Emission and Absorption Spectra - College Physics

1. In stellar astronomy, the strength of hydrogen absorption lines in a star's spectrum is used to determine stellar temperature. Cooler stars show strong hydrogen absorption in the Balmer series, while very hot stars show weak Balmer absorption. Which explanation best accounts for this temperature dependence?
   Answer: In very hot stars, most hydrogen is ionized, leaving fewer neutral atoms to produce Balmer absorption When you encounter questions about stellar spectra and temperature, think about ionization equilibrium - the balance between neutral atoms and ions at different temperatures. The Balmer series corresponds to electron transitions in neutral hydrogen atoms (specifically from higher energy levels down to n=2). For strong Balmer absorption to occur, you need abundant neutral hydrogen atoms in the appropriate excited states. At moderate stellar temperatures (around 10,000 K), conditions are ideal: there's enough thermal energy to populate the n=2 level, but not so much that hydrogen becomes completely ionized. In very hot stars (above ~25,000 K), thermal energy becomes so intense that most hydrogen atoms lose their electrons entirely, becoming ionized (H⁺). Since Balmer absorption requires neutral hydrogen atoms with electrons that can absorb photons, fewer neutral atoms means weaker absorption lines. This is why option B correctly explains the temperature dependence. Option A incorrectly suggests hot stars contain more hydrogen - stellar hydrogen abundance doesn't correlate with temperature this way, and saturation would actually strengthen lines, not weaken them. Option C mentions molecular dissociation, but hydrogen molecules dissociate at much lower temperatures than those affecting Balmer lines, and stellar hydrogen is primarily atomic anyway. Option D describes thermal broadening, which does occur at high temperatures, but broadened lines are still detectable - they don't disappear. Remember this key principle: stellar spectroscopy is fundamentally about ionization states. When analyzing absorption line strength versus temperature, always consider whether the atoms responsible for those lines can exist in the required ionization state at the given temperature.
2. A student observes the emission spectrum of hydrogen and notes that the Balmer series shows lines at 656 nm, 486 nm, 434 nm, and 410 nm. Based on this pattern, which wavelength would most likely represent the next line in this series?
   Answer: 397 nm, following the decreasing wavelength trend When you encounter hydrogen emission spectra questions, you're dealing with electron transitions between energy levels that follow predictable mathematical patterns. The Balmer series represents transitions from higher energy levels (n ≥ 3) down to the n = 2 level. Looking at the given wavelengths—656 nm, 486 nm, 434 nm, and 410 nm—you can see they correspond to the n = 3→2, n = 4→2, n = 5→2, and n = 6→2 transitions respectively. The wavelengths decrease as the energy gap increases for transitions from higher levels. Following this pattern, the next transition (n = 7→2) should produce a wavelength slightly shorter than 410 nm. Using the Rydberg equation, this calculates to approximately 397 nm, making choice A correct. Choice B (380 nm) represents a common misconception about the series limit. While the Balmer series does approach a limit around 365 nm as n approaches infinity, 380 nm isn't that limit value. Choice C (750 nm) falls outside the visible Balmer series entirely—wavelengths this long belong to the infrared Paschen series (transitions to n = 3). Choice D (364 nm) is actually closer to the true series limit, but the question asks for the "next line," meaning the n = 7→2 transition, not the theoretical limit. Remember: emission series follow decreasing wavelength patterns as you move to higher initial energy levels, and each series has a characteristic limit that the wavelengths approach but never quite reach.
3. A gas of hydrogen atoms is illuminated with white light. Dark absorption lines are observed in the transmitted spectrum at wavelengths corresponding to transitions ending at n = 2. Which statement best explains why no absorption lines are observed for transitions ending at n = 4?
   Answer: Transitions ending at n = 4 require electrons to start from higher energy levels that are not significantly populated at room temperature When analyzing atomic absorption spectra, you need to consider both the energy requirements for transitions AND the initial population of energy levels in the atoms. Absorption occurs when electrons jump from a lower energy level to a higher one by absorbing photons. For transitions ending at n = 4, electrons must start from even higher levels (n = 5, 6, 7, etc.) and absorb photons to reach n = 4. However, at room temperature, virtually all hydrogen atoms exist in the ground state (n = 1) due to the Maxwell-Boltzmann distribution - higher energy levels are exponentially less populated as energy increases. The observed absorption lines ending at n = 2 (the Balmer series) occur because electrons start from n = 1 and absorb visible light photons to jump to n = 2. Since ground state atoms are abundant, these absorptions are easily observed. Answer D correctly identifies that transitions ending at n = 4 require starting populations in n = 5+ levels, which are essentially empty at room temperature, making these absorptions undetectable. Answer A is wrong because n = 4 transitions actually involve lower energy photons than ground state to n = 4 transitions. Answer B misunderstands the process - electrons would need to start from higher levels (n > 4), not lower ones, to end at n = 4. Answer C incorrectly suggests instability; the n = 4 level is perfectly stable. Remember: atomic absorption depends on both photon energy AND initial state populations. Always consider the thermal distribution of electrons when predicting which transitions you'll observe.
4. A helium discharge tube produces emission lines at 587.6 nm and 667.8 nm. When white light is passed through cool helium gas, absorption lines are observed at these same wavelengths. If the intensity of the 587.6 nm emission line is much stronger than the 667.8 nm line, what can be predicted about the absorption lines?
   Answer: The 587.6 nm absorption line will be much darker than the 667.8 nm line This question tests your understanding of atomic spectroscopy and the relationship between emission and absorption processes in atoms. When you encounter problems about spectral lines, remember that emission and absorption are complementary processes involving the same energy transitions. The key principle here is that both emission and absorption involve identical electronic transitions between energy levels in helium atoms. When helium atoms emit light, electrons drop from higher to lower energy states, releasing photons at specific wavelengths. The intensity of each emission line depends on how many atoms undergo that particular transition, which relates to the population of electrons in the upper energy level and the transition probability. During absorption, the reverse process occurs: atoms absorb photons of the same wavelengths to promote electrons to higher energy states. The strength of an absorption line (how much light gets absorbed) depends on the same factors that determine emission intensity - the number of atoms available for that transition and the transition probability. Since the 587.6 nm emission line is much stronger, this indicates more atoms are involved in this transition, making the corresponding absorption line much darker. Choice A is correct because stronger emission correlates with stronger absorption for the same transition. Choice B reverses this relationship incorrectly. Choice C ignores the fundamental connection between emission and absorption processes. Choice D incorrectly suggests no relationship exists when they're actually directly related. Remember: emission and absorption spectra are mirror images of each other. Strong emission lines correspond to strong (dark) absorption lines because they involve the same atomic transitions.
5. A neon sign produces its characteristic red-orange glow through electron transitions in neon atoms. If this same neon gas is placed in front of a broad-spectrum light source (containing all visible wavelengths), which observation would be expected?
   Answer: Dark absorption lines will appear at the same wavelengths as the neon emission lines When you encounter questions about atomic emission and absorption, remember that these are complementary processes involving the same energy transitions. The key insight is that atoms can both emit and absorb photons at identical wavelengths corresponding to specific electron energy level differences. In a neon sign, electrons in neon atoms are excited to higher energy levels and emit characteristic red-orange light when they fall back to lower levels. When you place the same neon gas in front of a broad-spectrum light source, the atoms will absorb photons at precisely the same wavelengths they would emit. This creates dark absorption lines in the transmitted spectrum at those specific wavelengths - making option B correct. Option A is wrong because absorption removes light at specific wavelengths rather than enhancing it. The transmitted light will actually be dimmer in the red-orange region where neon absorbs. Option C incorrectly assumes the broad-spectrum source will cause more emission than the electrical excitation in a neon sign. While some emission might occur, the dominant effect you'd observe is absorption, not enhanced emission. Option D fails to recognize that the physical processes are different - in the sign, atoms are electrically excited and emit light, while in front of the broad-spectrum source, ground-state atoms absorb specific wavelengths. Remember this fundamental principle: atoms emit and absorb at the same characteristic wavelengths. This connection between emission and absorption spectra is crucial for understanding how we identify elements in stars and other distant objects.
6. A student measures the wavelengths of the first four lines in the Balmer series of hydrogen and plots $\frac{1}{λ}$ versus $\frac{1}{n^2}$ where n is the principal quantum number of the upper level. Based on the Rydberg formula, what should be observed in this plot?
   Answer: A straight line with y-intercept equal to $\frac{R}{4}$ where R is the Rydberg constant When analyzing spectroscopic data, you're often looking for linear relationships that reveal fundamental constants. The Balmer series represents hydrogen transitions from higher energy levels (n ≥ 3) down to n = 2. The Rydberg formula for hydrogen is: $\frac{1}{λ} = R\left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right)$ For the Balmer series, $n_f = 2$ (final state) and $n_i = n$ (initial state), so: $\frac{1}{λ} = R\left(\frac{1}{4} - \frac{1}{n^2}\right) = \frac{R}{4} - \frac{R}{n^2}$ Rearranging: $\frac{1}{λ} = -R \cdot \frac{1}{n^2} + \frac{R}{4}$ This is the equation of a straight line in the form y = mx + b, where plotting $\frac{1}{λ}$ versus $\frac{1}{n^2}$ gives a slope of -R and y-intercept of $\frac{R}{4}$. Answer A is wrong because the slope is actually -R, not +R. Answer B incorrectly suggests the slope equals $R\left(\frac{1}{4} - \frac{1}{n^2}\right)$, which is actually the entire expression for $\frac{1}{λ}$, not the slope. Answer C is wrong because the Rydberg formula produces a linear relationship, not a curve. Answer D correctly identifies that the y-intercept equals $\frac{R}{4}$, which occurs when $\frac{1}{n^2} = 0$ (n → ∞). Study tip: When you see spectroscopic series problems, immediately write out the Rydberg formula and rearrange it into y = mx + b form to identify slopes and intercepts for linear plots.
7. Two identical samples of hydrogen gas are prepared: Sample A at 300 K and Sample B at 3000 K. Both samples are excited with identical light sources and their emission spectra are recorded. How will the relative intensities of different emission lines compare between the two samples?
   Answer: Sample B will show stronger high-energy transitions and weaker low-energy transitions compared to Sample A When you encounter questions about emission spectra at different temperatures, you're dealing with how thermal energy affects the population of atomic energy levels according to the Boltzmann distribution. At higher temperatures, atoms have more thermal kinetic energy, leading to more collisions that can excite electrons to higher energy states. The Boltzmann distribution tells us that while lower energy states are still more populated than higher ones, the relative population of high-energy states increases dramatically with temperature. Since emission line intensity depends on how many atoms are in the excited state that produces that transition, higher temperatures favor stronger high-energy emission lines. Sample B at 3000 K (10 times hotter than Sample A) will have significantly more atoms in high-energy excited states, making high-energy transitions much more intense. Conversely, the relative intensity of low-energy transitions decreases because fewer atoms remain in the lower excited states—they've been promoted to higher levels by thermal collisions. Choice A is incorrect because temperature directly affects the population distribution among energy levels, so intensities cannot be identical. Choice C reverses the correct relationship—the cooler sample cannot have more high-energy transitions since it lacks sufficient thermal energy. Choice D is wrong because while high-energy lines become much stronger at high temperatures, low-energy transitions don't disappear entirely; they just become relatively weaker. Remember: higher temperature means more thermal energy available to populate higher energy states, shifting the emission spectrum toward higher-energy (shorter wavelength) lines.
8. The spectrum of light from a distant galaxy shows hydrogen absorption lines that are shifted to longer wavelengths compared to laboratory measurements. However, the pattern and relative spacing of the lines remain unchanged. What does this observation indicate about the hydrogen atoms in the galaxy?
   Answer: The hydrogen atoms are identical to those on Earth, but the galaxy is moving away from us This question tests your understanding of the Doppler effect and how it applies to astronomical observations. When you encounter spectral line shifts in astronomy problems, think about what causes wavelength changes while preserving the underlying atomic physics. The key insight is that hydrogen atoms are universal - they have the same energy level structure everywhere in the universe. When light from these atoms travels through expanding space or from a moving source, the wavelengths get stretched (redshifted) or compressed (blueshifted), but the relative spacing between spectral lines remains constant because the energy level differences in hydrogen atoms haven't changed. Answer B correctly identifies that the hydrogen atoms are identical to Earth's hydrogen, but the observed redshift indicates the galaxy is receding from us. This wavelength shift occurs due to the Doppler effect - either from the galaxy's motion or the expansion of space itself. Answer A is wrong because extreme conditions don't alter hydrogen's fundamental energy level structure. The atomic physics remains the same regardless of location. Answer C incorrectly attributes the shift to temperature. While temperature affects line intensity and broadening, it doesn't systematically shift all lines to longer wavelengths while preserving their relative spacing. Answer D misunderstands quantum mechanics. Different quantum states would change the relative intensities of various lines and could produce entirely different spectral patterns, not a uniform wavelength shift of the same pattern. Remember: when spectral patterns are preserved but uniformly shifted in wavelength, think motion and Doppler effect, not altered atomic physics.
9. An astronomer analyzes the spectrum of a distant star and observes both emission and absorption lines for hydrogen. The emission lines appear at wavelengths of 656 nm, 486 nm, and 434 nm, while absorption lines appear at 121 nm and 103 nm. What can be concluded about the different regions of the star producing these spectral features?
   Answer: The emission lines come from the hot stellar core, while absorption lines come from the cooler stellar atmosphere When analyzing stellar spectra, you need to understand how temperature affects the production of emission versus absorption lines. Stars have layered structures with hot cores and relatively cooler outer atmospheres, and each region produces different spectral signatures. The wavelengths given reveal important clues about the physical conditions in different stellar regions. The emission lines at 656 nm, 486 nm, and 434 nm correspond to the Balmer series of hydrogen, which forms when electrons drop from higher energy levels to n=2. These longer wavelengths in the visible spectrum are typically produced in hot, dense regions where atoms are highly excited and can emit photons freely. The absorption lines at 121 nm and 103 nm belong to the Lyman series, where electrons transition to n=1, producing shorter UV wavelengths. These form when cooler hydrogen atoms absorb specific wavelengths from continuous light passing through them. Answer A correctly identifies that emission lines originate from the hot stellar core, while absorption lines come from the cooler stellar atmosphere. Answer B reverses this relationship incorrectly. Answer C suggests temporal differences rather than spatial ones, missing the key point about different stellar regions. Answer D incorrectly assumes the lines come from different elements, when both sets are clearly hydrogen transitions. Remember this pattern: hot, dense stellar cores produce emission spectra as excited atoms release energy, while cooler outer atmospheres create absorption spectra as atoms absorb specific wavelengths from the core's continuous spectrum. Temperature determines whether you see emission or absorption features.
10. A neon gas discharge tube emits light at wavelengths of 640 nm, 585 nm, and 510 nm among others. If the same neon gas at room temperature is illuminated with white light containing all these wavelengths, which of the following best describes what will be observed in the absorption spectrum?
    Answer: Weak or no absorption at 640 nm, 585 nm, and 510 nm because most neon atoms are in the ground state In the discharge tube, neon atoms are excited to high energy levels and emit photons when returning to lower states. However, at room temperature, nearly all neon atoms are in their ground state. The emission lines at 640 nm, 585 nm, and 510 nm likely correspond to transitions between excited states (not involving the ground state), so ground-state atoms cannot absorb these specific photons. Absorption would only occur for wavelengths corresponding to transitions FROM the ground state. Choice A incorrectly assumes emission and absorption spectra are always identical. Choice B is wrong because noble gases can still absorb photons. Choice D incorrectly suggests temperature changes the fundamental energy level differences.
11. In a laboratory experiment, sodium vapor is illuminated with white light, and the absorption spectrum shows dark lines at 589.0 nm and 589.6 nm (the sodium D-doublet). When the same sodium vapor is excited electrically, bright emission lines appear at exactly the same wavelengths. However, when the sodium vapor is first heated to 800°C and then the absorption experiment is repeated, which of the following changes would most likely be observed?
    Answer: Additional absorption lines appear at shorter wavelengths corresponding to transitions from thermally populated excited states When you encounter questions about atomic spectra and temperature effects, think about how thermal energy affects the population of atomic energy levels and creates new transition possibilities. At room temperature, sodium atoms exist primarily in their ground state, so absorption occurs only from the ground state to excited states, producing the familiar D-doublet lines at 589.0 nm and 589.6 nm. However, when you heat the sodium vapor to 800°C, thermal energy becomes significant compared to the energy gaps between atomic levels. This thermal energy populates higher energy states that were previously unoccupied, following the Boltzmann distribution $N_i \propto e^{-E_i/kT}$. Once these excited states contain appreciable populations of atoms, new absorption transitions become possible. Atoms can now absorb photons to transition from these thermally populated excited states to even higher energy levels. Since these transitions start from higher initial energy states, they require less energy and correspond to shorter wavelengths than the original ground-state transitions. Answer choice A is wrong because thermal motion doesn't prevent specific wavelength absorption—it affects line broadening, not the fundamental absorption process. Choice B incorrectly suggests only broadening occurs; while some broadening does happen, the key change is the appearance of new lines. Choice C misunderstands thermal effects—atomic energy levels don't expand with temperature, and thermal expansion would be negligible compared to electronic energy scales. For spectroscopy problems, remember that temperature creates two main effects: line broadening due to increased atomic motion, and new absorption lines from thermally populated excited states at high temperatures.
12. A sodium vapor lamp emits characteristic yellow light at 589 nm. This emission occurs when electrons transition from an excited state to the ground state. If a beam of this yellow light is then passed through cool sodium vapor, what will be observed?
    Answer: A dark absorption line will appear at 589 nm in the transmitted light spectrum This question tests your understanding of atomic absorption and emission spectra, which are fundamental to how we identify elements in astronomy and chemistry. When you see problems involving light interacting with atoms of the same element that produced it, think about the reversibility of atomic transitions. Sodium atoms emit 589 nm yellow light when electrons fall from an excited state to the ground state. The key insight is that this process is reversible: if ground-state sodium atoms absorb photons of exactly 589 nm, their electrons will jump up to that same excited state. When the emitted light passes through cool sodium vapor (where atoms are mostly in the ground state), those atoms will selectively absorb the 589 nm photons, creating a dark absorption line at that precise wavelength in the transmitted spectrum. Looking at the wrong answers: (A) is incorrect because the sodium atoms will definitely interact with light of the exact energy they can absorb. (C) describes scattering, but this isn't a scattering phenomenon—it's specific absorption at discrete energy levels. The absorbed energy gets re-emitted in random directions, but the key observation is the missing wavelength in the forward beam. (D) mentions the Doppler effect, which only occurs when there's relative motion between source and observer, not simply from light passing through a medium. Remember this pattern: atoms absorb the same wavelengths they emit. This principle explains how we identify elements in distant stars—their absorption spectra reveal which elements are present in their atmospheres.
13. An electron in a hydrogen atom absorbs a photon and transitions from n = 2 to n = 4. Later, this electron emits two photons in sequence as it returns to the ground state. If the first emitted photon has wavelength λ₁ and the second has wavelength λ₂, which relationship must be satisfied?
    Answer: $\frac{1}{λ₁} + \frac{1}{λ₂} = \frac{1}{λ_{2→1}}$ When analyzing hydrogen atom transitions, the key principle is energy conservation. The energy of a photon relates to its wavelength by $E = \frac{hc}{λ}$, so shorter wavelengths correspond to higher energies. The electron absorbs energy to jump from n = 2 to n = 4, then returns to ground state (n = 1) by emitting two photons. There are two possible emission pathways: either 4→3→1 or 4→2→1. In both cases, energy conservation requires that the total energy emitted equals the energy difference between n = 4 and n = 1. Since $E = \frac{hc}{λ}$, we can write: $\frac{hc}{λ₁} + \frac{hc}{λ₂} = \frac{hc}{λ_{4→1}}$ Dividing by hc gives us: $\frac{1}{λ₁} + \frac{1}{λ₂} = \frac{1}{λ_{4→1}}$ The notation $λ_{2→1}$ in choice C represents the wavelength for a transition from n = 2 to n = 1, which is equivalent to $λ_{4→1}$ in terms of the final energy state reached. Choice A incorrectly relates the emitted photons to the absorbed photon (which went from n = 2 to n = 4, not n = 4 to n = 1). Choice B adds wavelengths directly, ignoring that energy is inversely proportional to wavelength. Choice D creates an arbitrary product relationship with no physical basis. Remember: in atomic transitions, always think in terms of energy conservation and the inverse relationship between photon energy and wavelength. When multiple photons are involved, add their energies (which means adding the reciprocals of their wavelengths).
14. A sample of hydrogen gas at low density is illuminated with a beam of photons, each having energy 12.1 eV. The hydrogen atoms are initially all in the ground state (n = 1, energy = -13.6 eV). What will be observed in the emission spectrum after illumination?
    Answer: Both Lyman and Balmer series will be observed When photons interact with hydrogen atoms, you need to consider both absorption and subsequent emission processes. The key is determining what energy levels the atoms can reach and what transitions become possible. The 12.1 eV photons can excite ground state hydrogen atoms ($E_1 = -13.6$ eV) to the $n = 3$ level, since $E_3 = -13.6/3^2 = -1.51$ eV, requiring $13.6 - 1.51 = 12.09$ eV for the transition. This matches closely with the available photon energy. The atoms cannot reach $n = 4$ because that would require 12.75 eV, which exceeds the photon energy. Once excited to $n = 3$, the atoms can emit photons through multiple pathways: they can transition directly back to $n = 1$ (producing Lyman series emission) or cascade down by first going to $n = 2$, then to $n = 1$. The $3 \rightarrow 2$ transition produces Balmer series emission, while the subsequent $2 \rightarrow 1$ transition contributes to the Lyman series. Option A is incorrect because atoms don't only transition directly to the ground state—they can cascade through intermediate levels. Option B is wrong because while some atoms transition $3 \rightarrow 2$ (Balmer), others will transition $2 \rightarrow 1$ (Lyman) as part of the cascade. Option D is incorrect because 12.1 eV photons have sufficient energy to excite the atoms from the ground state. Study tip: In atomic spectroscopy problems, always check what energy levels are accessible with the given photon energy, then consider all possible downward transitions—atoms rarely take just one path back to ground state.
15. In a laboratory experiment, students heat different metallic elements and observe their emission spectra. They notice that each element produces a unique pattern of colored lines. Which statement best explains why each element has a distinctive emission spectrum?
    Answer: Each element has a unique electronic structure with specific allowed energy levels When you encounter questions about atomic emission spectra, you're dealing with quantum mechanics and electron behavior within atoms. The key insight is that electrons can only exist at specific energy levels, and light emission occurs when electrons transition between these levels. Each element has a unique emission spectrum because of its distinctive electronic structure. Electrons in atoms occupy specific energy levels or orbitals, and these allowed energy levels are different for each element due to their unique nuclear charge and electron arrangements. When electrons absorb energy (like heat), they jump to higher energy levels. As they fall back down, they emit photons with energies exactly equal to the difference between energy levels. Since $E = hf$ (where h is Planck's constant and f is frequency), specific energy differences produce specific frequencies and wavelengths of light, creating the characteristic colored lines you observe. Option A is incorrect because density doesn't determine electron energy levels or light emission patterns. Option C fails because atomic mass has no direct relationship to the electronic energy level structure that governs emission spectra. Option D is wrong because while different temperatures affect the intensity of emission, the specific wavelengths emitted remain characteristic of each element regardless of temperature. Remember this key principle: emission spectra are like atomic fingerprints. Just as no two people have identical fingerprints, no two elements have identical emission spectra because each has a unique arrangement of electrons and energy levels. This is why astronomers can identify elements in distant stars just by analyzing their light.
16. A mercury vapor lamp produces a characteristic blue-green line at 436 nm when mercury atoms transition from an excited state (call it state E) to a lower excited state (call it state L). If this lamp is used to illuminate mercury vapor at room temperature, under what conditions would a dark absorption line appear at 436 nm?
    Answer: Only if the mercury vapor is first excited so that significant numbers of atoms populate state L Absorption requires atoms to be initially in the lower energy state of the transition. The 436 nm emission corresponds to state E → state L transition, so absorption of 436 nm photons requires atoms to be in state L to be excited to state E. At room temperature, mercury atoms are predominantly in the ground state, not in the excited state L. Only if the vapor is pre-excited (by electrical discharge, heating, or other means) to populate state L would absorption at 436 nm be observed. Choice A is incorrect because absorption requires population of the lower state. Choice C focuses on density rather than energy state population. Choice D incorrectly states that atoms cannot absorb wavelengths they emit.
17. An electron in a hydrogen atom transitions from the n = 4 energy level to the n = 2 energy level. If the energy of the n = 2 level is -3.4 eV and the energy of the n = 4 level is -0.85 eV, what type of electromagnetic radiation is emitted and what is its energy?
    Answer: Visible light with energy 2.55 eV When you encounter atomic energy level transitions, you're dealing with the emission or absorption of photons with specific energies. The key principle is energy conservation: the photon energy equals the difference between initial and final energy levels. To find the emitted photon energy, calculate the energy difference between the two levels. Since the electron drops from n = 4 to n = 2, you subtract the final energy from the initial energy: $E_{photon} = E_4 - E_2 = (-0.85 \text{ eV}) - (-3.4 \text{ eV}) = 2.55 \text{ eV}$. Now you need to identify what type of electromagnetic radiation has this energy. Photons with energies around 1.65-3.3 eV correspond to visible light wavelengths (roughly 380-750 nm). Looking at the wrong answers: Choice A incorrectly categorizes this as ultraviolet radiation. While the energy calculation (2.55 eV) is correct, UV radiation requires higher energies, typically above 3.1 eV. Choice C makes a calculation error, adding the energy levels instead of finding their difference (3.4 + 0.85 = 4.25 eV), and incorrectly identifies this as infrared, which has much lower energies (less than 1.65 eV). Choice D again gets the energy right but misidentifies the radiation type—X-rays have much higher energies, typically thousands of eV. Remember that the n = 4 to n = 2 transition in hydrogen is part of the famous Balmer series, which produces visible light. When you see hydrogen transitions ending at n = 2, think visible light spectrum.
18. In a laboratory, students observe the emission spectrum of hydrogen and notice that the Balmer series (transitions ending at n = 2) produces visible light, while the Lyman series (transitions ending at n = 1) produces ultraviolet light. What can be concluded about the Paschen series (transitions ending at n = 3)?
    Answer: It produces infrared radiation because the energy differences are smaller than those in the Balmer series When analyzing hydrogen emission spectra, you need to understand how energy differences between electron shells determine the type of electromagnetic radiation emitted. The key principle is that larger energy gaps produce higher-energy photons (shorter wavelengths), while smaller energy gaps produce lower-energy photons (longer wavelengths). The energy required to move an electron decreases as you move farther from the nucleus. Since the Lyman series involves transitions to n = 1 (closest to nucleus), these transitions release the most energy, producing high-energy UV radiation. The Balmer series transitions to n = 2, releasing moderate energy that corresponds to visible light wavelengths. For the Paschen series (transitions to n = 3), the energy differences are smaller than both Lyman and Balmer series because n = 3 is farther from the nucleus than n = 2 or n = 1. Smaller energy differences mean lower-energy photons, which correspond to longer wavelengths in the infrared region. This makes choice A correct. Choice B is wrong because X-rays require much larger energy differences than any hydrogen transition can provide. Choice C incorrectly suggests visible light production—infrared wavelengths are longer than visible light, not visible light itself. Choice D contradicts the energy-distance relationship by claiming UV radiation with shorter wavelengths. Study tip: Remember the pattern for hydrogen series: the lower the final energy level (n), the higher the energy released and the shorter the wavelength. Lyman (n=1) → UV, Balmer (n=2) → visible, Paschen (n=3) → infrared.
19. A spectroscopic analysis reveals that one star's absorption spectrum shows hydrogen lines shifted to shorter wavelengths by 0.3%, and helium lines also shifted to shorter wavelengths by 0.3%. A second star shows hydrogen lines shifted to longer wavelengths by 0.4%, but its helium lines are shifted to longer wavelengths by only 0.2%. What can be concluded about these observations?
    Answer: The first star is approaching Earth; the second star is receding with some additional effect affecting the helium lines The first star shows consistent shifts for both elements (0.3% shorter wavelengths), indicating it's approaching Earth due to the Doppler effect. The second star shows redshift for both elements but different amounts (0.4% vs 0.2%), suggesting the star is receding but additional effects beyond simple Doppler motion are present, affecting the elements differently. This could be due to gravitational effects, interstellar absorption, or complex stellar atmospheric conditions. Choice B incorrectly suggests both are approaching. Choice C incorrectly attributes the shifts to temperature and atmospheric effects rather than motion. Choice D is wrong because the inconsistent shifts in the second star cannot be explained by Doppler motion alone.
20. A student observes that when a particular gas is excited in a discharge tube, it emits light at wavelengths of 400 nm, 500 nm, and 600 nm. When the same gas at room temperature is illuminated with light containing wavelengths from 350 nm to 700 nm, absorption occurs only at 400 nm and 500 nm. Why is there no absorption at 600 nm?
    Answer: The 600 nm emission represents a transition between excited states rather than from an excited state to ground state The key insight is that emission can occur between any two energy levels, but absorption from room-temperature gas primarily occurs from the ground state. If 600 nm is emitted but not absorbed, this suggests the 600 nm photon corresponds to a transition between two excited states, neither of which is the ground state. Since atoms at room temperature are in the ground state, they cannot absorb 600 nm photons. The 400 nm and 500 nm wavelengths likely represent transitions from excited states down to the ground state, so ground-state atoms can absorb these wavelengths. Choice A incorrectly suggests longer wavelength means insufficient energy. Choice C incorrectly invokes selection rules. Choice D incorrectly attributes the effect to temperature dependence of absorption probability.

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