a star.
(d) A planetary nebula is made up primarily of gas.
(e) A planetary nebula is ejected into space by a low-mass star that is nearing the end of its life.
25. Stephanie observes a binary star system in which stars Oliver and Jasper are orbiting around a common
“center of mass”. Stephanie determines that the “center of mass” of the binary star system is much closer
to star Oliver than it is to star Jasper. From this, she can conclude that
(a) Star Oliver has a higher surface temperature than star Jasper.
(b) Star Oliver has a greater density than star Jasper.
(c) Star Oliver has a greater radius than star Jasper.
(d) Star Oliver contains more mass than star Jasper.
(e) Both choices (b) and (d) can be concluded from the given information.
26. The main reason that the strength of gravity at the surface of the star Sirius B (a white dwarf) is so
much stronger than the strength of gravity at the Earth’s surface is that
(a) Sirius B has a much smaller radius than Earth does.
(b) Sirius B contains much more mass than Earth does.
(c) Sirius B has a much larger radius than Earth does.
(d) Sirius B has a different chemical composition than Earth does.
(e) Sirius B is has a much higher temperature than Earth does.
27. When high-mass stars die, which types of objects do astronomers believe they leave behind?
(a) White dwarfs and neutron stars.
(b) Black holes and white dwarfs.
(c) Neutron stars and black holes.
(d) Red supergiant stars.
(e) Main-sequence stars.
28. T or
F
. The Orion Nebula is a vast region of space in which a thin, low-density gas is being heated
by hot, young stars. When Armando takes a spectrum of the hot, low-density, glowing gas of the
Orion Nebula, he will see a spectrum that is commonly referred to as a “continuous” spectrum.
29.
T
or F. Red light has a lower frequency than blue light does.
30. T or
F
. The Sun, like the Earth, is mainly made up of the elements iron and oxygen.
31.) The basic physical principle that astronomers believe explains why neutron stars are typically
“born” spinning very rapidly is known as
(a) the conservation of angular momentum.
(b) Fraunhofer’s Third Law of spectral analysis.
(c) Einstein’s mass-energy relation: E=mc2.
(d) the conservation of energy.
(e) Newton’s version of Kepler’s Third Law.
32. When a white dwarf experiences a typical “nova” explosion, what do astronomers believe has just
happened?
(a) A burst of fission of the element uranium on the surface of the white dwarf.
(b) A burst of fusion of the element iron on the surface of the white dwarf.
(c) A burst of fusion of the element hydrogen on the surface of the white dwarf.
(d) A brief episode of “neutron bombardment” of heavy atomic nuclei on the surface of the white dwarf.
(e) A thermonuclear runaway that began in the center of the white dwarf, and has now destroyed the
entire star.
33. Hydrostatic equilibrium refers to
(a) the fact that the Earth pulls on the moon with exactly the same force that the moon exerts on the Earth.
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(b) the fact that equal and opposite forces are exerted by the strong nuclear force and the electric force in
the atomic nucleus.
(c) the state of equilibrium that exists between the positively charged atomic nucleus and the negatively
charged electrons that “orbit” it.
(d) the condition of matter that exists when the temperature approaches 0 K.
(e) a state of equilibrium in a star in which the inward pull of gravity is just balanced by the outward
forces of gas and radiation pressure.
34. Matthew is creating waves in a pool of water by tapping his finger once per second on the water’s surface.
If Matthew wishes to increase the velocity with which the waves are traveling through the water, he should:
(a) tap the water more forcefully.
(b) increase the rate at which he’s tapping the water’s surface.
(c) generate waves with a longer wavelength.
(d) both choices given in answers (a) and (b) will increase the velocity of the waves.
(e) None of the above answers is correct, since nothing that Matthew can do will change the velocity
of the waves in the pool of water, since their velocity is completely determined by the medium
through which they are traveling.
35.
T
or F. When an electron “jumps” from the n= 2 to the n= 1 energy state in a hydrogen atom,
it emits a photon with a wavelength of λ= 1216 A. When an electron “jumps” from the n= 3 to the n = 1
̊
energy state in a hydrogen atom it emits a photon with a wavelength that is less than λ= 1216 A.
̊
36.
T
or F. The majority of the mass of an atom is contained in its nucleus
.
37. T or
F
. Scientists believe that nuclear fusion is occurring throughout the Sun, from its core region all the
way to its surface.
38.) The diagram below shows a hot, opaque source of light (produced, for instance, by a solid object or a
very dense gas) and a “cloud” of low-density gas nearby. Four different positions from which these
objects may be viewed are indicated by arrows
.
Which one of the following statements is FALSE?
(a) If you observe the spectrum of the hot source from position (1), you will see a continuous spectrum.
(b) If you observe the spectrum of the hot source from position (2), you will see a continuous spectrum with
dark lines (i.e., an absorption spectrum).
(c) If you observe the spectrum of the cloud of low-density gas from position (3), you will see a continuo
spectrum along with several distinct bright (or, “emission”) lines.
(d) The spectrum of the cloud of low-density gas observed from position (4) should be nearly the same as
the spectrum observed from position (3).
(e) Spectra obtained from positions (2), (3), or (4) should enable you to identify some of the chemical elements i
the cloud of low-density gas.
39. An atom contains 1 proton and 1 neutron in its nucleus. What kind of element is this atom?
(a) hydrogen.
(b) helium.
(c) oxygen.
(d) iron.
(e) It is not possible to tell what kind of element this atom is, without also being told how many electrons there
are orbiting the nucleus.
40.
T
or F. The more mass a star is born with, the more luminous it will be when it is a main-sequence star.
41. T or
F
. Fewer than 1% of stars exist in binary star systems.
42.) The end result of the proton-proton chain of nuclear fusion in the Sun is that hydrogen nuclei are converted
(a) antimatter and nothing else.
(b) pure energy, and nothing else.
(c) iron nuclei, photons, neutrinos, and positrons.
(d) helium nuclei, photons, neutrinos, and positrons.
(e) helium nuclei, neutrinos, and nothing else.
43. A sunspot represents
(a) a region of the Sun’s surface that is made out of nearly pure iron, unlike the rest of the Sun’s surface,
which is largely made out of hydrogen.
(b) a region in the Sun’s core in which nuclear fusion has stopped.
(c) a region in the Sun’s photosphere in which nuclear fusion has stopped.
(d) a region in the Sun’s photosphere that is significantly redshifted compared to the surrounding regions,
making it appear less bright.
(e) a region of the Sun’s surface that is cooler, and thus darker, than the surrounding regions.
44. The Hβ spectral line of the Balmer series of hydrogen has a wavelength of 4861 Angstroms for a hydrogen s
that is at rest. Of the following, which observed wavelength of this line in a star’s spectrum would indicate
that the star is approaching us (that is, moving towards us) at the greatest speed?
(a) 4851 Angstroms.
(b) 4860 Angstroms.
(c) 4861 Angstroms.
(d) 4871 Angstroms.
(e) None of the above. It is not possible to answer this question with the information given, since astronomers
can not tell how fast a star is moving towards them merely by examining the star’s spectrum.
45. Using a good pair of binoculars, you observe a section of the sky where there are stars of many different
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apparent brightnesses. You find one star that appears especially dim. From this observation, you
can immediately conclude that this star looks dim because:
(a) it is very far away.
(b) it has a very low luminosity.
(c) it is giving off most of its energy in the infrared region of the electromagnetic spectrum.
(d) there is lots of interstellar dust along the line-of-sight to this star.
(e) it could appear dim for more than one of the above reasons; there is no way to tell which answer
is right without more a more detailed study of the star.
46. Deep inside a red supergiant star near the end of its life are several concentric shells of different elements
around a central core. What are the general trends seen in these layers, going from outside inward?
(a) Element nuclei get heavier, temperature increases, and time for each layer to develop increases.
(b) Element nuclei get heavier, temperature decreases, and time for each layer to develop decreases.
(c) Element nuclei get heavier, temperature decreases, and time for each layer to develop increases.
(d) Element nuclei get heavier, temperature increases, and time for each layer to develop decreases.
(e) Element nuclei get lighter, temperature increases, and time for each layer to develop increases.
47.) The type of star that astronomers currently believe can explode as a “Type Ia” supernova is
(a) a red supergiant in a binary star system.
(b) a white dwarf in a binary star system.
(c) an isolated neutron star.
(d) a high-mass, main-sequence star in a binary star system.
(e) an isolated red giant star.
48.
T
or F. A white dwarf with a mass of 1.1 MSun has a smaller radius than a white dwarf with a mass of 0.7 MS
49. T or
F
. The planet Jupiter is technically considered to be a star since nuclear fusion does occur in its core,
although at a greatly reduced rate compared with the Sun.
50. Of the following, which color of star has the highest surface temperature?
(a) Yellow star.
(b) Orange star.
(c) Red star.
(d) Blue star.
(e) It is not possible to answer this question with the information given, since the color of a star tells us
which elements are in its atmosphere, not what its surface temperature is.
WEEKLY READING QUIZZES
1.) An atom contains 1 proton and 1 neutron in its nucleus. What kind of element is this atom?
(a) hydrogen
.
(b) helium.
(c) oxygen.
(d) iron.
(e) It is not possible to tell what kind of element this atom is, without also being told how many
electrons there are orbiting the nucleus.
(Solution: Consult Text, section 5.4.2 (in particular, Figure 5.15), as well as the Course Reader, slides
147 and 148. This is a straightforward definition: The kind of element an atom is is defined by the
number of protons it has in its nucleus. Hydrogen is, by definition, the simplest element, containing
just one proton in its nucleus; it does not matter how many neutrons or how many electrons there
happen to be, if there is one proton the element is hydrogen.)
2. Which of the following has the lowest mass?
(a) A neutron.
(b) The nucleus of a helium atom.
(c) A hydrogen atom.
(d) An electron.
(e) A proton.
(Solution:Consult Text, section 5.4.2, and Course Reader, slide 145. From the Text and Reader
slide, we learn that the neutron and proton are both much more massive than the electron (about
2,000 times more massive!). We also know that, by definition, the nucleus of a helium atom contains
at least two protons (and, often, two neutrons as well) and that a hydrogen atom contains
at least one proton (and, usually, one electron, if it is electrically neutral). So, since all
of the given choices contain at least the mass of one proton or one neutron, which have close to
2,000 times more mass than an electron does, the correct choice for the item with the lowest mass
must be an electron.)
3. At a distance of about 4 LY, Alpha Centauri is the closest star to the Earth (other than the Sun)
and is clearly visible to the unaided eye for observers located in the Southern hemisphere. If it were
located instead at a distance of 12 LY, its apparent brightness in our night sky would be
(a) 1/4 its present value.
(b) three times its present value.
(c) 1/9 its present value
(d) the same as its present value.
(e) 1/3 its present value.
(Solution: Consult Text, section 5.1.4, as well as the Course Reader, slides 151 and 152 as well as
the Mathematical Toolkit. As discussed in class, the apparent brightness of a light source decreases
as the inverse square of the distance to that source. That is: Apparent Brightness ∝ 1/Distance^
2
In this case, we have tripled the distance. Thus, using “The Short Method” for solving ratio problems
(see the Mathematical Toolkit for the longer approach, if you wish), the apparent brightness of this
source will be 1/3^2 = 1/9 its present value. Thus, the answer is: 1/9 its present value.)
4. A
s defined by astronomers, a standard candle is:
(a) An object of known volume.
(b) An object of known luminosity.
(c) An object of known mass.
(d) An object of known apparent brightness.
(e) An object of known density.
(Solution: Consult Text, section 5.1.4, as well as the Course Reader, slide 151. This one is a straight
definition: A standard candle is an object of known luminosity. As the text discusses, “standard
candles” will prove to be extremely useful to astronomers, since if you know the luminosity of a
source, and can measure its apparent brightness, you can derive its distance — and distance, in
general, is one of the toughest things to determine in astronomy!)
5. In the early days of our Universe (i.e.,∼14 billion years ago), the only elements that scientists b
existed were
(a) oxygen, hydrogen, and carbon.
(b) oxygen, hydrogen, and iron.
(c) hydrogen, helium, and lithium.
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(d) hydrogen, helium, and iron.
(e) cadmium and zinc.
(Solution: Consult Course Reader, slide 150. This question is a rarity for a reading quiz: The answer
was only discussed in lecture and written on a slide (i.e., Slide 150), and is not (yet) found in the
reading! But, it is such an important fact that in the early days of our Universe (i.e.,∼14 billion
years ago), the only elements that scientists believe existed were hydrogen, helium, and lithium
that it merited its own slide, and its own reading quiz question. Note that, as discussed in class,
these three elements are the simplest elements on the periodic table of elements, containing just one,
two, and three protons (H, He, and Li, respectively). The Universe apparently liked “simple” things
early on!)
6. An artist who likes working with sources of light decides to make a modern sculpture out of electrified
glass tubes that contain very thin (i.e., not dense) neon gas. When the sculpture is finished, and the
electricity is turned on, the tubes glow with a rich red color (just like the neon “gas discharge tube”
you saw in class). What we are seeing is the result of:
(a) a continuous spectrum.
(b) an emission spectrum.
(c) a white spectrum.
(d) an absorption spectrum.
(e) a Fraunhofer line.
(Solution: Consult Text, section 5.3. In class, we looked at neon gas discharge tubes, and noticed
that they produced a bright-line, or emission spectrum. This artist is doing exactly what we did (as
stated by the question). Thus, the rich red color that is being produced by the glowing neon is the
result of an emission spectrum. Of course, if you were to disperse the light from the neon tube
with a prism, you would see that the “rich red” color is, in fact, made up of many different colors,
which combine to produce the color red!)
7. A hot, glowing lump of coal is taken out of a fireplace. When you look at the glowing coal through
a glass prism, the spectrum that you see is
(a) an absorption spectrum
(b) an emission spectrum.
(c) a continuous spectrum.
(d) a bright-line spectrum.
(e) both choices (b) and (d) are correct, since “bright-line spectrum” and “emission spectrum” are
the same thing.
(Solution: Consult Text, section 5.3, and Course Reader, slides 157, 165 and 167B. A lump of hot
coal is a hot, luminous, opaque solid, which we know always produces a continuous spectrum
Note that the fact a “bright-line spectrum” and an “emission spectrum” are, in fact the same thing
does not make choice (e) correct, since neither of them are the correct answer (i.e., a continuous
spectrum)!)
8. An absorption spectrum:
(a) is produced when a thin (i.e., not dense) gas is heated and starts to glow.
(b) is a continuous “rainbow” of all the colors – red, orange, yellow, green, blue, indigo, violet – no
colors are missing.
(c) appears as a pattern or series of bright lines, consisting of only certain, specific, colors.
(d) consists of a series or pattern of dark lines superimposed upon the continuous spectrum of a
source.
(e) is produced by a hot, solid, opaque object, like the filament in a lightbulb.
(Solution: Consult Text, section 5.3, and Course Reader, slides 160, 161, 162, 163, 164, 165, 167,
and 167B. As discussed in class, an absorption spectrum is what occurs when you view a source of
continuous light through a thin (i.e., not dense) gas, which is located along the line of sight. It is the
thin gas that absorbs out of the continuous spectrum particular “colors”, or, wavelengths, producing
the “dark lines” in the otherwise continuous spectrum. So, an absorption spectrum consists of a
series or pattern of dark lines superimposed upon the continuous spectrum of a source.
All of the other choices describe the production of either a pure, continuous spectrum or and emission
(or, “bright-line”) spectrum.)
9. One of the great triumphs of spectroscopy was when astronomers identified a new element in the
spectrum of the Sun; this element was only later found on Earth. Today, this element is called:
(a) Oxygen.
(b) Hydrogen.
(c) Solarium.
(d) Lithium.
(e) Helium.
(Solution: Consult Text, section 5.3.3. As discussed also in class, in the later part of the 19th
century, scientists had found a pattern of absorption lines in the Sun’s spectrum that did not seem to
correspond to those produced by any known gas on the Earth. Thus, they named this “hypothetical
element in the Sun” helium, after Helios, the Greek name for the God of the Sun. Thus, the correct
choice is: Helium.)
10. As we’ve discussed in class, the Orion Nebula is a vast region of space in which a thin, low-density gas
is being heated by hot, young stars. When Victoria takes a spectrum of the hot, low-density, glowing
gas of the Orion Nebula, she will see a spectrum that is commonly referred to as a “continuous” spectrum.
(a) True.
(b) False.
(Solution: Consult Text, section 5.3.3 (especially Figure 5.21), and Course Reader, slide 165 and 168
and the associated discussion in lecture. So, here you are given an astronomical application of the
power of spectroscopy: The ability to figure out what type of object something is by simply getting a
spectrum of it. As discussed in class, when astronomers first got a spectrum of the Orion Nebula and
saw a emission spectrum, all debate was ended: It was a thin, low-density gas being heated. Since
a heated, thin, low density gas always yields an emission spectrum, it is false that it will produce
a continuous spectrum — such a spectrum is only produced by a heated, opaque solid, liquid, or
high-density gas.)
1. The apparent brightnesses of stars in general tells us nothing about their distances (that is, we cannot
assume that stars that appear dimmer in our sky are farther away). In order for apparent brightness
to be a good indicator of distance, all stars would have to:
(a) have the same luminosity.
(b) exhibit exactly the same Doppler shift of their spectral lines.
(c) have exactly the same radius.
(d) be at least 100 LY away from us, so that the changing distance from Earth, due to its annual
orbit around the Sun, becomes negligible.
(e) not have any exoplanets in orbit around them.
(Solution: Consult Text, section 17.1.2, and Course Reader, slide 151. This question focuses on the
very important point that apparent brightness is simply how bright an object appears to be in the
sky, whereas luminosity is actually the true measure of how much light (energy) an object gives off
each second. Of all of the choices given, the only one that has any bearing at all on this question is
luminosity – if all stars had the same luminosities, then they would be so-called “standard candles”,
and their apparent brightnesses would be excellent indicators of their distances through the inverse-
square relation of light propagation (review the Powerpoint slide!). So, we conclude that if, and
only if, all stars had the same luminosity would their apparent brightnesses be an indicator of their
distances. So, the correct answer is: have the same luminosity.)
2. The light that allows you to see this very interesting reading quiz is made up of waves.
In these waves, the distance between successive crests is called the:
(a) Wavelength.
(b) Frequency.
(c) Velocity.
(d) Crestal longitude.
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(e) Dispersion.
(Solution: Consult Text, section 5.1 (especially Figure 5.4); Course Reader, slide 172; and Course
Reader, Required Reading: “Light Waves”. This one is a straight definition: Wavelength is defined
to be the distance between any two identical points on two successive waves.Thus, the distance
between successive crests is measure of the waves’ wavelength.)
3. Water waves are striking a shoreline with a frequency of 4 waves per second. If the waves have a
wavelength of 2 meters per wave, then the speed of the waves is
(a) 2 meters per second.
(b) 0.5 meters per second.
(c) 1/8 meters per second.
(d) 8 meters per second.
(e) 4 meters per second.
(Solution: Consult Text, section 5.1.2; Course Reader, slides 172 and 172; and Course Reader,
Required Reading: “Light Waves”. As discussed in class, waves are characterized by 3 fundamental
properties: velocity (v), frequency (f), and wavelength (λ), which are related by the equation:
v=λf
So, if you are given any two of the quantities, you can solve for the third one. In this case, you are
told that f= 4 waves/second and that λ= 2 meters/wave.
Thus, plugging these values into the above formula yields:
v= 2 meters/wave × 4 waves/second
or
v= 8 meters/second
Thus, the speed of the waves is 8 meters per second.)
4. You are creating waves in a shallow pan of water by gently tapping your finger on the water’s surface
once per second. If you increase the rate at which you tap the water’s surface to two times per second,
which of the following will happen?
(a) The wavelength of the resulting waves will be smaller than it was before.
(b) The frequency of the resulting waves will be lower than it was before.
(c) The wavelength of the resulting waves will be longer than it was before.
(d) The wavelength, frequency, and velocity of the waves will all remain unchanged.
(e) Both of the effects described in choices (b) and (c) will happen.
(Solution: Consult Text, section 5.1.2; Course Reader, slides 172 and 173; and Course Reader,
Required Reading: “Light Waves”. As stressed in class, for a given medium (water in this case),
increasing the frequency of waves (by, e.g., tapping the surface of water more times per second) will
decrease the waves’ wavelength; this relationship can be seen explicitly by considering the equation
v=λf, for the case where v remains constant (which it always does, for a given medium). In this
question, we are increasing the frequency from 1 Hz to 2 Hz. Thus, the wavelength of the resulting
waves will be smaller than it was before. (It will actually be cut in half, although the question
does not ask for that specific an answer.) A key component of this question is to recognize that the
speed of the waves is unchanged by the change in frequency. Thus, the correct choice is that
the wavelength of the resulting waves will be smaller than it was before.)
5. The scientist who worked out the mathematics of the connections between electricity, magnetism,
and light in the 19th century was:
(a) Gustav Kirchoff.
(b) Christian Doppler.
(c) Josef Fraunhofer.
(d) Isaac Newton.
(e) James Clerk Maxwell.
(Solution: Consult Text, section 5.1.1, and Course Reader, slide 174. While all of the scientists listed
as possible answers contributed to our understanding of light, the specific one who worked out the
mathematics of the connections between electricity, magnetism, and light in the 19th century was
James Clerk Maxwell.)
6. The wavelength of the H-alpha emission line of hydrogen is 6,563 Angstroms.
What color is this spectral line?
(a) Violet.
(b) Blue.
(c) Green.
(d) Yellow.
(e) Red.
(Solution: Consult Course Reader, slides 180 and 181, as well as the past two weeks’ lectures! This
question was a bit unusual, in that it primarily tested how closely you have been following the
lectures given in class, where this particular subject was repeatedly gone over. Your introduction to
this began in the class when you looked through a prism (actually, a “dispersion grating”) and saw
the bright-line spectrum of hydrogen with your own eyes, and witnessed a bright red line, along with
an aqua and a deep blue line. In a following lecture, you were told that that red line is referred to
as the “Hα” line (see slide 180). Finally, you were told that, of all of the hundreds (thousands) of
spectral lines that have been identified, the only one you needed to specifically know, was the Hα
line. So, while not specifically discussed very much (yet) in the textbook, it was a key focus of the
lectures. Thus, the answer to this is that the H-alpha emission line is red in color.)
7. What is the approximate range of wavelengths for the visible part of the electromagnetic spectrum?
(a) 1 to 1000 nanometers.
(b) 4000 to 7000 Angstroms.
(c) All wavelengths smaller than 400 nanometers.
(d) 400 to 700 nanometers.
(e) Both choices (b) and (d) are correct.
(Solution: Consult Text, sections 5.2.1 and 5.3.1 (especially figures 5.9 and 5.10), as well as Appendix
D; consult Course Reader, slides 173 and 174. This one was stressed in class: Visible light is the
range of electromagnetic waves that can be detected by the human eye, and spans the region from
4000 to 7000 Angstroms in the electromagnetic spectrum. And, since 400 to 700 nanometers is the
same as 4000 to 7000 Angstroms, it means that both choices (b) and (d) are correct in this
case!)
8. The Hβ spectral line of the Balmer series of hydrogen has a wavelength of 4861 Angstroms for a
hydrogen source that is at rest. Of the following, which observed wavelength of this line in a star’s
spectrum would indicate that the star is approaching us (that is, moving towards us) at the greatest
speed?
(a) 4851 Angstroms.
(b) 4860 Angstroms.
(c) 4861 Angstroms.
(d) 4871 Angstroms.
(e) None of the above. It is not possible to answer this question with the information given, since
astronomers can not tell how fast a star is moving towards them merely by examining the star’s
spectrum.
(Solution: Consult Text, section 5.6.2; Course Reader, Required Reading: “Light Waves” (subsection
“The Doppler Effect: Light from a Moving Source”); and Course Reader, slides 182 – 187. Since this
star is coming towards us, it means that its spectral lines will appear to be blueshifted to us. That
means that its spectral absorption lines will appear in the spectrum of the star at bluer, or shorter,
wavelengths than they would if the star were at rest. From the Doppler shift formula given in the
Text as well as the required reading, we see that the faster a star is approaching us, the greater its
blueshift will be, and the lower the observed wavelengths of its spectral lines will appear. Thus, for
this question, we want the answer that represents the smallest wavelength. Of the choices given, the
shortest wavelength is 4851 Angstroms. Thus, 4851 Angstroms is the correct answer!)
9. Approximately how many planets have now been discovered and confirmed to be orbiting stars other
than the Sun?
a) None; we’ve yet to discover and confirm a single planet orbiting a star other than our Sun.
(b) About 10.
(c) About 100.
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(d) About 5,000.
(e) About one billion.
(Solution: Consult...the lectures! While the textbook section 21.4, and the Course Reader slides 188 –
194 both discuss exoplanets in some detail, neither tells you how many we currently have discovered,
since this number is constantly changing! However, during last week’s lecture, you were told that the
current number is in the thousands. Thus, of the numbers given, the best choice is about 5,000.)
10. The bright star Altair is located at a distance of about 5 parsecs from Earth. What parallax would
you expect to be observed for this star?
(a) 0.2 arcsec.
(b) 0.2 seconds.
(c) 0.2 degrees.
(d) 5 arcsec.
(e) 60 arcsec.
(Solution: Consult Text, section 19.2.3, and Course Reader, slides 199 – 202. This is a math problem.
We know that parallax is related to distance through the formula:
D= 1/p ,
where D is the distance to a star measured in parsec and p is the measured parallax of the star, in
arcsec. So, here we know that D = 5 parsec; thus we have, solving for p :
p= 1/5 = 0.2 arcsec
Thus, the correct answer is that the parallax expected is 0.2 arcsec.)
1. The H-alpha spectral line of the Balmer series of hydrogen has a wavelength of 6563 Angstroms for
a hydrogen source that is at rest. Of the following, which observed wavelength of this spectral line
in a star’s spectrum would indicate that the star is receding from us (i.e., moving away from us) at
the greatest speed?
(a) 6530 Angstroms.
(b) 6540 Angstroms.
(c) 6520 Angstroms.
(d) 6563 Angstroms.
(e) 6600 Angstroms.
(Solution: Consult Text, section 5.6.2, and Course Reader, slides 182 – 187. Since this star is moving
away from us, it means that its spectral lines will appear to be redshifted to us. That means that
its spectral absorption lines will appear in the spectrum of the star at redder, or longer, wavelengths
than they would if the star were at rest. From the Doppler shift formula given in the Text as well
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as the statement at the bottom of Slide 186 (i.e., “the greater the blue- (red-) shift, the greater the
radial velocity of the object”), we see that the faster a star is receding from us, the greater its redshift
will be, and the higher the observed wavelengths of its spectral lines will appear to us. Thus, for this
question, we want the answer that represents the greatest/largest wavelength. Of the choices given,
the greatest wavelength is 6600 Angstroms. Thus, 6600 Angstroms is the correct answer!)
2. Compared to a photon that has a wavelength equal to 8000 Angstroms, a photon having a wavelength
equal to 4000 Angstroms has
(a) 1/2 the energy.
(b) double the energy.
(c) exactly the same energy.
(d) 1/2 the frequency.
(e) both choices (b) and (d) are correct.
(Solution: Consult Text, sections 5.1 and 5.4, and Course Reader, slides 204 & 205. First, let’s see
what we’re given. We’re told that we have two photons, and one of them has twice the wavelength
of the other. OK. To figure out the correct answer(s) now relies on understanding two formulas
contained in the text and Reader, that were both discussed in class. The first formula relates the
wavelength, frequency, and speed of photons, and is found in section 5.1 of the text:
c=λf
Rewriting this to solve for frequency yields:
f=c/λ
Since c is the speed of light (a constant), we see that f and λ are inversely related. That means that if
you cut the wavelength in half, you double the frequency of the photon. So, we know that compared
to a photon that has a wavelength equal to 8000 Angstrom, a photon having a wavelength equal to
4000 Angstromhas twice the frequency. However, we don’t find that choice among the answers! So,
there’s one more thing to do here. The second formula that we need to use relates the energy of a
photon to its frequency, and is found in section 5.4 of your text:
E=hf, where h is a constant. From this formula, we see that E and f are directly related.
This means that if you double the frequency of a photon, you double the energy. Since in this question
the 4000 Angstrom photon has twice the frequency compared with the 8000 Angstrom photon, it therefore
has double the energy. Note that this is the only correct choice among those listed!)
3. When an electrically neutral atom loses one or more electrons, it is said to be:
(a) Ionized.
(b) In its ground state.
(c) Excited.
(d) Blueshifted.
(e) Redshifted.
(Solution: Consult Text, section 5.5.3, and Course Reader, slide 214. This one is a straight definition.
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An ionized atom is one that has lost one or more electrons.)
4. The temperature of the Sun’s photosphere (i.e., the Sun’s “surface”) is most nearly equal to:
(a) 100 K.
(b) 6,000 K.
(c) 6,000,000 K.
(d) 15,000,000 K.
(e) 6×10^9K.
(Solution: Consult Text, sections 5.2.3, 15.1.3, and Table 15.1; consult Course Reader, slide 221.
This is a “straight fact” question about one of the few numbers that I actually want you to know,
at least in a general sense (i.e., you don’t need to know the exact number, just be able to pick it out
from a group of very different numbers, as in the problem here). The actual answer is given in Table
15.1, as well as the caption to Figure 15.5 of your text, and from several Powerpoint slides discussed
in class (and reprinted in your Reader). From all of these places, we learn that the temperature of
the solar photosphere is about 5800 K. Thus, the best answer of the ones given is 6,000 K.)
5. The reason that sunspots appear darker than the regions surrounding them is:
(a) Sunspots represent shadows cast by nearby mountainous regions on the Sun’s surface.
(b) Sunspots represent regions on the Sun’s surface that are cooler than the surrounding gases.
(c) Sunspots arise from gases falling into the sun, and thus the light they emit is strongly redshifted
towards the “dark” end of the visible spectrum.
(d) Sunspots are composed of nearly pure iron, which does not emit much visible light.
(e) Sunspots represent lakes of water on the Sun’s surface, which do not emit much light.
(Solution: Consult Text, sections 15.1.3 and 15.2.1; and Course Reader, slide 221. The temperature
of the Sun’s photosphere is typically about 5,800 K, whereas the temperature at the center of a large
sunspot is typically about 3,800 K, or about 2,000 degrees cooler. Thus, sunspots appear darker than
the regions surrounding them because they represent regions on the Sun’s surface that are
cooler than the surrounding gases.)
6. Stars ‘A’ and ‘B’ have identical radii, and are both located the same distance from Earth. Star ‘A’
has a surface temperature of 6,000 K, while star ‘B’ has a surface temperature of 12,000 K. For an
observer located on Earth, which of the following statements is TRUE? (Carefully read all choices
before deciding on which is the best answer!)
a) The two stars’ continuous spectra will peak at the same wavelength
(b) Star ‘A’ will appear bluer in color than star ‘B’.
(c) Star ‘A’ will appear fainter in the sky than star ‘B’ will.
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(d) Star ‘B’ will appear bluer in color than star ‘A’.
(e) Both choices (c) and (d) are true.
(Solution: Consult Text section 5.2; and Course Reader, slides 205, 222, and 227. The textbook
section gives an explicit mathematical formula relating the temperature of a hot, opaque object to
its color and surface brightness. In class, we simply said that for a given object, the hotter it is, the
bluer its color and the brighter it will appear. Since here we have two stars of exactly the same size
(i.e., the same surface area) and at exactly the same distance from us, the hotter star will appear
both bluer and brighter than the cooler star witll.Thus,
both choices (c) and (d) are true,making (e) the best choice.)
7. The second most abundant element in the Sun, by mass, is:
(a) Iron.
(b) Hydrogen.
(c) Carbon.
(d) Helium
.
(e) Oxygen.
(Solution: Consult Text, section 15.1.1, especially Table 15.2, as well as Course Reader, slide 235. As
discussed in class, the sun consists primarily of hydrogen (most abundant) and helium (second most
abundant), with just a trace of other elements. Thus, the second most abundant element in the Sun,
by mass, is helium.)
8. Who “pays the bill” for the energy generated by nuclear fusion in the Sun? In other words, from
where does all of the energy pouring out of the Sun ultimately come?
(a) Heavy nuclei are breaking apart into lighter nuclei, releasing energy.
(b) A little bit of mass is lost in each nuclear reaction, and is turned into energy – that is, the Sun
is (slowly) losing mass!
(c) Material is falling into the Sun and being vaporized to produce energy.
(d) American taxpayers pay this bill, as they do so many others.
(e) The Sun is spinning more slowly as time goes on; rotation energy is being lost.
(Solution: Consult text, section 16.2 (especially section 16.2.1), and Course Reader, slides 237 – 241
and 243. As discussed in class, the Sun produces energy through the conversion of mass into energy
through nuclear fusion. About 4 million tons of matter is converted into energy each second; thus,
the sun is losing mass (but very little, compared to its total mass). Thus, who pays the bill for the
energy that is produced? The Sun does! A little bit of mass is lost in each nuclear reaction,
and is turned into energy – that is, the Sun is (slowly) losing mass!
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9. According to the Einstein’s famous formula
E=mc^2,
(a) energy can travel much faster than light (in fact, its speed can approach the speed of light
squared).
(b) a little bit of mass can be converted into a substantial quantity of energy.
(c) mass has to travel at the speed of light before it can produce any energy.
(d) when two masses collide, a lot of light is always produced.
(e) Albert Einstein was a tremendous tap-dancer.
(Solution: Consult Text, section 16.2, specifically subsection 16.2.1; consult Course Reader, slides
237 – 243. This question puts Einstein’s formula in its simplest terms – the equation simply tells us
that mass is a form of energy. How much energy?? A LOT: The conversion rate is that you take
the mass, and multiply it by a HUGE number (i.e., the speed of light squared), to get how much
energy it contains. So, this equation states that a little bit of mass can be converted into a
substantial quantity of energy. )
10. In what part of the sun is the fusion of hydrogen occurring today?
(a) Only in the corona.
(b) Only in the central, “core” regions of the sun.
(c) Pretty much throughout the sun.
(d) Only right near the photosphere.
(e) None of the above is correct, since fusion of hydrogen is not occurring anywhere in the sun
anymore!
(Solution: Consult Text, section 16.2.4, and Course Reader, slide 241. Nuclear fusion only occurs in
the Sun where the temperature is sufficiently high that the hydrogen nuclei (protons) are traveling
fast enough for them to occassionally collide and stick together – i.e., fuse – to form helium.
The temperature is only high enough for this to happen in the center, or core, regions of the star.
As
stated in the book: “Calculations show that nearly all of the Sun’s energy is generated within about
150,000 km of its core, or within less than 10% of its total volume.” Thus nuclear fusion of hydrogen
occurs only in the central, “core” regions of the sun. )
1. Which one of the following is NOT a property of neutrinos?
(a) Neutrinos rarely interact with matter.
(b) Neutrinos come in 3 types.
(c) Neutrinos can transform from one type into another.
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(d) Neutrinos travel at nearly the speed of light.
(e) Neutrinos are extremely massive particles, estimated (although not known precisely) to have a
mass greater than 100 times the mass of a proton.
(Solution: Consult Text,§16.2, 16.4.2; Reader, slides 245 & 246. The answer to this question comes
straight out of the lectures, course slides, and reading.In particular, slide #246, which lists the 5
basic properties of neutrinos, four of which are listed as potential (incorrect) answers to the question;
the only choice that is NOT a property of neutrinos is that neutrinos are extremely massive
particles, estimated (although not known precisely) to have a mass greater than 100
times the mass of a proton. While it is TRUE that the precise mass of a neutrino is not known,
it is known that they have very little mass, less than 1/500,000 the mass of an electron!)
2. When four hydrogen nuclei fuse to form a helium nucleus through the “proton-proton chain”, the
mass of the resulting helium nucleus:
(a) is about 0.7% less than the sum of the masses of the four hydrogen nuclei.
(b) is about 43% less than the sum of the masses of the four hydrogen nuclei.
(c) is about 43% greater than the sum of the masses of the four hydrogen nuclei.
(d) is about 0.7% greater than the sum of the masses of the four hydrogen nuclei.
(e) None of the above is correct: We all know that matter can neither be created nor destroyed,
so the helium nucleus has a mass exactly equal to the sum of the masses of the four hydrogen
nuclei.
(Solution: Consult Text,§16.2.6, as well as 16.2.5. This is whole essence of nuclear fusion — light
atomic nuclei fuse together to form heavier nuclei, and in the process convert a small amount of
mass into pure energy. The specific amount of mass that is converted when hydrogen nuclei fuse into
helium nuclei is roughly 0.71%. Thus, the best choice is that the mass of the resulting helium nucleus
is about 0.7% less than the sum of the masses of the four hydrogen nuclei.)
3. When a neutrino is produced in the center of the Sun, roughly how long does it take that neutrino
to reach the surface of the Sun (i.e., how long does it take the neutrino to “get out” of the Sun)?
(a) About 14 billion years.
(b) About 100,000 years.
(c) About 2 seconds.
(d) About 8 minutes.
(e) None of the above is correct, since neutrinos never actually make it to the surface of the Sun;
they are destroyed almost instantly due to collisions with positrons.
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(Solution:Consult Text, §16.4.2; Course Reader, slide 247. Neutrinos travel at very close to the
speed of light and since the Sun is transparent to them (i.e., they do not interact with the material
of the Sun), they zip right out, making it all the way from the center of the Sun to its surface in
about 2 seconds. Thus, as emphasized in class, neutrinos really provide our only direct view into
what is happening in the Sun’s core right now. )
4. When a positron and an electron collide, they produce:
(a) a hydrogen nucleus.
(b) a deuterium nucleus.
(c) a neutron.
(d) pure energy in the form of photons.
(e) a neutrino.
(Solution: Consult Text,§16.2.2; Course Reader, slide 244. This one tests whether you understand
what happens when a particle and its antiparticle collide. The answer is that when they collide,
the original particles are annihilated, and substantial amounts of energy in the form of photons are
produced. Since a positron is the anti-particle to an electron, when a positron and an electron collide,
they produce pure energy in the form of photons. )
5. Roughly how long does it take for the energy from photons created by hydrogen fusion at the center
of the Sun to reach the surface of the Sun?
(a) About 2 seconds.
(b) About 3 or 4 days.
(c) About 10 billion years.
(d) About 1,000,000 years.
(e) About 8 minutes.
(Solution: Consult Text,§16.2.5; Course Reader, slide 247). As stated in the text and gone over in
class, it takes a considerable amount of time for the energy of the photons produced through nuclear
fusion in the sun’s core to finally make it to the Sun’s surface. The rough amount of time is given
as being between 100,000 and 1,000,000 years. Thus, of the choices given, the best is that it takes
about 1,000,000 years for the energy from photons created by hydrogen fusion at the center of the
Sun to reach the surface of the Sun. )
6. The Sun is an enormous ball of gas. Left to itself, a ball of so many gas atoms should collapse under
its own tremendous gravity. Why is our Sun not collapsing?
(a) The sun is, in fact, slowly collapsing at a rate of about 250 feet per century.
(b) King Kong lives in the center of the Sun, and holds up the outer layers.
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(c) Neutrinos from the core exert an enormous pressure on the layers of the Sun as they travel
outward, and keep our star from collapsing.
(d) The gravity of the planets around the Sun pulls its material outward, preventing collapse.
(e) Nuclear fusion in the core keeps the temperature high, and the resulting gas pressure and
radiation pressure inside the Sun exactly balance gravity.
(Solution: Consult Text, §16.3.2; Course Reader, slides 250, 251, and 252. Why isn’t our Sun
collapsing? Well, something must be holding it up! From the text, as well as our in-class discussion,
that “something” is gas pressure and radiation pressure. Thus, our sun is not collapsing since
nuclear fusion in the core keeps the temperature high, and the resulting gas pressure
and radiation pressure inside the Sun exactly balance gravity. )
7. Interstellar gas consists primarily of
(a) neon.
(b) mercury.
(c) oxygen.
(d) hydrogen.
(e) iron.
(Solution:Consult Text,§20.1.0 and Reader, slides 258 and 259.As discussed in class and as
described in the brief introductory section to Chapter 20 of the text that was assigned: Interstellar
gas consists primarily of hydrogen. )
8. Many names used by astronomers are misleading or outdated. A good example is the term
planetary nebula, which astronomers use to refer to:
(a) a shell of gas ejected by, and expanding away from, an extremely hot low-mass star that is
nearing the end of its life.
(b) the swirling disk of material out of which planets will form around a young star.
(c) an entire “cluster” of thousands of stars all close together which, through a small telescope,
looks like thousands of individual planets.
(d) the remains of an exploded high-mass star.
(e) a region of gas and dust in the interstellar medium out of which new stars are forming.
(Solution:Consult Text,§22.4.3; Course Reader, slides 270→275. As stated in the book and
defined in class, a planetary nebula is a shell of gas ejected by, and expanding away from, an
extremely hot low-mass star that is nearing the end of its life. This process has nothing
to do with planets; the name came about since through the crude telescopes that were available in
the 1800’s, some astronomers believed that the “fuzzy patches” they were seeing surrounding stars
in fact represented regions where planets might be forming around a new star. As we now know,
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planetary nebulae are actually associated with the very last part of life for low-mass stars, not with
their births.)
9. Which one of the following statements about the life of a star with a mass equal to that of our Sun
is TRUE?
(a) Before the star dies, it will fuse dozens of different kinds of elements in its core.
(b) The star will become a “red giant” star roughly 10 million years after it was born.
(c) As the star is dying, a considerable amount of its mass will be lost into space.
(d) After the star is no longer a main sequence star, there is no further fusion of hydrogen anywhere
in the star.
(e) Fusion will end in the core of this star once iron has been produced there.
(Solution: Consult Text, §22.4.3 and 22.4.4. Our Sun is a low-mass star, and thus will go through
the “planetary nebula” phase, during which it will cast off up to 25% of its mass into space (and
may lose even more during other stages of its evolution). Thus, it is certainly a true statement that
as the star is dying, a considerable amount of its mass will be lost into space. All other
statements listed are false: I.e., it will not fuse “dozens” of different kinds of elements (only hydrogen
and helium get fused in the cores of low-mass stars like our Sun); it will not become a “red giant” in
∼10 million years (it’s 1,000 times longer – 10 billion years); it will not cease all fusion once it is no
longer a main-sequence star (it will continue fusing hydrogen in a shell surrounding the core region;
it will also begin fusing helium in the core; and fusion will end in the core after carbon and oxygen
are produced, not iron.)
10. What is the heaviest element (that is, the one with the most protons) that gets produced in the core
of a “high-mass” star during the star’s life?
(a) Hydrogen.
(b) Helium.
(c) Iron.
(d) Gold.
(e) Lead.
(Solution: Consult Text,§22.5.1; Course Reader, slide 275. A “high-mass” star is one that is born
with greater than eight times the mass of the Sun. As discussed in class and in the reading, these
massive stars are able to achieve core temperatures far in excess of the “low-mass” stars, and thus are
able to fuse “heavier” elements than carbon and oxygen. The are “stopped” only by nuclear physics:
Elements heavier than iron (i.e., more than 26 protons in the nucleus) require energy to form via
nuclear fusion. Thus, this type of fusion does not occur, and the heavies element that gets produced
in the core of a “high-mass” star during the star’s life is iron.)
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1. What is true about the location of the “center of mass” of a binary star system that consists of two
stars with different masses?
(a) The center of mass is located at the exact mid-point of the line connecting the centers of the
two stars.
(b) The center of mass is located closer to the lower mass star.
(c) The center of mass is located closer to the higher mass star.
(d) The center of mass is located at the exact center of the star with the lower mass.
(e) The center of mass is located at the exact center of the star with the higher mass.
(Solution:Consult Text,§18.2.1; Course Reader, slides 280 & 281.For a binary star system, the
center of mass represents the point between the two stars about which they both orbit.
For two stars of exactly the same mass, this point will be located at the point between them that is exactly
equal distance from both stars. For two stars of unequal mass, however, the center of mass is
located closer to the higher mass star. Figure 18.5 of your text gives a nice representation of
the situation.)
2. Which of the following is the correct life sequence for a low-mass star?
(a) Main sequence, white dwarf, red giant.
(b) White dwarf, red giant, main sequence.
(c) Main sequence, red giant, white dwarf.
(d) Red giant, main sequence, white dwarf.
(e) White dwarf, main sequence, red giant.
(Solution: Consult Text,§22.1 and 23.1, as well as Course Reader, slides 258 — 264, and 286.
A low-mass star is defined as a star that is born with less than 8 times the mass of the Sun. These stars
start life as main-sequence stars, then, after exhausting the hydrogen in their core, puff up to become
red-giant stars, then “blow off” a planetary nebula, before finally having their core regions compress
and become a white dwarf. Thus, the correct life sequence for a low-mass star is main sequence,
red giant, white dwarf.)
3. In terms of the mass of the Sun (i.e., M
Sun
), what is the maximum mass that astronomers believe a
white dwarf can have?
(a) About 0.1 MSun
(that is,∼1/10 the mass of the Sun).
(b) About 1.0 MSun.
(c) About 1.4 MSun.
(d) About 8.0 MSun
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.
(e) None of the above answers is correct, since astronomers do not believe there is any theoretical
upper limit to a white dwarf’s mass.
(Solution:Consult Text,§23.1.4; Course Reader, slide 285.A white dwarf is the final end state
in the life of an isolated, low-mass star.It is supported by a quantum-mechanical force known as
electron-degeneracy pressure. The astrophysicist Subrahmanyan Chandrasekhar first worked out that
the maximum mass that such a star can have is about 1.
4 MSun, which has come to be known as the“Chandrasekhar limit”; beyond this limit, electron
degeneracy pressure is incapable of supporting the star.)
4. Deep inside a red supergiant star near the end of its life are several concentric shells of different
elements around a central core. What are the properties of these layers, going from outside inward?
(a) Element nuclei get heavier, temperature decreases, and time for each layer to develop decreases.
(b) Element nuclei get heavier, temperature decreases, and time for each layer to develop increases.
(c) Element nuclei get heavier, temperature increases, and time for each layer to develop decreases.
(d) Element nuclei get lighter, temperature increases, and time for each layer to develop increases.
(e) Element nuclei get heavier, temperature increases, and time for each layer to develop increases.
(Solution: Consult Text,§23.2.1 and 23.3.1; Course Reader, slides 293 – 295, 299 – 300. As a high-
mass star (i.e., the type that becomes a red supergiant) nears the end of its life, it fuses progressively
heavier elements in its core, with the results of previous nuclear fusion of lighter elements arranged in
a shell-like structure surrounding it; see Figure 23.6 of the text for a diagram. The core temperature
continually increases during this stage. The amount of time spent fusing each progressively heavier
element in the core is less than the previous fusion; if you look at Table 23.2, you can see that for
the star that produced SN 1987A (i.e., a 20 solar mass star), hydrogen fusion lasts about 8 million
years, whereas silicon fusion lasts only a few days; you can also see here just how much the central
temperature increases with each new round of fusion. Thus, the properties of these layers, going from
outside inward, are that element nuclei get heavier, temperature increases, and time for
each layer to develop decreases.)
5. What is so “catastrophic” about electrons and protons being forced together in the iron core during
the first moments of a core-collapse supernova explosion?
(a) The combining of protons with electrons produces the element helium, a very “heavy” element,
whose gravity causes the core of the star to collapse.
(b) When protons and electrons combine, they “annihilate” and release pure energy, which blows
up the star immediately.
(c) It is the electrons, now being destroyed, which were providing much of the core’s pressure
support against gravitational collapse.
(d) The energy released when this happens makes anti-neutrons combine with neutrons which re-
leases a tremendous amount of energy.
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(e) This question has no correct answer since electrons and protons can never be “forced together”,
even in the extreme conditions that exist at the start of a core-collapse supernova.
(Solution: Consult Text,§23.2.2; Course Reader, slide 294. The force largely responsible for supporting
the iron core of a massive star against gravitational collapse is electron degeneracy pressure. When
the electrons and protons are squeezed so much that they combine to form neutrons (+ neutrinos),
suddenly, all of that support that was being provided by the electrons disappears, and the iron core
is no longer supported against collapse. Thus, the thing that is so catastrophic about electrons and
protons being forced together in the iron core during the first moments of a core-collapse supernova
explosion is that it is the electrons, now being destroyed, which were providing much of
the core’s pressure support against gravitational collapse.)
6. Professor Leonard claims to have found a main-sequence star one-half the mass of the Sun that is in
the process of collapsing to form a black hole. This would be
(a) no big deal, since a star with one-half the Sun’s mass always ends up as a black hole.
(b) unusual, because hardly any stars form with one-half the Sun’s mass.
(c) impossible, because black holes are only produced when two very massive stars collide.
(d) impossible, because such a low-mass star would end up as a neutron star.
(e) extremely surprising, because such a low-mass star would lack sufficiently strong gravity to crush
it into a black hole.
(Solution:Consult Text,§23.2, and Table 23.1; Course Reader, slide 295. In order for Nature to
produce a black hole out of a star, it is thought that the star (or, at least, what remains of it after it
has experienced mass-loss during its life, and explosion upon its death) must have a mass of at least
three times the mass of the Sun (most likely, it started life with greater than 8 – or, even, 40 – times
the mass of the sun). Since this bumbling professor has found a star with only one-half the mass of
the Sun (AND, it isn’t even dying, since it’s a main-sequence star!), to claim that it is forming a black
hole would be extremely surprising, because such a low-mass star would lack sufficiently
strong gravity to crush it into a black hole.)
7. Which one of the following is not considered to be one of the “compact objects” that can be left
behind after a star dies?
(a) A neutron star.
(b) A red supergiant.
(c) A black hole.
(d) A white dwarf.
(e) None of the above answers is correct, since all of these are compact objects that can be left
behind after a star dies.
(Solution: Consult Text,§23.1, 23.2, and 23.5; Course Reader, slides 297 & 298. So, in class and
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in your text, you learned that “low-mass stars” will always leave behind a white dwarf; “high-mass
stars” will leave behind either a neutron star or a black hole. On the other hand, a “red supergiant”
is a brief phase towards the end of the life of a high-mass star in which it becomes extremely bloated
(grows in radius). Thus, of the choices given, only a red supergiant is (a) decidedly not
a “compact object”, and (b) is not a terminal stage in the life of a high-mass star (that is, the star still
has longer to live), and so it is clearly not a “compact object” that is left behind after a star dies!)
8. What mechanism do we believe to be responsible for the fact that most neutron stars are “born”
spinning very rapidly?
(a) The young neutron star absorbs the spin of planets that are sucked into them during the collapse
of the star.
(b) Angular momentum is conserved as the slowly rotating star collapses down to the very small
size of the neutron star.
(c) The neutron star accretes matter from a close companion star and gets “spun up”.
(d) It is a consequence of all the energy that is converted to matter during “core-collapse”.
(e) None of the above choices is correct; young neutron stars typically take over a year to rotate
even once!
(Solution:Consult Text,§23.4.2; Course Reader, slide 303. Like a spinning ice skater pulling her
arms towards her body, a rotating, collapsing star will “spin up”. The technical explanation for this
effect is that angular momentum is conserved as the slowly rotating star collapses down
to the very small size of the neutron star.)
9. Which of the following statements about the Crab Nebula is FALSE?
(a) Inside the nebula, astronomers have discovered many newly formed, high-mass stars (i.e., stars
with greater than eight times the mass of the Sun).
(b) The neutron star inside the Crab Nebula shows clear evidence of slowing down just a little bit
in its rotation over time.
(c) It is part of what remains of a supernova explosion first seen on Earth in 1054 AD.
(d) In addition to pulses of radio energy, we can observe pulses of visible light and X-rays from the
Crab Nebula’s “pulsar”.
(e) The Crab Nebula glows with radiation at many wavelengths, and its overall energy output is
more than 100,000 times that of the Sun.
(Solution: Consult Text,§23.4; Course Reader, slide 306. The Crab Nebula is the shattered remnant
of a once-massive star that blew itself apart as a supernova in 1054 AD. Inside, we have detected a
pulsar, emitting both radio and visible light (as well as X-ray). This pulsar is spinning very rapidly
– about 30 times per second – but is also slowing down by a tiny amount over time; this rotational
energy that is being lost is believed to be powering the enormous power output of the nebula, which
is currently putting out more energy than 100,000 Suns. So, four of the five statements that are given
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are demonstrably true; this leaves just one as the possible “false” choice. And so, by the process
of elimination, the one choice that must be FALSE, is: Inside the nebula, astronomers have
discovered many newly formed, high-mass stars (i.e., stars with greater than eight times
the mass of the Sun). That is, the Crab nebula is not currently an active region of star formation.)
10. Astronomers currently believe that a Type Ia supernova
(a) occurs when a white dwarf accumulates matter from a companion star and steadily increases
its mass until a critical density is reached at (or near) its center, triggering a thermonuclear
runaway that completely incinerates the star.
(b) results from the core-collapse, and subsequent envelope ejection, of a high-mass star (i. e., one
born with an initial mass greater than about 8 MSun).
(c) occurs when an isolated white dwarf (i.e., one without a “companion star”) cools down to a
critical temperature (about 100 K), at which point it suddenly collapses down to the size of a
small city, and ejects its “envelope”.
(d) occurs when a neutron star collides with a white dwarf.
(e) occurs when a white dwarf accumulates matter from its own planetary nebula, gains mass, and
is induced to explode.
(Solution: Consult Text,§23.5.2. As discussed in class and in the text, astronomers currently believe
that a Type Ia supernova occurs when a white dwarf accumulates matter from a companion
star and steadily increases its mass until a critical density is reached at (or near) its
center, triggering a thermonuclear runaway that completely incinerates the star. It
ultimate results from the death of a low-mass, not a high-mass, star.)
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