IB Physics Quiz: Apply Fusion And Stars
20 questions · exam conditions
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Apply Fusion And StarsQuestion 1 of 20

What is the defining characteristic of a star on the main sequence of the Hertzsprung-Russell diagram?

It is fusing helium into carbon in its core.
It is in the process of contracting from a nebula.
It is fusing hydrogen into helium in its core.
It has exhausted all nuclear fuel in its core.
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IB Physics Quiz

IB Physics Quiz: Apply Fusion And Stars

Practice Apply Fusion And Stars in IB Physics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on Apply Fusion And Stars, giving you a quick way to practice the rules, question types, and explanations that matter most for IB Physics.

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Question 1

What is the defining characteristic of a star on the main sequence of the Hertzsprung-Russell diagram?

  1. It is fusing helium into carbon in its core.
  2. It is in the process of contracting from a nebula.
  3. It is fusing hydrogen into helium in its core. (correct answer)
  4. It has exhausted all nuclear fuel in its core.
Explanation: The main sequence is the long, stable phase of a star's life. During this phase, the star's energy is generated by the nuclear fusion of hydrogen into helium in its core. This process maintains the hydrostatic equilibrium that supports the star against gravity. Stars leave the main sequence when the hydrogen fuel in the core is depleted.

Question 2

Star X has a parallax angle of 0.050 arcseconds. Star Y is at the same distance from Earth as Star X, but has a luminosity that is 16 times greater than Star X. What is the ratio of the radius of Star Y to the radius of Star X (RY/RXR_Y/R_X), assuming both stars have the same surface temperature?

  1. 2
  2. 4 (correct answer)
  3. 8
  4. 16
Explanation: The parallax information is extra detail to confirm the distances are equal. Luminosity (L) is related to radius (R) and surface temperature (T) by the Stefan-Boltzmann law, L=σAT4=σ(4πR2)T4L = \sigma A T^4 = \sigma (4\pi R^2) T^4. So, L∝R2T4L \propto R^2 T^4. If the temperatures are the same (TX=TYT_X = T_Y), then L∝R2L \propto R^2. The ratio of luminosities is LY/LX=(RY/RX)2L_Y/L_X = (R_Y/R_X)^2. We are given LY=16LXL_Y = 16L_X, so 16=(RY/RX)216 = (R_Y/R_X)^2. Taking the square root gives RY/RX=4R_Y/R_X = 4.

Question 3

Star P has a surface temperature of TT and a radius of RR. Star Q has a surface temperature of 2T2T and a radius of R/4R/4. What is the ratio of the luminosity of Star Q to the luminosity of Star P (LQ/LPL_Q/L_P)?

  1. 1/4
  2. 1 (correct answer)
  3. 4
  4. 16
Explanation: Luminosity is given by the Stefan-Boltzmann law, L=4πR2σT4L = 4\pi R^2 \sigma T^4. We can write the ratio LQ/LP=(RQ/RP)2(TQ/TP)4L_Q/L_P = (R_Q/R_P)^2 (T_Q/T_P)^4. We are given RQ=R/4R_Q = R/4 and TQ=2TT_Q = 2T. Substituting these values gives LQ/LP=((R/4)/R)2((2T)/T)4=(1/4)2(2)4=(1/16)×16=1L_Q/L_P = ((R/4)/R)^2 ((2T)/T)^4 = (1/4)^2 (2)^4 = (1/16) \times 16 = 1. The stars have the same luminosity.

Question 4

The stellar parallax method is limited to measuring the distances of relatively nearby stars. What is the fundamental reason for this limitation?

  1. The Earth's orbital diameter provides too short a baseline for the triangulation of very distant stars. (correct answer)
  2. Light from distant stars is too redshifted for accurate positional measurement.
  3. The gravitational pull of the Sun affects the path of light from distant stars, distorting their apparent position.
  4. Atmospheric interference prevents the measurement of the very small angles associated with distant stars.
Explanation: Stellar parallax uses the Earth's orbit as a baseline to measure the apparent shift in a star's position. For very distant stars, this shift (the parallax angle) becomes extremely small, as the baseline (about 2 AU) is insignificant compared to the star's distance. Eventually, the angle becomes too small to be measured with sufficient precision against the background of even more distant objects, even with space-based telescopes. While atmospheric interference is a problem (B), the fundamental limit is the baseline length relative to the distance.

Question 5

Star Alpha has a parallax angle that is twice as large as the parallax angle of Star Beta. Both stars have the same apparent brightness. What is the ratio of the luminosity of Alpha to the luminosity of Beta (Lα/LβL_\alpha/L_\beta)?

  1. 1/4 (correct answer)
  2. 1/2
  3. 2
  4. 4
Explanation: Distance (d) is inversely proportional to the parallax angle (p), so d∝1/pd \propto 1/p. Since pα=2pβp_\alpha = 2p_\beta, the distance to Alpha is half the distance to Beta: dα=dβ/2d_\alpha = d_\beta / 2. Luminosity (L) is related to apparent brightness (b) and distance by L=4πd2bL = 4\pi d^2 b. Since their apparent brightnesses are the same (bα=bβb_\alpha = b_\beta), luminosity is proportional to the square of the distance: L∝d2L \propto d^2. Therefore, the ratio of luminosities is Lα/Lβ=(dα/dβ)2=((dβ/2)/dβ)2=(1/2)2=1/4L_\alpha/L_\beta = (d_\alpha/d_\beta)^2 = ((d_\beta/2)/d_\beta)^2 = (1/2)^2 = 1/4.

Question 6

A red giant star and a main-sequence star have the same luminosity. Which statement correctly compares their surface temperatures and radii?

  1. The red giant has a higher surface temperature and a larger radius.
  2. The red giant has a higher surface temperature and a smaller radius.
  3. The red giant has a lower surface temperature and a smaller radius.
  4. The red giant has a lower surface temperature and a larger radius. (correct answer)
Explanation: Red giants are characterized by their low surface temperatures (making them appear reddish) and very high luminosities. Main-sequence stars can have a wide range of properties. If a red giant has the same luminosity as a main-sequence star, the main-sequence star must be very hot and massive (e.g., a blue star). From the relationship L∝R2T4L \propto R^2 T^4, if LL is the same for both but the red giant's temperature TT is much lower, its radius RR must be much larger to compensate.

Question 7

The distance to Star Kepler is 100 pc and the distance to Star Tycho is 25 pc. The two stars have the same luminosity. What is the ratio of the apparent brightness of Tycho to the apparent brightness of Kepler (bTycho/bKeplerb_{Tycho}/b_{Kepler})?

  1. 1/16
  2. 1/4
  3. 4
  4. 16 (correct answer)
Explanation: Apparent brightness (b) is related to luminosity (L) and distance (d) by the inverse square law: b=L/(4πd2)b = L / (4\pi d^2). Since the luminosities are the same, b∝1/d2b \propto 1/d^2. The ratio of the apparent brightnesses is bTycho/bKepler=(dKepler/dTycho)2b_{Tycho}/b_{Kepler} = (d_{Kepler}/d_{Tycho})^2. Plugging in the values: bTycho/bKepler=(100 pc/25 pc)2=(4)2=16b_{Tycho}/b_{Kepler} = (100\text{ pc}/25\text{ pc})^2 = (4)^2 = 16.

Question 8

A star is observed to have a parallax angle of 2.42×10−72.42 \times 10^{-7} radians. Given that 11 radian is approximately 2.06×1052.06 \times 10^5 arcseconds, what is the approximate distance to this star in parsecs (pc)?

  1. 8.5 \times 10^{11} pc
  2. 50 pc
  3. 4.1 \times 10^6 pc
  4. 20 pc (correct answer)
Explanation: First, convert the parallax angle from radians to arcseconds. parcsec=prad×(2.06×105 arcsec/rad)p_{\text{arcsec}} = p_{\text{rad}} \times (2.06 \times 10^5 \text{ arcsec/rad}). parcsec=(2.42×10−7)×(2.06×105)≈0.050p_{\text{arcsec}} = (2.42 \times 10^{-7}) \times (2.06 \times 10^5) \approx 0.050 arcseconds. The distance in parsecs is the reciprocal of the parallax angle in arcseconds: dpc=1/parcsecd_{\text{pc}} = 1 / p_{\text{arcsec}}. So, dpc=1/0.050=20d_{\text{pc}} = 1 / 0.050 = 20 pc.

Question 9

During the helium flash in a low-mass star (M<2M⊙M < 2 M_{\odot}), the core temperature rises from 1.0×1081.0 \times 10^8 K to 3.0×1083.0 \times 10^8 K. If the triple-alpha reaction rate has a temperature dependence of T40T^{40}, by what factor does the helium fusion rate increase during this temperature rise?

  1. The helium fusion rate increases by a factor of 340≈1.2×10193^{40} \approx 1.2 \times 10^{19} (correct answer)
  2. The helium fusion rate increases by a factor of 2740≈2.3×105727^{40} \approx 2.3 \times 10^{57}
  3. The helium fusion rate increases by a factor of 940≈6.8×10389^{40} \approx 6.8 \times 10^{38}
  4. The helium fusion rate increases by a factor of 120≈4.1×1082120 \approx 4.1 \times 10^{82}
Explanation: The reaction rate scales as T40T^{40}, so the rate increases by (Tf/Ti)40=(3.0×108/1.0×108)40=340≈1.2×1019(T_f/T_i)^{40} = (3.0 \times 10^8/1.0 \times 10^8)^{40} = 3^{40} \approx 1.2 \times 10^{19}. B incorrectly cubes the temperature ratio before applying the exponent. C uses the square of the temperature ratio. D appears to use an incorrect temperature dependence exponent.

Question 10

A Type Ia supernova occurs when a white dwarf accretes material from a companion star and approaches the Chandrasekhar limit. If the white dwarf's initial mass is 0.8M⊙0.8 M_{\odot} and it accretes material at a rate of 10−8M⊙10^{-8} M_{\odot} per year, approximately how long will the accretion process continue before carbon ignition occurs? (Use Chandrasekhar limit = 1.4M⊙1.4 M_{\odot})

  1. Approximately 6×1076 \times 10^7 years, assuming steady accretion and no mass loss through nova explosions (correct answer)
  2. Approximately 2×1082 \times 10^8 years, accounting for the decreasing accretion efficiency as the white dwarf approaches the mass limit
  3. Approximately 8×1088 \times 10^8 years, including the effects of periodic nova outbursts that expel some accreted material
  4. Approximately 1.5×1091.5 \times 10^9 years, considering the gradual decrease in companion star mass transfer rate over time
Explanation: The required mass gain is 1.4M⊙−0.8M⊙=0.6M⊙1.4 M_{\odot} - 0.8 M_{\odot} = 0.6 M_{\odot}. At an accretion rate of 10−8M⊙10^{-8} M_{\odot} per year, the time required is 0.6M⊙/10−8M⊙/year=6×1070.6 M_{\odot} / 10^{-8} M_{\odot}/\text{year} = 6 \times 10^7 years. This is the basic calculation assuming steady accretion. B incorrectly assumes efficiency changes significantly. C overestimates nova mass loss effects. D incorrectly assumes the mass transfer rate decreases substantially over this timescale.

Question 11

A Population III star with zero metallicity and mass 200M⊙200 M_{\odot} has a main sequence lifetime of approximately 3×1063 \times 10^6 years. If this star produces and disperses 50M⊙50 M_{\odot} of newly synthesized heavy elements (metals) into the surrounding medium upon its death, what is the average rate of metal enrichment in units of solar masses per year during the star's lifetime?

  1. The average metal production rate is approximately 8.7×10−2M⊙8.7 \times 10^{-2} M_{\odot} per year considering only the final explosive nucleosynthesis phase
  2. The average metal production rate is approximately 3.3×10−4M⊙3.3 \times 10^{-4} M_{\odot} per year when calculated over the star's total lifetime
  3. The average metal production rate is approximately 2.1×10−3M⊙2.1 \times 10^{-3} M_{\odot} per year during the active nucleosynthesis phases
  4. The average metal production rate is approximately 1.7×10−5M⊙1.7 \times 10^{-5} M_{\odot} per year throughout the star's main sequence lifetime (correct answer)
Explanation: When you encounter stellar nucleosynthesis problems, focus on the fundamental relationship between total production and time duration. Population III stars are the universe's first generation of stars with zero initial metal content, making their metal production particularly significant for cosmic chemical evolution. To find the average metal production rate, you need to divide the total metals produced by the star's entire lifetime. The star produces 50M⊙50 M_{\odot} of heavy elements over its 3×1063 \times 10^6 year main sequence lifetime. The calculation is straightforward: 50M⊙3×106 years=1.67×10−5M⊙\frac{50 M_{\odot}}{3 \times 10^6 \text{ years}} = 1.67 \times 10^{-5} M_{\odot} per year, which rounds to 1.7×10−5M⊙1.7 \times 10^{-5} M_{\odot} per year. Option A incorrectly assumes metal production occurs only during the final explosive phase, dramatically overestimating the rate by using a much shorter timeframe. Option B contains a calculation error, likely from incorrect unit conversion or mathematical mistakes. Option C assumes metals are produced only during "active nucleosynthesis phases," but this is misleading since nucleosynthesis occurs throughout the star's lifetime at varying rates. The key insight is that while most heavy elements are indeed produced and ejected during the supernova explosion, the question asks for the average rate over the star's lifetime. This is a common type of problem in stellar astrophysics where you must distinguish between instantaneous processes and time-averaged quantities. Remember: for average rate calculations in stellar physics, always use the total timespan unless explicitly told otherwise, even when the physical process is concentrated in a brief final phase.

Question 12

A red giant star is losing mass at a rate of 10−7M⊙10^{-7} M_{\odot} per year through stellar winds. The escaping gas has an average speed of 2020 km/s. If the star's photospheric radius is 100R⊙100 R_{\odot}, what is the approximate number density of particles in the stellar wind at a distance of 1000R⊙1000 R_{\odot} from the star's center? (Assume the wind is spherically symmetric and consists primarily of hydrogen atoms.)

  1. Approximately 2.1×1082.1 \times 10^{8} particles per cubic meter at the specified distance from the stellar center
  2. Approximately 3.7×1093.7 \times 10^{9} particles per cubic meter at the specified distance from the stellar center (correct answer)
  3. Approximately 1.5×10111.5 \times 10^{11} particles per cubic meter at the specified distance from the stellar center
  4. Approximately 6.2×10126.2 \times 10^{12} particles per cubic meter at the specified distance from the stellar center
Explanation: Mass loss rate = ρvA=ρv(4πr2)\rho v A = \rho v (4\pi r^2). At r=1000R⊙=7×1011r = 1000 R_{\odot} = 7 \times 10^{11} m: ρ=M˙4πr2v=10−7×2×10304π(7×1011)2×2×104=6.1×10−18\rho = \frac{\dot{M}}{4\pi r^2 v} = \frac{10^{-7} \times 2 \times 10^{30}}{4\pi (7 \times 10^{11})^2 \times 2 \times 10^4} = 6.1 \times 10^{-18} kg/m³. Number density = ρ/(mH)=6.1×10−18/(1.67×10−27)≈3.7×109\rho/(m_H) = 6.1 \times 10^{-18}/(1.67 \times 10^{-27}) \approx 3.7 \times 10^9 particles/m³. A uses incorrect radius conversion. C neglects the r2r^2 dependence. D uses photospheric radius instead of the specified distance.

Question 13

In a neutron star merger, the r-process nucleosynthesis creates elements heavier than iron through rapid neutron capture. If a nucleus captures neutrons at a rate of 102310^{23} neutrons per second and each neutron capture increases the mass number by 1, starting from 56Fe^{56}Fe, approximately how long does it take to produce 200Hg^{200}Hg assuming no beta decay occurs during this timescale?

  1. Approximately 5.2×10−125.2 \times 10^{-12} seconds, which approaches the timescale for competing beta decay processes
  2. Approximately 2.3×10−182.3 \times 10^{-18} seconds, confirming that neutron capture outpaces beta decay during the merger
  3. Approximately 7.8×10−157.8 \times 10^{-15} seconds, allowing multiple neutron captures before any significant beta decay
  4. Approximately 1.4×10−211.4 \times 10^{-21} seconds, which is much shorter than typical beta decay timescales (correct answer)
Explanation: When you encounter r-process nucleosynthesis problems, you're dealing with extremely rapid nuclear reactions where neutron capture rates determine how quickly heavy elements form. The key is recognizing that this is a simple rate calculation. To find the time needed, you need to determine how many neutrons must be captured and divide by the capture rate. Starting from 56Fe^{56}Fe (mass number 56) to reach 200Hg^{200}Hg (mass number 200), the nucleus must capture 200−56=144200 - 56 = 144 neutrons. With a capture rate of 102310^{23} neutrons per second, the time required is: t=144 neutrons1023 neutrons/second=1.44×10−21 secondst = \frac{144 \text{ neutrons}}{10^{23} \text{ neutrons/second}} = 1.44 \times 10^{-21} \text{ seconds} This matches answer choice D (approximately 1.4×10−211.4 \times 10^{-21} seconds). Answer A (5.2×10−125.2 \times 10^{-12} seconds) is far too large, suggesting a calculation error or confusion about the capture rate. Answer B (2.3×10−182.3 \times 10^{-18} seconds) is also too large by several orders of magnitude. Answer C (7.8×10−157.8 \times 10^{-15} seconds) similarly overestimates the time required. The physical reasoning in choice D is also correct: this timescale is indeed much shorter than typical beta decay processes (which occur on microsecond to millisecond timescales), which is precisely why the r-process can build up neutron-rich nuclei before they decay. For r-process problems, always identify the mass difference first, then apply the given rate directly. The extreme speeds involved are what make this process unique in stellar environments.

Question 14

In the CNO cycle, the net reaction converts four hydrogen nuclei into one helium nucleus. If a 5M⊙5 M_{\odot} star derives 80% of its energy from the CNO cycle and 20% from the pp-chain during its main sequence phase, and the star's luminosity is 100L⊙100 L_{\odot}, approximately how many carbon-12 nuclei participate in CNO reactions each second throughout the entire star?

  1. Approximately 1.1×10371.1 \times 10^{37} carbon-12 nuclei participate in CNO reactions per second
  2. Approximately 2.8×10382.8 \times 10^{38} carbon-12 nuclei participate in CNO reactions per second
  3. Approximately 5.6×10395.6 \times 10^{39} carbon-12 nuclei participate in CNO reactions per second (correct answer)
  4. Approximately 1.4×10411.4 \times 10^{41} carbon-12 nuclei participate in CNO reactions per second
Explanation: CNO energy output = 0.8×100L⊙=0.8×100×3.8×1026=3.04×10280.8 \times 100 L_{\odot} = 0.8 \times 100 \times 3.8 \times 10^{26} = 3.04 \times 10^{28} W. Each He-4 formation releases 26.7 MeV = 4.27×10−124.27 \times 10^{-12} J. Rate of He-4 formation = 3.04×1028/(4.27×10−12)=7.1×10393.04 \times 10^{28}/(4.27 \times 10^{-12}) = 7.1 \times 10^{39} per second. In the CNO cycle, each C-12 nucleus is regenerated, so the number of C-12 nuclei participating equals the He-4 formation rate, approximately 5.6×10395.6 \times 10^{39} per second. A uses only pp-chain energy. B underestimates by a factor of ~20. D overcounts by including intermediate CNO isotopes.

Question 15

During silicon burning in a massive star core, the sequence 28Si+α→32S+γ^{28}Si + \alpha \rightarrow ^{32}S + \gamma followed by 32S+α→36Ar+γ^{32}S + \alpha \rightarrow ^{36}Ar + \gamma occurs rapidly. If the core temperature is 3×1093 \times 10^9 K and the silicon-28 abundance decreases from 30% to 5% of the core mass in 24 hours, what does this timescale suggest about the star's immediate evolutionary future?

  1. The star will continue silicon burning for approximately 1000 years before beginning iron peak nucleosynthesis in a controlled manner
  2. The star will exhaust all nuclear fuel within days and undergo core collapse when iron peak elements dominate the core composition (correct answer)
  3. The star will transition to a stable helium-burning phase as silicon depletion triggers a temporary halt in energy production
  4. The star will experience thermal pulses as silicon burning becomes thermally unstable due to the rapid consumption rate
Explanation: The extremely rapid silicon consumption (25% of core mass in 24 hours) indicates the star is in the final stages of nuclear burning. Silicon burning produces iron peak elements (Fe, Ni, Co) which cannot undergo further exothermic fusion. Once the core becomes dominated by these elements (within days at this consumption rate), no further energy generation is possible, leading to core collapse and supernova. A vastly overestimates the remaining lifetime. C incorrectly suggests a return to helium burning. D confuses this with lower-mass star behavior.

Question 16

Two stars of equal mass but different metallicity undergo core hydrogen fusion. Star A has metallicity Z = 0.02 and Star B has metallicity Z = 0.001. Which statement best describes the expected difference in their evolutionary timescales and fusion processes?

  1. Star A will have a longer main sequence lifetime because higher metallicity increases the efficiency of the CNO cycle relative to the pp-chain
  2. Star B will have a longer main sequence lifetime because lower metallicity reduces opacity, allowing more efficient energy transport and lower core temperatures
  3. Star A will have a shorter main sequence lifetime because higher metallicity increases opacity, requiring higher core temperatures and faster fusion rates (correct answer)
  4. Both stars will have identical main sequence lifetimes because metallicity only affects post-main sequence evolution through stellar wind mass loss
Explanation: Higher metallicity increases stellar opacity, which impedes radiative energy transport. This forces the core to reach higher temperatures to maintain hydrostatic equilibrium, leading to faster fusion rates and shorter main sequence lifetimes. A is incorrect because higher CNO efficiency would actually shorten, not lengthen, the lifetime. B incorrectly states that lower opacity leads to lower core temperatures. D is wrong because metallicity significantly affects main sequence evolution, not just post-MS phases.

Question 17

A main-sequence star is in hydrostatic equilibrium. If the rate of fusion in its core were to temporarily increase due to a random fluctuation, what would be the subsequent response of the core?

  1. The core temperature and pressure would increase, causing the core to expand and cool, which in turn reduces the fusion rate. (correct answer)
  2. The increased energy output would be immediately radiated away from the surface, causing the core to cool and contract, increasing the fusion rate further.
  3. The gravitational force would increase to counteract the higher radiation pressure, compressing the core and leading to a stable, higher fusion rate.
  4. The core would expand rapidly, but the fusion rate would remain constant, establishing a new equilibrium at a larger stellar radius.
Explanation: A main-sequence star is regulated by a negative feedback loop. An increase in the fusion rate leads to a higher temperature and greater outward radiation pressure. This pressure causes the core to expand. According to the ideal gas law, this expansion causes the core to cool and become less dense. The lower temperature and density then lead to a decrease in the fusion rate, restoring the star to equilibrium. This process is how stars remain stable for billions of years.

Question 18

A star is located in the lower-left region of the Hertzsprung-Russell diagram. What properties are characteristic of this star?

  1. High luminosity and high surface temperature.
  2. Low luminosity and high surface temperature. (correct answer)
  3. High luminosity and low surface temperature.
  4. Low luminosity and low surface temperature.
Explanation: The Hertzsprung-Russell (HR) diagram plots luminosity (or absolute magnitude) on the y-axis (increasing upwards) versus surface temperature (or spectral class) on the x-axis (increasing to the left). The lower-left region, therefore, corresponds to stars with low luminosity (bottom part of the diagram) and high surface temperature (left part of the diagram). These stars are known as white dwarfs.

Question 19

Which statement correctly describes the balance of forces that maintains the stability of a main-sequence star, known as hydrostatic equilibrium?

  1. The inward force of electrostatic attraction is balanced by the outward force of the strong nuclear interaction.
  2. The inward force of gravity is balanced by the outward centrifugal force from the star's rapid rotation.
  3. The inward force of gravity is balanced by the outward pressure created by hot gas and radiation from nuclear fusion. (correct answer)
  4. The inward pressure from interstellar gas is balanced by the outward pressure from the star's magnetic field.
Explanation: Hydrostatic equilibrium is the state of balance in a star where the inward pull of gravity is exactly counteracted by the outward push of pressure. This outward pressure has two main components: gas pressure from the thermal motion of particles in the hot plasma, and radiation pressure from the photons generated by nuclear fusion in the core. The other options describe forces that are either incorrect, not primary, or act in the wrong direction for this equilibrium.

Question 20

Two stars, Betelgeuse (a red supergiant) and Rigel (a blue supergiant), have approximately the same luminosity. What can be concluded about their relative surface temperatures and radii?

  1. Betelgeuse has the same temperature as Rigel, but a much larger radius.
  2. Betelgeuse has a higher temperature and a much smaller radius than Rigel.
  3. Betelgeuse has a lower temperature, but their radii are approximately equal.
  4. Betelgeuse has a lower temperature and a much larger radius than Rigel. (correct answer)
Explanation: The color of a star indicates its surface temperature. Betelgeuse is red, indicating a low surface temperature (around 3,500 K). Rigel is blue, indicating a very high surface temperature (over 12,000 K). Luminosity is related to radius and temperature by L∝R2T4L \propto R^2 T^4. For their luminosities (L) to be approximately equal, if Betelgeuse has a much lower temperature (T), its radius (R) must be significantly larger to compensate for the T4T^4 term. Therefore, Betelgeuse is much larger than Rigel.