Astronomy Quiz: Spectral Types And Composition
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Spectral Types And CompositionQuestion 1 of 20

The spectrum of Star A shows very sharp, narrow absorption lines. The spectrum of Star B, which has the same spectral type and luminosity class as Star A, shows the same absorption lines but they are all significantly broadened. Which of the following is the most plausible primary explanation for the difference between the two spectra?

The light from Star B has passed through a turbulent gas cloud that has smeared out its spectral features.
Star B has a much higher abundance of heavy elements, which creates more overlapping spectral lines.
Star B is moving towards the observer at a much higher radial velocity than Star A.
Star B is rotating rapidly, causing different parts of its surface to have different line-of-sight velocities.
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Astronomy Quiz

Astronomy Quiz: Spectral Types And Composition

Practice Spectral Types And Composition in Astronomy with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Spectral Types And Composition, giving you a quick way to practice the rules, question types, and explanations that matter most for Astronomy.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

The spectrum of Star A shows very sharp, narrow absorption lines. The spectrum of Star B, which has the same spectral type and luminosity class as Star A, shows the same absorption lines but they are all significantly broadened. Which of the following is the most plausible primary explanation for the difference between the two spectra?

  1. The light from Star B has passed through a turbulent gas cloud that has smeared out its spectral features.
  2. Star B has a much higher abundance of heavy elements, which creates more overlapping spectral lines.
  3. Star B is moving towards the observer at a much higher radial velocity than Star A.
  4. Star B is rotating rapidly, causing different parts of its surface to have different line-of-sight velocities. (correct answer)
Explanation: When you encounter questions about spectral line broadening, think systematically about what physical processes can affect the width and shape of absorption lines in stellar spectra. The key insight here is that both stars have identical spectral types and luminosity classes, meaning they have the same surface temperature, chemical composition, and evolutionary stage. This eliminates intrinsic differences and points toward motion-related effects. Star B's broadened lines are best explained by rapid rotation (D). When a star spins quickly, different parts of its visible surface move at different velocities relative to us—one limb approaches while the other recedes. This creates a range of Doppler shifts across the stellar disk. Light from the approaching side is slightly blue-shifted, while light from the receding side is red-shifted. When combined, these shifts spread each absorption line across a broader wavelength range, creating the observed broadening. Option A is incorrect because turbulent gas clouds would affect both stars similarly if they're in the same region, and interstellar broadening typically has different characteristics. Option B fails because the problem states both stars have the same spectral type, indicating identical compositions—plus, heavy element abundance affects line strength, not just width. Option C represents a misunderstanding of Doppler shift: uniform radial velocity shifts the entire spectrum but doesn't broaden individual lines. Remember this pattern: when comparing similar stars with different spectral line widths, rotational broadening is often the culprit. Rapid rotation is common in young, massive stars and creates this distinctive "smearing" effect across all absorption features.

Question 2

An astronomer observes the spectrum of a main-sequence star and finds no discernible absorption lines from the hydrogen Balmer series (e.g., H-alpha, H-beta). The astronomer concludes that the star must be deficient in hydrogen. Why is this conclusion likely premature and possibly incorrect?

  1. The hydrogen could be in molecular form (H2), which does not have absorption lines in the visible spectrum.
  2. The star's rotational velocity could be so high that the Balmer lines are broadened into non-existence while leaving other lines visible.
  3. The star could be either very hot (ionizing hydrogen) or very cool (leaving electrons in the ground state), both of which result in weak Balmer lines. (correct answer)
  4. The star could be very old and have fused all of its atmospheric hydrogen into helium through shell burning.
Explanation: When analyzing stellar spectra, you need to understand that the visibility of absorption lines depends not just on elemental abundance, but critically on the physical conditions in the star's atmosphere. The hydrogen Balmer series forms when electrons transition between the second energy level and higher levels, but this requires hydrogen atoms to have electrons in the excited n=2 state. The correct answer is C because stellar temperature dramatically affects hydrogen's ionization state and electron distribution. In very hot stars (O and B types), intense radiation ionizes most hydrogen atoms completely, leaving few electrons available for Balmer transitions. Conversely, in very cool stars (M dwarfs), nearly all electrons remain in the ground state (n=1), making Balmer absorption weak since few electrons occupy the n=2 level needed for these transitions. Intermediate-temperature stars like the Sun show strong Balmer lines because conditions are just right for significant n=2 populations. Answer A is incorrect because while H₂ molecules do exist in cool stellar atmospheres, they would dissociate into atomic hydrogen at stellar temperatures, and this wouldn't explain the complete absence of Balmer lines. Answer B is wrong because extremely high rotation would broaden all absorption lines equally, not selectively eliminate just the Balmer series while preserving others. Answer D is incorrect because stellar fusion occurs in cores, not atmospheres, and shell burning doesn't affect surface composition in main-sequence stars. Remember: weak or absent spectral lines don't automatically indicate elemental deficiency. Always consider how temperature and pressure affect atomic physics before concluding about stellar composition.

Question 3

The spectrum of a star shows a peculiar profile for its hydrogen lines: each line consists of a broad absorption feature with a narrower, sharp emission peak seemingly superimposed in its center. What is the most likely physical model for this system?

  1. A binary system where one star produces absorption lines and a companion produces emission lines that are perfectly aligned by chance.
  2. A single, non-rotating star with an extremely hot core and a very cool, low-density outer atmosphere.
  3. A rapidly rotating hot star is surrounded by a cooler, dense, rotating disk of gas in the star's equatorial plane. (correct answer)
  4. The star is located behind a distant emission nebula, and the nebula's emission lines happen to fall within the star's absorption lines due to a coincidental radial velocity.
Explanation: When you encounter spectral line profiles that combine both absorption and emission features, you're looking at evidence of complex stellar environments with multiple gas components at different temperatures and velocities. The described profile—broad absorption with a narrow emission peak at the center—is a classic signature of a Be star system. In these systems, a rapidly rotating hot star ejects material that forms a dense, rotating circumstellar disk in the equatorial plane. The star itself produces broad absorption lines due to its high rotation speed (Doppler broadening). Meanwhile, the cooler, denser disk gas produces narrow emission lines at the same wavelengths because it's being excited by the star's intense radiation. This creates the characteristic "shell spectrum" where emission appears superimposed on absorption. Option A fails because perfect alignment of absorption and emission lines from two separate stars would be an extraordinary coincidence, and binary orbital motion would cause the lines to shift relative to each other over time. Option B is physically unrealistic—a hot core cannot produce absorption lines while a cool outer atmosphere produces emission; absorption forms in cooler gas overlying hotter regions. Option D requires an implausible coincidence that a background nebula's emission lines would precisely match the star's absorption lines and have exactly the right radial velocity. When you see complex spectral profiles combining absorption and emission, think about multi-component systems. Be stars with circumstellar disks are among the most common examples you'll encounter in stellar spectroscopy problems.

Question 4

An astronomer first observes a star's spectrum with a low-resolution spectrograph, which reveals a single, broad absorption feature around 589 nm. They then re-observe the star with a high-resolution spectrograph. This new spectrum reveals that the single feature is actually two distinct, narrow absorption lines very close together (at 589.0 nm and 589.6 nm). What is the most significant new piece of information revealed by the high-resolution spectrum?

  1. The star's radial velocity can now be measured for the first time.
  2. The element responsible can be definitively identified as sodium, which has a well-known doublet at these wavelengths. (correct answer)
  3. The star's surface temperature can be determined more accurately from the line separation.
  4. The star can be identified as a spectroscopic binary, with each star contributing one of the lines.
Explanation: When you encounter spectroscopy questions involving resolution changes, focus on what new spectral details become visible and what information those details provide. The key insight here is recognizing the sodium doublet. The wavelengths 589.0 nm and 589.6 nm correspond precisely to sodium's characteristic D-lines (D₂ and D₁), which are among the most famous absorption features in stellar spectroscopy. This doublet arises from sodium's electronic structure—specifically transitions from the 3p to 3s energy levels, where spin-orbit coupling splits the 3p level into two slightly different energies. The high-resolution spectrum reveals this splitting, allowing definitive identification of sodium as the absorbing element. Choice A is incorrect because radial velocity can be measured from any absorption line by detecting Doppler shifts—the low-resolution spectrum would have been sufficient for this measurement. Choice C is wrong because line separation in doublets reflects atomic physics (spin-orbit coupling), not stellar temperature. While temperature affects line strength, the 0.6 nm separation between these lines is an intrinsic property of sodium atoms. Choice D misinterprets the situation—in a spectroscopic binary, you'd expect to see Doppler-shifted versions of the same lines, not the precise wavelengths of a known atomic doublet. Study tip: Memorize the major atomic doublets, especially sodium's D-lines at 589 nm. When you see two closely spaced lines at characteristic wavelengths, think atomic structure rather than binary motion or temperature effects. The precision of the wavelength match is your clue to elemental identification.

Question 5

An astronomer obtains a spectrum of a Type Ia supernova near its maximum brightness and a spectrum of a Type II supernova, also near its maximum brightness. What is the most fundamental compositional difference an astronomer would expect to see between these two spectra?

  1. The Type Ia spectrum will be dominated by absorption lines of heavy elements like iron, while the Type II spectrum will show only helium lines.
  2. The Type Ia spectrum will lack hydrogen lines, while the Type II spectrum will show prominent, broad hydrogen lines. (correct answer)
  3. The Type Ia spectrum will be a pure emission spectrum, while the Type II spectrum will be a pure absorption spectrum.
  4. The Type Ia spectrum will be significantly blueshifted due to a higher explosion velocity, while the Type II spectrum will be redshifted.
Explanation: When analyzing supernova spectra, you're examining the fundamental differences in how these stellar explosions occur and what material gets ejected. The key distinction lies in the progenitor stars and explosion mechanisms. Type Ia supernovae result from white dwarf stars that accrete material from a companion until they reach a critical mass and explode completely. Since white dwarfs are the remnants of stars that have already shed their outer hydrogen layers, the explosion ejects material that's predominantly carbon, oxygen, and heavier elements produced during the thermonuclear explosion. Consequently, Type Ia spectra lack hydrogen absorption lines. Type II supernovae occur when massive stars (at least 8 solar masses) exhaust their nuclear fuel and undergo core collapse. These stars retain their hydrogen-rich outer envelopes throughout their evolution, so when they explode, the spectra show prominent, broad hydrogen lines from this ejected hydrogen-rich material. Looking at the wrong answers: Choice A incorrectly suggests Type II spectra show only helium lines, when they're actually dominated by hydrogen. Choice C misrepresents the spectral types entirely—both supernovae produce absorption spectra, not pure emission or absorption. Choice D confuses velocity differences with compositional differences; while explosion velocities do vary, the question specifically asks about compositional differences, not Doppler shifts. Remember this key pattern: Type Ia means "no hydrogen" (white dwarf origin), while Type II means "hydrogen present" (massive star with intact envelope). This compositional signature is the most reliable way to distinguish these supernova types spectroscopically.

Question 6

The spectrum of a high-redshift quasar is observed. At wavelengths shorter than the quasar's intrinsic Lyman-alpha emission line, a dense series of numerous, narrow absorption lines is seen, often called the 'Lyman-alpha forest.' What is the accepted explanation for this forest of lines?

  1. Each narrow line is Lyman-alpha absorption from an intergalactic cloud of hydrogen at a different, lower redshift between the quasar and Earth. (correct answer)
  2. The quasar is ejecting numerous small, high-velocity gas clouds, and each cloud produces an absorption line at a different blueshift.
  3. The light from the quasar is passing through the atmosphere of a single, extremely large galaxy cluster that is rich in complex molecules.
  4. Quantum fluctuations in the vacuum of space between the quasar and Earth absorb photons at discrete, resonant wavelengths.
Explanation: The Lyman-alpha forest is a key piece of evidence for the large-scale structure of the universe. The light from a distant quasar acts as a background source. As this light travels towards Earth, it passes through many otherwise invisible clouds of intergalactic hydrogen gas. Each cloud is at a different distance and thus has a different cosmological redshift. Each cloud imposes a Lyman-alpha absorption line on the quasar's spectrum, with the amount of redshift corresponding to the cloud's distance. The result is a 'forest' of absorption lines at wavelengths shorter (bluer) than the quasar's own Lyman-alpha emission line.

Question 7

In a laboratory, a dense, hot tungsten filament lamp producing a continuous spectrum is placed behind a sealed glass container of cool, low-density sodium gas. An observer uses a spectroscope to analyze the light. Which of the following correctly describes the spectra seen from two different positions?

  1. Both observers see the same emission spectrum, as the gas is energized by the lamp's light and glows at its characteristic wavelengths.
  2. Both observers see the same absorption spectrum, as the gas removes the same wavelengths of light regardless of viewing angle.
  3. An observer looking directly at the lamp through the gas sees an absorption spectrum; an observer looking at the gas from the side (90 degrees) sees an emission spectrum. (correct answer)
  4. An observer looking through the gas sees a continuous spectrum, while the observer at the side sees nothing because the cool gas does not emit light.
Explanation: When analyzing stellar spectra, you need to understand how light interacts with gas depending on your viewing geometry. This setup demonstrates three fundamental types of spectra and how observation angle affects what you see. The tungsten lamp produces a continuous spectrum containing all wavelengths. When this light passes through the cool sodium gas, sodium atoms absorb specific wavelengths corresponding to their electron energy transitions, creating dark absorption lines in the continuous spectrum. However, those absorbed photons don't disappear—the sodium atoms re-emit them in all directions. An observer looking directly through the gas (lamp → gas → observer) sees the continuous spectrum with sodium absorption lines removed. An observer positioned at 90 degrees sees only the re-emitted light from the sodium atoms, which appears as bright emission lines against a dark background. Choice A is wrong because the gas produces emission lines only when viewed from the side, not absorption lines, and the spectra differ by position. Choice B incorrectly assumes both observers see absorption spectra—the side observer sees emission lines instead. Choice D is wrong because cool gas does emit light through re-emission of absorbed photons, making it visible from the side. The correct answer is C: the direct observer sees absorption lines (continuous spectrum with missing wavelengths), while the side observer sees emission lines (bright lines against darkness). Remember this geometry rule: absorption spectra appear when looking through gas toward a continuous source, while emission spectra appear when viewing the gas from the side after it has absorbed energy.

Question 8

An analysis of a star's absorption spectrum reveals prominent, wide bands identified as belonging to titanium oxide (TiO). The presence and strength of these features strongly suggest which of the following about the star?

  1. The star is a pre-main-sequence object, and the molecular features originate in its surrounding protoplanetary disk.
  2. The star is exceptionally rich in the elements titanium and oxygen compared to other stars.
  3. The star is surrounded by a dense, cold interstellar cloud which is imprinting molecular absorption onto the starlight.
  4. The star has a very low surface temperature, allowing molecules to form and remain stable in its atmosphere. (correct answer)
Explanation: When you encounter stellar spectra questions, focus on the relationship between temperature and molecular formation. Stellar atmospheres reveal crucial information about physical conditions through their absorption features. The presence of strong titanium oxide (TiO) bands tells you this star has an extremely cool atmosphere. Molecules like TiO can only form and survive in stellar atmospheres when temperatures drop below approximately 4,000 K. At higher temperatures, the intense thermal energy breaks apart molecular bonds, leaving only atomic absorption lines. These cool stars are classified as M-type red giants or red dwarfs, where the low surface temperature creates an environment where complex molecules can exist stably in the stellar photosphere. Option A incorrectly suggests the TiO comes from a protoplanetary disk. While disks can show molecular features, the question specifically states these are absorption features in the star's spectrum, indicating they originate in the stellar atmosphere itself. Option B misinterprets the spectral evidence - strong TiO bands don't mean unusual titanium/oxygen abundance, but rather the right temperature conditions for molecule formation. Even stars with normal elemental composition show prominent TiO if they're cool enough. Option C wrongly attributes the molecular features to intervening interstellar material, but interstellar clouds typically show much narrower absorption lines and different molecular species than what's observed in cool stellar atmospheres. Remember this key principle: molecular absorption bands in stellar spectra are temperature indicators. Strong molecular features always point to cool stellar atmospheres where molecules can survive the thermal environment.

Question 9

A spatially unresolved binary system consists of an O-type main-sequence star and an M-type supergiant. If an astronomer takes a single optical spectrum of the combined light from this system, what is the most likely appearance of the spectrum?

  1. A continuous spectrum with absorption lines that are an average of the two stellar types, appearing similar to a G-type star spectrum.
  2. A pure emission spectrum, as the intense radiation from the O-type star excites the gas in the M-type star's extended atmosphere.
  3. The spectrum of the O-type star only, as its much higher luminosity completely overwhelms the light from the M-type star at all optical wavelengths.
  4. A continuous spectrum that is strong in the blue with He II absorption lines, combined with a spectrum that is strong in the red with TiO molecular absorption bands. (correct answer)
Explanation: When analyzing the combined spectrum of a binary system, you need to consider how light from both stars contributes to what we observe. Since the system is spatially unresolved, we see the sum of both stellar spectra. The correct approach is to recognize that each star contributes its characteristic spectral features based on its stellar type and luminosity. An O-type main-sequence star is extremely hot (30,000-50,000 K), producing strong blue/UV continuum emission and helium II absorption lines. An M-type supergiant is much cooler (3,000-4,000 K) but has an enormous surface area, making it quite luminous in red wavelengths and showing prominent titanium oxide (TiO) molecular bands. Answer D correctly describes this additive effect: you'll see the O-star's blue-dominated continuum with He II lines superimposed on the M-supergiant's red-dominated continuum with TiO bands. Answer A is wrong because stellar spectra don't average—they add together, preserving distinct features from each star. Answer B incorrectly assumes the system becomes an emission nebula; while the O-star is hot, the M-supergiant isn't a gas cloud but a distinct stellar photosphere. Answer C underestimates the M-supergiant's contribution; despite the O-star's high surface brightness, the supergiant's enormous size makes it very luminous, especially at red wavelengths where the O-star is relatively dim. Remember: in composite spectra, look for features from both components. The relative strength depends on each star's luminosity at different wavelengths, not just overall brightness.

Question 10

An optical spectrum of a distant galaxy shows a relatively flat continuum with several narrow stellar absorption features. Superimposed on this is a very strong, broad emission line corresponding to the H-alpha transition. Which of the following is the most likely cause of this prominent feature?

  1. An active galactic nucleus (AGN) where gas is moving at high speeds around a supermassive black hole. (correct answer)
  2. A massive burst of star formation creating numerous H II regions throughout the galaxy.
  3. The combined light from millions of red giant stars whose atmospheres are expanding.
  4. A gravitational lensing event by a foreground object, which selectively amplifies hydrogen emission.
Explanation: The width of a spectral line is related to the range of velocities of the emitting gas along the line of sight (Doppler broadening). The extremely broad emission lines seen in some galaxies are a key signature of Active Galactic Nuclei (AGN). They originate from the 'broad-line region,' where clouds of gas orbit the central supermassive black hole at very high speeds (thousands of km/s), creating a large Doppler broadening. Star formation regions produce narrow emission lines, red giants have absorption spectra, and gravitational lensing amplifies all light without changing spectral features.

Question 11

The absorption spectrum of a star is dominated by lines from singly ionized helium (He II) and multiply ionized heavy elements (e.g., O III, N IV), while lines from neutral helium (He I) and neutral metals are very weak or absent. What is the most direct inference about this star?

  1. The star has an extremely high surface temperature, sufficient to ionize most atoms in its atmosphere. (correct answer)
  2. The star has an unusual composition, consisting almost entirely of helium and other heavy elements.
  3. The star is very young and has not yet fused lighter elements into heavier ones in its core.
  4. The star's powerful magnetic field is stripping electrons from atoms in its atmosphere through non-thermal processes.
Explanation: The ionization state of atoms in a star's atmosphere is primarily determined by the surface temperature. High levels of ionization, such as the presence of He II (which requires 54.4 eV to create from He I) and even more highly ionized elements like O III and N IV, are only possible at very high temperatures (above 30,000 K). This is characteristic of hot O-type stars. While composition plays a role, the dominant factor determining which spectral lines are visible is temperature.

Question 12

A spatially unresolved binary system consists of an O-type main-sequence star and an M-type supergiant. If an astronomer takes a single optical spectrum of the combined light from this system, what is the most likely appearance of the spectrum?

  1. A continuous spectrum with absorption lines that are an average of the two stellar types, appearing similar to a G-type star spectrum.
  2. A pure emission spectrum, as the intense radiation from the O-type star excites the gas in the M-type star's extended atmosphere.
  3. The spectrum of the O-type star only, as its much higher luminosity completely overwhelms the light from the M-type star at all optical wavelengths.
  4. A continuous spectrum that is strong in the blue with He II absorption lines, combined with a spectrum that is strong in the red with TiO molecular absorption bands. (correct answer)
Explanation: When analyzing the combined spectrum of a binary system, you need to consider how light from both stars contributes to what we observe. Since the system is spatially unresolved, we see the sum of both stellar spectra. The correct approach is to recognize that each star contributes its characteristic spectral features based on its stellar type and luminosity. An O-type main-sequence star is extremely hot (30,000-50,000 K), producing strong blue/UV continuum emission and helium II absorption lines. An M-type supergiant is much cooler (3,000-4,000 K) but has an enormous surface area, making it quite luminous in red wavelengths and showing prominent titanium oxide (TiO) molecular bands. Answer D correctly describes this additive effect: you'll see the O-star's blue-dominated continuum with He II lines superimposed on the M-supergiant's red-dominated continuum with TiO bands. Answer A is wrong because stellar spectra don't average—they add together, preserving distinct features from each star. Answer B incorrectly assumes the system becomes an emission nebula; while the O-star is hot, the M-supergiant isn't a gas cloud but a distinct stellar photosphere. Answer C underestimates the M-supergiant's contribution; despite the O-star's high surface brightness, the supergiant's enormous size makes it very luminous, especially at red wavelengths where the O-star is relatively dim. Remember: in composite spectra, look for features from both components. The relative strength depends on each star's luminosity at different wavelengths, not just overall brightness.

Question 13

The spectrum of Star A shows very sharp, narrow absorption lines. The spectrum of Star B, which has the same spectral type and luminosity class as Star A, shows the same absorption lines but they are all significantly broadened. Which of the following is the most plausible primary explanation for the difference between the two spectra?

  1. The light from Star B has passed through a turbulent gas cloud that has smeared out its spectral features.
  2. Star B has a much higher abundance of heavy elements, which creates more overlapping spectral lines.
  3. Star B is moving towards the observer at a much higher radial velocity than Star A.
  4. Star B is rotating rapidly, causing different parts of its surface to have different line-of-sight velocities. (correct answer)
Explanation: When you encounter questions about spectral line broadening, think systematically about what physical processes can affect the width and shape of absorption lines in stellar spectra. The key insight here is that both stars have identical spectral types and luminosity classes, meaning they have the same surface temperature, chemical composition, and evolutionary stage. This eliminates intrinsic differences and points toward motion-related effects. Star B's broadened lines are best explained by rapid rotation (D). When a star spins quickly, different parts of its visible surface move at different velocities relative to us—one limb approaches while the other recedes. This creates a range of Doppler shifts across the stellar disk. Light from the approaching side is slightly blue-shifted, while light from the receding side is red-shifted. When combined, these shifts spread each absorption line across a broader wavelength range, creating the observed broadening. Option A is incorrect because turbulent gas clouds would affect both stars similarly if they're in the same region, and interstellar broadening typically has different characteristics. Option B fails because the problem states both stars have the same spectral type, indicating identical compositions—plus, heavy element abundance affects line strength, not just width. Option C represents a misunderstanding of Doppler shift: uniform radial velocity shifts the entire spectrum but doesn't broaden individual lines. Remember this pattern: when comparing similar stars with different spectral line widths, rotational broadening is often the culprit. Rapid rotation is common in young, massive stars and creates this distinctive "smearing" effect across all absorption features.

Question 14

An analysis of a star's absorption spectrum reveals prominent, wide bands identified as belonging to titanium oxide (TiO). The presence and strength of these features strongly suggest which of the following about the star?

  1. The star is a pre-main-sequence object, and the molecular features originate in its surrounding protoplanetary disk.
  2. The star is exceptionally rich in the elements titanium and oxygen compared to other stars.
  3. The star is surrounded by a dense, cold interstellar cloud which is imprinting molecular absorption onto the starlight.
  4. The star has a very low surface temperature, allowing molecules to form and remain stable in its atmosphere. (correct answer)
Explanation: When you encounter stellar spectra questions, focus on the relationship between temperature and molecular formation. Stellar atmospheres reveal crucial information about physical conditions through their absorption features. The presence of strong titanium oxide (TiO) bands tells you this star has an extremely cool atmosphere. Molecules like TiO can only form and survive in stellar atmospheres when temperatures drop below approximately 4,000 K. At higher temperatures, the intense thermal energy breaks apart molecular bonds, leaving only atomic absorption lines. These cool stars are classified as M-type red giants or red dwarfs, where the low surface temperature creates an environment where complex molecules can exist stably in the stellar photosphere. Option A incorrectly suggests the TiO comes from a protoplanetary disk. While disks can show molecular features, the question specifically states these are absorption features in the star's spectrum, indicating they originate in the stellar atmosphere itself. Option B misinterprets the spectral evidence - strong TiO bands don't mean unusual titanium/oxygen abundance, but rather the right temperature conditions for molecule formation. Even stars with normal elemental composition show prominent TiO if they're cool enough. Option C wrongly attributes the molecular features to intervening interstellar material, but interstellar clouds typically show much narrower absorption lines and different molecular species than what's observed in cool stellar atmospheres. Remember this key principle: molecular absorption bands in stellar spectra are temperature indicators. Strong molecular features always point to cool stellar atmospheres where molecules can survive the thermal environment.

Question 15

An astronomer observes the spectrum of a main-sequence star and finds no discernible absorption lines from the hydrogen Balmer series (e.g., H-alpha, H-beta). The astronomer concludes that the star must be deficient in hydrogen. Why is this conclusion likely premature and possibly incorrect?

  1. The hydrogen could be in molecular form (H2), which does not have absorption lines in the visible spectrum.
  2. The star's rotational velocity could be so high that the Balmer lines are broadened into non-existence while leaving other lines visible.
  3. The star could be either very hot (ionizing hydrogen) or very cool (leaving electrons in the ground state), both of which result in weak Balmer lines. (correct answer)
  4. The star could be very old and have fused all of its atmospheric hydrogen into helium through shell burning.
Explanation: When analyzing stellar spectra, you need to understand that the visibility of absorption lines depends not just on elemental abundance, but critically on the physical conditions in the star's atmosphere. The hydrogen Balmer series forms when electrons transition between the second energy level and higher levels, but this requires hydrogen atoms to have electrons in the excited n=2 state. The correct answer is C because stellar temperature dramatically affects hydrogen's ionization state and electron distribution. In very hot stars (O and B types), intense radiation ionizes most hydrogen atoms completely, leaving few electrons available for Balmer transitions. Conversely, in very cool stars (M dwarfs), nearly all electrons remain in the ground state (n=1), making Balmer absorption weak since few electrons occupy the n=2 level needed for these transitions. Intermediate-temperature stars like the Sun show strong Balmer lines because conditions are just right for significant n=2 populations. Answer A is incorrect because while H₂ molecules do exist in cool stellar atmospheres, they would dissociate into atomic hydrogen at stellar temperatures, and this wouldn't explain the complete absence of Balmer lines. Answer B is wrong because extremely high rotation would broaden all absorption lines equally, not selectively eliminate just the Balmer series while preserving others. Answer D is incorrect because stellar fusion occurs in cores, not atmospheres, and shell burning doesn't affect surface composition in main-sequence stars. Remember: weak or absent spectral lines don't automatically indicate elemental deficiency. Always consider how temperature and pressure affect atomic physics before concluding about stellar composition.

Question 16

In a laboratory, a dense, hot tungsten filament lamp producing a continuous spectrum is placed behind a sealed glass container of cool, low-density sodium gas. An observer uses a spectroscope to analyze the light. Which of the following correctly describes the spectra seen from two different positions?

  1. Both observers see the same emission spectrum, as the gas is energized by the lamp's light and glows at its characteristic wavelengths.
  2. Both observers see the same absorption spectrum, as the gas removes the same wavelengths of light regardless of viewing angle.
  3. An observer looking directly at the lamp through the gas sees an absorption spectrum; an observer looking at the gas from the side (90 degrees) sees an emission spectrum. (correct answer)
  4. An observer looking through the gas sees a continuous spectrum, while the observer at the side sees nothing because the cool gas does not emit light.
Explanation: When analyzing stellar spectra, you need to understand how light interacts with gas depending on your viewing geometry. This setup demonstrates three fundamental types of spectra and how observation angle affects what you see. The tungsten lamp produces a continuous spectrum containing all wavelengths. When this light passes through the cool sodium gas, sodium atoms absorb specific wavelengths corresponding to their electron energy transitions, creating dark absorption lines in the continuous spectrum. However, those absorbed photons don't disappear—the sodium atoms re-emit them in all directions. An observer looking directly through the gas (lamp → gas → observer) sees the continuous spectrum with sodium absorption lines removed. An observer positioned at 90 degrees sees only the re-emitted light from the sodium atoms, which appears as bright emission lines against a dark background. Choice A is wrong because the gas produces emission lines only when viewed from the side, not absorption lines, and the spectra differ by position. Choice B incorrectly assumes both observers see absorption spectra—the side observer sees emission lines instead. Choice D is wrong because cool gas does emit light through re-emission of absorbed photons, making it visible from the side. The correct answer is C: the direct observer sees absorption lines (continuous spectrum with missing wavelengths), while the side observer sees emission lines (bright lines against darkness). Remember this geometry rule: absorption spectra appear when looking through gas toward a continuous source, while emission spectra appear when viewing the gas from the side after it has absorbed energy.

Question 17

A spectral line of neutral helium (He I) is known to have a rest wavelength of λ0=587.56\lambda_0 = 587.56 nm. In the spectrum of a distant star, this same line is observed at a wavelength of λobs=587.72\lambda_{obs} = 587.72 nm. What can be inferred about the star?

  1. Its atmosphere is composed primarily of an unknown element, as the observed wavelength does not exactly match that of helium.
  2. It contains neutral helium in its atmosphere and is approaching Earth at approximately 82 km/s.
  3. It contains neutral helium in its atmosphere and is receding from Earth at approximately 82 km/s. (correct answer)
  4. It is a very hot star where all helium has been ionized, and the observed line is actually from a different element.
Explanation: When you encounter spectral line problems, you're dealing with the Doppler effect - the change in wavelength due to relative motion between source and observer. The key insight is whether the observed wavelength is longer (redshifted) or shorter (blueshifted) than the rest wavelength. Here, the helium line has shifted from 587.56 nm to 587.72 nm - an increase of 0.16 nm. This redshift indicates the star is moving away from us. Using the Doppler formula for radial velocity: v=c×Δλλ0v = c \times \frac{\Delta\lambda}{\lambda_0} Where Δλ=587.72−587.56=0.16\Delta\lambda = 587.72 - 587.56 = 0.16 nm. Substituting: v=(3×108 m/s)×0.16587.56=81.6 km/sv = (3 \times 10^8 \text{ m/s}) \times \frac{0.16}{587.56} = 81.6 \text{ km/s} This confirms answer C is correct - the star contains neutral helium and is receding at approximately 82 km/s. Answer A is wrong because small wavelength shifts are expected due to Doppler motion, not unknown elements. The line identification as helium is still valid. Answer B incorrectly interprets the redshift as blueshift. If the star were approaching, we'd see a shorter wavelength, not longer. Answer D misunderstands the physics. A redshifted helium line doesn't indicate ionization - it indicates motion. Ionized helium would produce entirely different spectral lines at different wavelengths. Remember: redshift means recession, blueshift means approach. Always check whether the observed wavelength is longer or shorter than the rest wavelength before calculating velocity direction.

Question 18

An astronomer obtains a spectrum of an unresolved point source. The spectrum consists of a faint, nearly featureless continuous spectrum with a single, exceptionally strong and narrow emission line of doubly ionized oxygen ([O III]) at 500.7 nm. Which of the following objects is the most likely source?

  1. A normal main-sequence G-type star similar to the Sun.
  2. A planetary nebula, which consists of a hot central star ionizing a surrounding shell of low-density gas. (correct answer)
  3. A reflection nebula, where starlight is scattered by a cloud of interstellar dust.
  4. A brown dwarf, whose cool atmosphere is characterized by molecular absorption bands.
Explanation: When you encounter a spectrum question in astronomy, focus on matching the observed spectral features to the physical conditions that produce them. The key clues here are the faint continuum combined with an exceptionally strong, narrow [O III] emission line. Planetary nebulae are perfect factories for producing exactly this type of spectrum. The hot central white dwarf (with surface temperatures around 50,000-100,000 K) emits intense ultraviolet radiation that ionizes the surrounding gas shell ejected during the star's red giant phase. This creates doubly ionized oxygen (O²⁺) in the low-density nebular gas. When these ions recombine and de-excite, they produce the characteristic bright [O III] emission lines. The "forbidden" nature of these transitions (indicated by the brackets) occurs readily in the extremely low-density environment of nebulae, where collisions are rare. The faint continuum comes from the hot central star, which appears dim because white dwarfs are small. Option A is wrong because main-sequence stars like the Sun show absorption spectra with many stellar absorption lines, not strong emission features. Option C fails because reflection nebulae only scatter existing starlight without creating new emission lines - they show the spectrum of the illuminating star. Option D is incorrect since brown dwarfs have cool temperatures (under 3,000 K) that produce molecular absorption bands, not high-energy emission lines requiring extreme ionization. Remember: strong forbidden emission lines like [O III] are signatures of hot, low-density ionized gas - immediately think planetary nebulae, H II regions, or other ionized nebular objects.

Question 19

An astronomer first observes a star's spectrum with a low-resolution spectrograph, which reveals a single, broad absorption feature around 589 nm. They then re-observe the star with a high-resolution spectrograph. This new spectrum reveals that the single feature is actually two distinct, narrow absorption lines very close together (at 589.0 nm and 589.6 nm). What is the most significant new piece of information revealed by the high-resolution spectrum?

  1. The star's radial velocity can now be measured for the first time.
  2. The element responsible can be definitively identified as sodium, which has a well-known doublet at these wavelengths. (correct answer)
  3. The star's surface temperature can be determined more accurately from the line separation.
  4. The star can be identified as a spectroscopic binary, with each star contributing one of the lines.
Explanation: When you encounter spectroscopy questions involving resolution changes, focus on what new spectral details become visible and what information those details provide. The key insight here is recognizing the sodium doublet. The wavelengths 589.0 nm and 589.6 nm correspond precisely to sodium's characteristic D-lines (D₂ and D₁), which are among the most famous absorption features in stellar spectroscopy. This doublet arises from sodium's electronic structure—specifically transitions from the 3p to 3s energy levels, where spin-orbit coupling splits the 3p level into two slightly different energies. The high-resolution spectrum reveals this splitting, allowing definitive identification of sodium as the absorbing element. Choice A is incorrect because radial velocity can be measured from any absorption line by detecting Doppler shifts—the low-resolution spectrum would have been sufficient for this measurement. Choice C is wrong because line separation in doublets reflects atomic physics (spin-orbit coupling), not stellar temperature. While temperature affects line strength, the 0.6 nm separation between these lines is an intrinsic property of sodium atoms. Choice D misinterprets the situation—in a spectroscopic binary, you'd expect to see Doppler-shifted versions of the same lines, not the precise wavelengths of a known atomic doublet. Study tip: Memorize the major atomic doublets, especially sodium's D-lines at 589 nm. When you see two closely spaced lines at characteristic wavelengths, think atomic structure rather than binary motion or temperature effects. The precision of the wavelength match is your clue to elemental identification.

Question 20

The light from a background star is observed to be passing through a very dense, cold interstellar molecular cloud. In the visible spectrum, the star is completely obscured. However, astronomers are able to obtain a spectrum in the near-infrared, which shows a series of sharp, dense absorption lines. What is the most likely origin of these infrared absorption lines?

  1. Electronic transitions of hydrogen and helium atoms in the cloud that are too weak to be seen in the visible.
  2. Vibrational and rotational transitions of molecules, such as carbon monoxide (CO) and water (H2O), within the cold cloud. (correct answer)
  3. The combined, redshifted light from numerous background galaxies shining through the cloud.
  4. Absorption lines from the star's own photosphere, which are only able to penetrate the cloud at infrared wavelengths.
Explanation: When you encounter questions about interstellar clouds and spectroscopy, focus on how different physical conditions produce characteristic spectral signatures at specific wavelengths. Dense, cold molecular clouds create ideal conditions for molecular formation and produce distinctive infrared absorption spectra. In these frigid environments (typically 10-50 K), molecules like CO, H₂O, and other compounds undergo vibrational and rotational transitions that occur at infrared wavelengths. These molecular motions require less energy than electronic transitions, making them observable in the infrared even when visible light is completely blocked by dust. The "sharp, dense" nature of the lines described is characteristic of these well-defined molecular transitions. Option A is incorrect because electronic transitions of hydrogen and helium typically occur at visible and UV wavelengths, not infrared, and wouldn't suddenly become visible just because the observation shifts to infrared. Option C misunderstands the scenario entirely—redshifted galaxy light wouldn't create absorption lines, and the question specifically describes absorption, not emission from background sources. Option D fails because if the cloud completely obscures visible light, it would also block the star's infrared light; stellar photospheric lines couldn't penetrate through such dense material. The key insight is that molecular clouds have a "fingerprint" in the infrared due to their cold temperatures and molecular composition. Remember this pattern: when you see cold, dense interstellar material combined with infrared observations showing sharp absorption lines, think molecular vibrational and rotational transitions—this is a fundamental tool astronomers use to study interstellar chemistry.