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

When a star like the Sun evolves into a red giant, how do its core temperature and surface temperature change compared to its main-sequence phase?

The core becomes cooler, and the surface becomes hotter.
The core becomes hotter, and the surface becomes cooler.
Both the core and the surface become cooler.
Both the core and the surface become hotter.
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IB Physics Quiz

IB Physics Quiz: Understand Fusion And Stars

Practice Understand 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.

What this quiz covers

This quiz focuses on Understand 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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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

When a star like the Sun evolves into a red giant, how do its core temperature and surface temperature change compared to its main-sequence phase?

  1. The core becomes cooler, and the surface becomes hotter.
  2. The core becomes hotter, and the surface becomes cooler. (correct answer)
  3. Both the core and the surface become cooler.
  4. Both the core and the surface become hotter.
Explanation: After exhausting hydrogen in its core, the core, now composed of helium, contracts under gravity. This contraction increases the core's temperature and density. The heating of the core ignites hydrogen fusion in a shell surrounding it. This shell fusion is very rapid and generates more energy than the core fusion did, causing the star's outer layers to expand significantly. This expansion causes the surface to cool, giving the star a reddish appearance. Thus, the core becomes hotter while the surface becomes cooler.

Question 2

An astronomer measures the stellar parallax of a star to be 0.050 arcseconds. What is the distance to this star?

  1. 0.050 pc
  2. 5.0 pc
  3. 20 pc (correct answer)
  4. 50 pc
Explanation: The formula for calculating distance in parsecs (pc) from the parallax angle in arcseconds (arcsec) is d(pc) = 1 / p(arcsec). Given p = 0.050 arcsec, the distance is d = 1 / 0.050 = 20 pc. A is the value of the parallax angle itself. B and D are common calculation errors.

Question 3

Why is the method of stellar parallax only effective for determining the distances to relatively nearby stars?

  1. The parallax angle becomes too large to be measured accurately for distant stars.
  2. The Earth's atmospheric distortion makes the small angular shift of distant stars impossible to resolve.
  3. The parallax angle for distant stars is too small to be measured with sufficient precision against the background. (correct answer)
  4. Distant stars are often part of galaxies whose own motion masks the parallax effect.
Explanation: Stellar parallax is the apparent shift in a star's position as observed from two different points in Earth's orbit. The parallax angle (p) is inversely proportional to the distance (d) to the star (p = 1/d). For very distant stars, d is very large, making p extremely small. Eventually, the angle becomes smaller than the resolving power of our best telescopes, making it indistinguishable from zero or measurement noise. C correctly identifies this fundamental limitation. A is the opposite of what occurs. B mentions atmospheric distortion, which is a challenge for all ground-based measurements but not the fundamental distance limitation. D is incorrect; galactic motion is a different phenomenon.

Question 4

Star X and Star Y have the same luminosity. The parallax angle of Star X is four times larger than the parallax angle of Star Y. What is the ratio of the apparent brightness of Star X to that of Star Y (b_X / b_Y)?

  1. 1/16
  2. 1/4
  3. 4
  4. 16 (correct answer)
Explanation: First, relate parallax and distance: d = 1/p. If p_X = 4 * p_Y, then d_Y = 4 * d_X. Second, use the inverse square law for brightness: b = L / (4πd²). Since luminosity L is the same for both stars, the ratio of their brightnesses is b_X / b_Y = (L / d_X²) / (L / d_Y²) = d_Y² / d_X². Substitute the distance relationship: b_X / b_Y = (4d_X)² / d_X² = 16d_X² / d_X² = 16. A is the inverse ratio. B and C are common errors from forgetting to square the distance ratio.

Question 5

Two main-sequence stars, X and Y, have the same apparent brightness when viewed from Earth. Star Y is known to be twice as far away as Star X. The surface temperature of Star Y is half the surface temperature of Star X. What is the ratio of the radius of Star Y to the radius of Star X (R_Y / R_X)?

  1. 2
  2. 4
  3. 8 (correct answer)
  4. 16
Explanation: This requires a multi-step solution. 1. Find the luminosity ratio: Since apparent brightness b = L / (4πd²) and b_X = b_Y, we have L_X / d_X² = L_Y / d_Y². Given d_Y = 2d_X, this means L_Y / L_X = (d_Y / d_X)² = 2² = 4. So, Star Y is 4 times more luminous than Star X. 2. Use the Stefan-Boltzmann law: L ∝ R²T⁴. We can write the ratio as L_Y / L_X = (R_Y / R_X)² * (T_Y / T_X)⁴. 3. Substitute the known values: We found L_Y / L_X = 4, and we are given T_Y / T_X = 0.5. So, 4 = (R_Y / R_X)² * (0.5)⁴. 4. Solve for the radius ratio: 4 = (R_Y / R_X)² * (1/16). This gives (R_Y / R_X)² = 4 * 16 = 64. Taking the square root, R_Y / R_X = 8.

Question 6

Star Altair has an apparent brightness b and a parallax of 0.20 arcseconds. Star Vega has the same apparent brightness b but a parallax of 0.10 arcseconds. What is the ratio of the luminosity of Vega to the luminosity of Altair (L_Vega / L_Altair)?

  1. 0.25
  2. 0.50
  3. 2.0
  4. 4.0 (correct answer)
Explanation: First, find the relationship between the distances. Distance d is inversely proportional to parallax p (d = 1/p). Since p_Altair = 2 * p_Vega, it means d_Vega = 2 * d_Altair. Second, use the relationship between apparent brightness b, luminosity L, and distance d: b = L / (4πd²). Since the apparent brightnesses are equal (b_Altair = b_Vega), we have L_Altair / d_Altair² = L_Vega / d_Vega². Rearranging for the ratio gives L_Vega / L_Altair = (d_Vega / d_Altair)². Substituting the distance relationship gives L_Vega / L_Altair = (2 * d_Altair / d_Altair)² = 2² = 4.0. C is the ratio of the distances, not luminosities. B is the ratio of the parallaxes. A is the inverse squared ratio.

Question 7

A star's spectrum is analyzed, and its peak emission wavelength is found to be in the infrared region of the electromagnetic spectrum. Where would this star most likely be located on a Hertzsprung-Russell diagram?

  1. In the upper-left, among the hot, luminous O-type stars.
  2. In the lower-right, among the cool, dim M-type stars. (correct answer)
  3. In the lower-left, among the hot but dim white dwarfs.
  4. On the main sequence, in the same region as an A-type star.
Explanation: According to Wien's displacement law (λ_max * T = constant), a longer peak wavelength (λ_max) corresponds to a lower surface temperature (T). Infrared radiation has a longer wavelength than visible light. Therefore, the star must be cool. On the H-R diagram, cool stars are located on the right side. The lower-right is the location of cool, dim stars (red dwarfs, M-type main sequence stars). The upper-right is for cool but luminous red giants. Since M-type stars are the most common, lower-right is the most likely location. A, C, and D all describe hotter stars with peak emissions at shorter wavelengths (visible or UV).

Question 8

Immediately after a Sun-like star exhausts the hydrogen in its core, it begins to leave the main sequence. What is the primary source of the increased luminosity that causes it to expand into a subgiant and then a red giant?

  1. The sudden ignition of helium fusion into carbon within the core.
  2. Gravitational potential energy released by the rapid contraction of the inert helium core.
  3. The fusion of hydrogen into helium in a shell surrounding the contracting, inert helium core. (correct answer)
  4. The radioactive decay of heavy elements created during its main-sequence lifetime.
Explanation: As the hydrogen-depleted core contracts and heats up, the layer of hydrogen just outside the core reaches the temperature and pressure required for fusion. This is known as hydrogen shell burning. This shell fusion is actually more energetic and produces a higher luminosity than the core fusion did during the main sequence phase, pushing the outer layers of the star outwards and causing it to expand. A (the 'helium flash') occurs later in the evolution of a low-mass star. B contributes to the heating of the shell but is not the primary energy output. D is not a significant energy source.

Question 9

A red giant star and a main-sequence star can have the same surface temperature. However, the red giant has a much greater luminosity. According to the Stefan-Boltzmann law, what is the primary reason for this difference in luminosity?

  1. The red giant is composed of heavier elements that undergo more energetic fusion reactions per unit area.
  2. The red giant has a significantly larger radius, resulting in a much greater energy-emitting surface area. (correct answer)
  3. The main-sequence star has a greater density, which suppresses the rate of energy emission from its surface.
  4. The main-sequence star must be much further away, so its apparent brightness is lower, reducing its luminosity.
Explanation: The Stefan-Boltzmann law states that a star's luminosity is proportional to its surface area and the fourth power of its surface temperature (L = σAT⁴). Since A = 4πR², this means L ∝ R²T⁴. If the temperatures (T) of the two stars are the same, the only way the red giant can be vastly more luminous (larger L) is if its radius (R) is much larger, giving it a much greater surface area (A). B correctly identifies this. A is incorrect; while fusion processes differ, the surface area is the direct cause in the Stefan-Boltzmann law. C is incorrect. D confuses apparent brightness with luminosity, which is an intrinsic property.

Question 10

A massive star with initial mass 25M⊙25 M_{\odot} undergoes core collapse when its iron core reaches 1.4M⊙1.4 M_{\odot}. If the core radius decreases from 3000 km to 12 km during collapse, what is the approximate ratio of the final gravitational binding energy to the initial binding energy of the core?

  1. The final binding energy is approximately 250 times larger than the initial binding energy (correct answer)
  2. The final binding energy is approximately 750 times larger than the initial binding energy
  3. The final binding energy is approximately 1250 times larger than the initial binding energy
  4. The final binding energy is approximately 2500 times larger than the initial binding energy
Explanation: Gravitational binding energy scales as U ∝ M²/R. Since the mass remains constant at 1.4 solar masses, the ratio depends only on radius: U_final/U_initial = R_initial/R_final = 3000 km/12 km = 250. This enormous increase in binding energy (from  1046~10^46 J to  2.5×1048~2.5×10^48 J) represents the energy available for the supernova explosion when the core rebounds.

Question 11

In a 0.8M⊙0.8 M_{\odot} red giant star, the core temperature is 8×1078 \times 10^7 K, just below the helium flash threshold. The core is supported by electron degeneracy pressure. Which statement best describes why helium fusion has not yet begun?

  1. The Coulomb barrier for helium-4 nuclei is too high, requiring temperatures above 1.2×1081.2 \times 10^8 K for significant tunneling (correct answer)
  2. The triple-alpha process requires three simultaneous collisions, making it improbable at current particle densities
  3. The core density is insufficient to overcome the quantum mechanical exclusion principle for helium nuclei
  4. The stellar luminosity is too low to maintain the thermal energy needed for sustained helium burning
Explanation: The primary barrier to helium fusion is the Coulomb barrier. Helium-4 nuclei have charge +2e, making their Coulomb barrier much higher than for hydrogen fusion. The triple-alpha process requires temperatures ~1×10^8 K to 1.5×10^8 K for significant reaction rates. At 8×10^7 K, quantum tunneling through the Coulomb barrier is still too improbable. While the triple-alpha process does require three-body interactions (B), the temperature threshold is the limiting factor, not collision frequency.

Question 12

During the helium flash phase of a low-mass star's evolution, the core temperature rises to approximately 10810^8 K. Which statement best explains why the helium fusion rate increases so dramatically during this event?

  1. The triple-alpha process has a much stronger temperature dependence than hydrogen fusion, approximately T40T^{40}
  2. The stellar core contracts rapidly, increasing both density and collision frequency simultaneously
  3. The Coulomb barrier for helium nuclei decreases significantly at these elevated temperatures
  4. Electron degeneracy pressure prevents normal thermal expansion, creating runaway heating conditions (correct answer)
Explanation: The helium flash occurs because the core is electron-degenerate, meaning pressure doesn't depend on temperature. When helium fusion begins, the energy released heats the core but cannot cause expansion (since pressure is independent of temperature in degenerate matter). This creates a runaway process where more heating leads to more fusion without the normal self-regulating expansion. While the triple-alpha process does have strong temperature dependence ( T40~T^40), option D identifies the fundamental physical cause of the runaway behavior.

Question 13

A Type Ia supernova occurs when a white dwarf accretes matter from a companion star. If the white dwarf has an initial mass of 1.2M⊙1.2 M_{\odot} and accretes 0.15M⊙0.15 M_{\odot} of hydrogen-rich material, which factor most directly determines whether explosive carbon fusion will occur?

  1. The rate of hydrogen burning on the surface determines the pressure conditions in the core
  2. The total mass approaches the Chandrasekhar limit, causing core compression and carbon ignition temperature (correct answer)
  3. The gravitational binding energy of the accreted material provides the activation energy for carbon fusion
  4. The helium flash from accumulated helium ash triggers a compression wave that ignites carbon
Explanation: Type Ia supernovae occur when the white dwarf mass approaches the Chandrasekhar limit (~1.4 solar masses). At 1.2 + 0.15 = 1.35 solar masses, the increased gravitational pressure compresses the carbon-oxygen core until temperatures reach ~5×10^8 K, sufficient for carbon fusion. The surface hydrogen burning (A) and gravitational binding energy (C) are insufficient, and helium flashes (D) don't directly trigger carbon ignition in this scenario.

Question 14

Which statement best explains why nuclear fusion in a star's core requires extremely high temperatures and densities?

  1. High temperatures are required to provide the high pressures needed to contain the plasma, and high densities ensure nuclei are close enough to fuse.
  2. High temperatures provide nuclei with sufficient kinetic energy to overcome electrostatic repulsion, and high densities increase the probability of collisions. (correct answer)
  3. High temperatures create the energetic photons that initiate the fusion reactions, and high densities are required for the strong nuclear force to be effective.
  4. High temperatures are needed to ionize hydrogen into a plasma state, and high densities provide the gravitational force to bind the resulting nuclei.
Explanation: Nuclear fusion requires two positively charged nuclei to get close enough for the strong nuclear force to bind them. They must overcome the strong electrostatic (Coulomb) repulsion. High temperatures correspond to high average kinetic energy of the nuclei, allowing them to approach each other despite this repulsion. High density increases the frequency of collisions between nuclei, making fusion reactions more likely to occur at a sustainable rate. B correctly identifies both of these critical roles. A confuses the roles of temperature and pressure. C is incorrect; photons are a product, not an initiator. D describes ionization, which is necessary but not the primary reason for the extreme conditions.

Question 15

Star X has a mass of 10 solar masses. Star Y has a mass of 0.5 solar masses. Both are on the main sequence. Which statement and reason correctly compares their main-sequence lifetimes?

  1. Star X will have a much longer lifetime because it has significantly more hydrogen fuel to burn.
  2. Star Y will have a much longer lifetime because its lower mass results in a much lower rate of fusion. (correct answer)
  3. Both stars will have similar lifetimes because the rate of fusion scales linearly with the amount of available fuel.
  4. It is impossible to compare their lifetimes without knowing their surface temperatures and radii.
Explanation: A star's main-sequence lifetime is determined by the ratio of its fuel (mass) to its rate of fuel consumption (luminosity). While more massive stars have more fuel, their luminosity increases much more rapidly with mass (approximately L ∝ M³.⁵). Therefore, the high-mass Star X burns through its fuel incredibly quickly, leading to a short lifetime. The low-mass Star Y has less fuel but consumes it at a very slow rate, resulting in a much longer lifetime. B provides the correct comparison and reason. A states a common misconception. C is incorrect as the relationship is not linear. D is incorrect; mass is the primary determinant of a main-sequence star's properties and lifetime.

Question 16

The main fusion process in a star like the Sun is the proton-proton chain. What is the net result of one full cycle of this chain?

  1. Four protons fuse to form one helium-4 nucleus, two positrons, two neutrinos, and release energy. (correct answer)
  2. Two protons and two neutrons fuse to form one helium-4 nucleus, releasing energy.
  3. Six protons interact to produce one helium-4 nucleus, two remaining protons, and energy.
  4. Four protons fuse to form two deuterium nuclei, two positrons, and two neutrinos, releasing energy.
Explanation: The net equation for the proton-proton chain shows that four hydrogen nuclei (protons) are consumed to produce one helium-4 nucleus. To conserve charge and lepton number, two positrons (e⁺) and two electron neutrinos (νₑ) are also emitted. The mass of the products is less than the mass of the reactants, and this mass difference is converted into energy (gamma rays and kinetic energy of the products). A correctly lists all net inputs and outputs. B is incorrect as neutrons are not initial reactants. C is confusing; while six protons may be involved in the intermediate steps of some branches, the net consumption is four protons. D describes only the first step of the chain, not the net result.

Question 17

Star A has a surface temperature T and radius R. Star B has a surface temperature of 2T and a radius of 0.5R. What is the ratio of the luminosity of Star B to the luminosity of Star A (L_B / L_A)?

  1. 2
  2. 4 (correct answer)
  3. 8
  4. 16
Explanation: Luminosity L is given by the Stefan-Boltzmann law, L = σ(4πR²)T⁴, so L is proportional to R²T⁴. To find the ratio L_B / L_A, we can write L_B / L_A = (R_B² T_B⁴) / (R_A² T_A⁴). Substitute the given values: R_B = 0.5R_A and T_B = 2T_A. The ratio becomes L_B / L_A = ((0.5R_A)² (2T_A)⁴) / (R_A² T_A⁴) = (0.25R_A² * 16T_A⁴) / (R_A² T_A⁴) = 0.25 * 16 = 4. A is a possible miscalculation. C is another common error (e.g., 0.5*16). D would be correct if the radius was the same.

Question 18

What is the defining characteristic of all stars that lie on the main sequence of the Hertzsprung-Russell diagram?

  1. The star is in hydrostatic equilibrium, with gravitational forces balanced by thermal and radiation pressure.
  2. The star generates energy primarily through the fusion of hydrogen into helium in its core. (correct answer)
  3. The star has exhausted the hydrogen in its core and is now fusing helium into carbon.
  4. The star's luminosity is directly proportional to the square of its surface temperature.
Explanation: The main sequence is specifically defined as the stage in a star's life where it is stably fusing hydrogen into helium in its core. This is the longest phase of a star's life. A is a condition for any stable star (including giants), not just main-sequence stars, so it is not the defining characteristic. C describes a star that has evolved off the main sequence into a horizontal branch or red giant phase. D is incorrect; the relationship between luminosity and temperature for main-sequence stars is more complex because radius also changes with mass along the sequence.

Question 19

A white dwarf is the stable stellar remnant of a low-mass star. With no ongoing nuclear fusion, what force prevents it from collapsing further under its own immense gravity?

  1. The outward thermal pressure from its extremely high residual temperature.
  2. The electrostatic repulsion between the closely packed atomic nuclei.
  3. A quantum mechanical pressure from degenerate electrons that resists further compression. (correct answer)
  4. The outward pressure from the slow fusion of carbon into heavier elements.
Explanation: A white dwarf is supported against gravitational collapse by electron degeneracy pressure. This is a quantum mechanical effect, arising from the Pauli exclusion principle, which states that no two electrons can occupy the same quantum state. As the star collapses, electrons are forced into higher energy levels, creating a pressure that is independent of temperature and halts the collapse. A is incorrect; thermal pressure is insufficient. B is incorrect; electrostatic repulsion is also insufficient. D is incorrect as no fusion occurs in a typical white dwarf.

Question 20

Considering the entire life cycle of a star, in which stage does it spend the vast majority (around 90%) of its lifetime?

  1. In the red giant branch, while it is fusing helium in its core.
  2. During the initial protostar phase, as gravitational collapse is an extremely gradual process.
  3. In the white dwarf phase, as the process of cooling takes many billions of years.
  4. On the main sequence, while it is fusing hydrogen in its core. (correct answer)
Explanation: A star's 'lifetime' is typically defined by its active, fusion-powered phases. The main sequence, where a star fuses hydrogen into helium in its core, is by far the longest and most stable phase, accounting for about 90% of its life. This is because hydrogen is the most abundant fuel and core hydrogen burning is a relatively slow, self-regulating process. The later stages, like the red giant phase (A), are much more rapid. While the white dwarf cooling phase (C) is very long, it is a remnant phase after fusion has ceased. The protostar phase (D) is also much shorter than the main-sequence lifetime.