College Physics Quiz: Fission Fusion And Nuclear Decay
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Fission Fusion And Nuclear DecayQuestion 1 of 20

A uranium-235 nucleus absorbs a slow neutron and splits into barium-141 and krypton-92, releasing additional neutrons. If the mass of the original U-235 nucleus plus neutron is 236.053 u and the combined mass of the fission products is 235.867 u, what is the energy released in this fission reaction?

173 MeV
186 MeV
201 MeV
158 MeV
194 MeV
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College Physics Quiz

College Physics Quiz: Fission Fusion And Nuclear Decay

Practice Fission Fusion And Nuclear Decay in College 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 Fission Fusion And Nuclear Decay, giving you a quick way to practice the rules, question types, and explanations that matter most for College Physics.

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Question 1

A uranium-235 nucleus absorbs a slow neutron and splits into barium-141 and krypton-92, releasing additional neutrons. If the mass of the original U-235 nucleus plus neutron is 236.053 u and the combined mass of the fission products is 235.867 u, what is the energy released in this fission reaction?

  1. 173 MeV (correct answer)
  2. 186 MeV
  3. 201 MeV
  4. 158 MeV
  5. 194 MeV
Explanation: When you encounter nuclear fission problems, you're applying Einstein's mass-energy equivalence principle. The key insight is that when nuclear reactions occur, small amounts of mass are converted into large amounts of energy according to E=mc2E = mc^2. To find the energy released, you need to calculate the mass defect—the difference between the initial mass and final mass of all particles involved. Here, the initial mass (U-235 nucleus + neutron) is 236.053 u, and the final mass (all fission products including Ba-141, Kr-92, and additional neutrons) is 235.867 u. The mass defect is: Δm=236.053−235.867=0.186 u\Delta m = 236.053 - 235.867 = 0.186 \text{ u} Converting this to energy using the standard conversion factor (1 u = 931.5 MeV/c²): E=0.186×931.5=173.3 MeVE = 0.186 \times 931.5 = 173.3 \text{ MeV} This confirms answer A (173 MeV) is correct. Answer B (186 MeV) represents a common error where students multiply the mass defect by 1000 instead of 931.5, confusing unit conversions. Answer C (201 MeV) likely comes from incorrectly calculating the mass defect or using wrong conversion factors. Answer D (158 MeV) might result from arithmetic errors in the mass difference calculation. Remember: for nuclear physics problems, always identify what's conserved (mass-energy) versus what's not (individual masses). The mass defect calculation is your starting point, and the conversion factor 931.5 MeV per atomic mass unit is essential to memorize for these problems.

Question 2

In the fusion reaction 2H+3H→4He+n^2H + ^3H → ^4He + n, deuterium and tritium combine to form helium-4 and a neutron. If the binding energy per nucleon is 1.11 MeV for deuterium, 2.83 MeV for tritium, and 7.07 MeV for helium-4, what is the Q-value (energy released) for this reaction?

  1. 17.6 MeV (correct answer)
  2. 14.3 MeV
  3. 21.2 MeV
  4. 12.5 MeV
  5. 19.1 MeV
Explanation: Nuclear fusion reactions release energy when light nuclei combine to form heavier, more stable nuclei. The key to solving these problems is understanding that the Q-value equals the difference in total binding energy between products and reactants. To find the Q-value, calculate the total binding energy for each nucleus by multiplying the binding energy per nucleon by the mass number. For the reactants: deuterium has 1.11 MeV/nucleon×2=2.22 MeV1.11 \text{ MeV/nucleon} \times 2 = 2.22 \text{ MeV}, and tritium has 2.83 MeV/nucleon×3=8.49 MeV2.83 \text{ MeV/nucleon} \times 3 = 8.49 \text{ MeV}, giving a total of 10.71 MeV10.71 \text{ MeV}. For the products: helium-4 has 7.07 MeV/nucleon×4=28.28 MeV7.07 \text{ MeV/nucleon} \times 4 = 28.28 \text{ MeV}, while the neutron contributes zero binding energy. The Q-value is the difference: 28.28−10.71=17.57 MeV28.28 - 10.71 = 17.57 \text{ MeV}, which rounds to 17.6 MeV17.6 \text{ MeV}. Choice B (14.3 MeV) likely results from calculation errors or using incorrect binding energy values. Choice C (21.2 MeV) suggests adding binding energies incorrectly or confusing the direction of the energy calculation. Choice D (12.5 MeV) might come from subtracting in the wrong direction or making arithmetic mistakes with the binding energy totals. Remember that in fusion problems, you're looking for the energy "gained" when nucleons become more tightly bound. Always multiply binding energy per nucleon by mass number first, then subtract reactants from products to find the released energy.

Question 3

Carbon-14 has a half-life of 5,730 years and undergoes beta-minus decay to nitrogen-14. A wooden artifact contains 25% of the original C-14 activity. Assuming the wood stopped absorbing C-14 when the tree died, approximately how old is the artifact?

  1. 11,460 years (correct answer)
  2. 8,595 years
  3. 17,190 years
  4. 14,325 years
  5. 5,730 years
Explanation: When you encounter radioactive decay problems, you're working with exponential decay where the amount of substance decreases by half every half-life period. The key relationship is N=N0(12)t/t1/2N = N_0 \left(\frac{1}{2}\right)^{t/t_{1/2}}, where N is the current amount, N₀ is the original amount, t is time elapsed, and t₁/₂ is the half-life. Since the artifact contains 25% of its original C-14 activity, we have NN0=0.25=14\frac{N}{N_0} = 0.25 = \frac{1}{4}. Notice that 14=(12)2\frac{1}{4} = \left(\frac{1}{2}\right)^2, meaning the sample has undergone exactly 2 half-lives of decay. With C-14's half-life of 5,730 years, the artifact's age is 2×5,730=11,4602 \times 5,730 = 11,460 years. Looking at the wrong answers: B) 8,595 years represents 1.5 half-lives, which would leave about 35% of the original activity, not 25%. C) 17,190 years equals 3 half-lives, leaving only 12.5% of the original activity. D) 14,325 years represents 2.5 half-lives, leaving approximately 18% of the original activity. A) 11,460 years correctly corresponds to 2 complete half-lives, reducing the activity to exactly 25%. Study tip: Memorize that each half-life reduces the substance by half: 100% → 50% → 25% → 12.5% → 6.25%. When you see common fractions like 1/4, 1/8, or 3/4, immediately think about how many half-lives they represent rather than using the exponential formula every time.

Question 4

Which statement correctly explains why fusion reactions require extremely high temperatures to occur spontaneously?

  1. High temperatures provide kinetic energy to overcome the Coulomb repulsion between positively charged nuclei (correct answer)
  2. High temperatures are needed to break the strong nuclear bonds within each nucleus before fusion
  3. High temperatures increase the binding energy per nucleon, making fusion energetically favorable
  4. High temperatures reduce the mass of the nuclei, decreasing the energy barrier for fusion
  5. High temperatures create the neutrons necessary for most fusion reactions to proceed
Explanation: Nuclear fusion questions test your understanding of the energy barriers that prevent atomic nuclei from combining. When you encounter fusion problems, focus on the fundamental forces at play and what conditions allow nuclei to overcome natural repulsion. Choice A correctly identifies the key challenge in fusion. Since all atomic nuclei carry positive charge, they experience strong electrostatic repulsion (Coulomb force) when brought close together. This repulsion follows an inverse square law - as nuclei approach each other, the repulsive force increases dramatically. At extremely high temperatures (millions of Kelvin), nuclei gain enough kinetic energy to overcome this "Coulomb barrier" and get close enough for the attractive strong nuclear force to take over, enabling fusion. Choice B misunderstands the fusion process. Fusion doesn't require breaking bonds within individual nuclei - instead, it involves combining separate nuclei. The strong nuclear force holds nucleons together within each nucleus and doesn't need to be overcome for fusion to occur. Choice C incorrectly suggests that temperature directly affects binding energy per nucleon. While fusion releases energy because the products have higher binding energy per nucleon than reactants, temperature doesn't change these fundamental nuclear properties - it only provides kinetic energy for the reaction. Choice D contains a physics misconception. Temperature doesn't reduce nuclear mass or change the energy barrier height. The barrier remains constant; high temperature simply gives nuclei enough kinetic energy to surmount it. Remember: fusion problems almost always come down to overcoming electrostatic repulsion between positive charges through high kinetic energy from extreme temperatures.

Question 5

In a nuclear reactor, the average number of neutrons from each fission that cause additional fissions is called the multiplication factor k. Which statement about reactor control is correct?

  1. k > 1 results in an increasing reaction rate, k = 1 maintains constant power, k < 1 causes the reaction to decrease (correct answer)
  2. k > 1 maintains constant power output, while k < 1 increases the reaction rate exponentially
  3. k must always equal exactly 2.4 to sustain a chain reaction since this is the average neutrons per fission
  4. k < 1 indicates supercritical operation where the reactor power increases without bound
  5. k = 1 represents subcritical operation requiring external neutron sources to maintain the reaction
Explanation: When you encounter nuclear reactor physics problems, focus on the multiplication factor k as the key parameter that determines reactor behavior. Think of k as representing how many "new" fissions result from each current fission - it's the ratio that controls whether the chain reaction grows, shrinks, or stays steady. The correct answer is A because it accurately describes the three reactor states. When k > 1 (supercritical), each fission produces more than one subsequent fission, so the reaction rate increases exponentially. When k = 1 (critical), each fission produces exactly one subsequent fission, maintaining steady power output. When k < 1 (subcritical), each fission produces less than one subsequent fission, causing the reaction rate to decrease and eventually stop. Option B reverses the relationships completely - k > 1 definitely doesn't maintain constant power, and k < 1 decreases rather than increases reaction rates. Option C confuses k with the total neutrons produced per fission. While fission does produce about 2.4 neutrons on average, most are absorbed by non-fuel materials or leak out. The multiplication factor k specifically counts only those neutrons that cause additional fissions. Option D incorrectly states that k < 1 indicates supercritical operation, when it actually indicates subcritical operation where power decreases. Remember this simple pattern: k greater than 1 means growing reaction, k equal to 1 means steady reaction, k less than 1 means shrinking reaction. The key insight is that k measures the neutrons that successfully cause new fissions, not the total neutrons produced.

Question 6

A Geiger counter detects 1600 counts per minute from a radioactive sample. If the sample's half-life is 30 minutes, what will be the count rate after 90 minutes?

  1. 200 counts per minute (correct answer)
  2. 400 counts per minute
  3. 100 counts per minute
  4. 800 counts per minute
  5. 533 counts per minute
Explanation: Radioactive decay problems test your understanding of exponential decay and half-life calculations. When you see a half-life question, remember that the activity decreases by exactly half during each half-life period, regardless of the starting amount. To solve this, first determine how many half-lives occur in 90 minutes. Since the half-life is 30 minutes, you have 90 minutes30 minutes=3\frac{90 \text{ minutes}}{30 \text{ minutes}} = 3 half-lives. Starting with 1600 counts per minute, after each half-life the count rate becomes:
  • After 1 half-life (30 min): 1600×12=8001600 \times \frac{1}{2} = 800 counts/min
  • After 2 half-lives (60 min): 800×12=400800 \times \frac{1}{2} = 400 counts/min
  • After 3 half-lives (90 min): 400×12=200400 \times \frac{1}{2} = 200 counts/min
Alternatively, use the formula: N(t)=N0×(12)nN(t) = N_0 \times \left(\frac{1}{2}\right)^{n} where nn is the number of half-lives. This gives 1600×(12)3=1600×18=2001600 \times \left(\frac{1}{2}\right)^3 = 1600 \times \frac{1}{8} = 200 counts/min. Choice A (200 counts/min) is correct. Choice D (800 counts/min) represents the count rate after only one half-life, not three. Choice B (400 counts/min) is the rate after two half-lives. Choice C (100 counts/min) would require four half-lives (120 minutes), not three. Remember: each half-life cuts the activity in half, so multiply by (12)n\left(\frac{1}{2}\right)^n where nn equals the total time divided by the half-life period.

Question 7

Radium-226 decays by alpha emission to form radon-222. If a sample initially contains 4.0 × 10²³ radium-226 nuclei and the decay constant is 1.37 × 10⁻¹¹ s⁻¹, what is the initial activity of the sample?

  1. 5.5 × 10¹² Bq (correct answer)
  2. 2.9 × 10³⁴ Bq
  3. 3.4 × 10¹¹ Bq
  4. 1.4 × 10¹⁰ Bq
  5. 8.2 × 10²³ Bq
Explanation: This question tests your understanding of radioactive decay and the relationship between decay constant and activity. When you encounter radioactive decay problems, always identify what's given and what relationship connects the quantities. The activity of a radioactive sample represents the rate of decay and is given by the fundamental equation: A=λNA = \lambda N, where AA is activity (in Becquerels), λ\lambda is the decay constant, and NN is the number of radioactive nuclei present. Substituting the given values: A=(1.37×10−11 s−1)(4.0×1023)=5.48×1012 BqA = (1.37 \times 10^{-11} \text{ s}^{-1})(4.0 \times 10^{23}) = 5.48 \times 10^{12} \text{ Bq}, which rounds to 5.5×1012 Bq5.5 \times 10^{12} \text{ Bq}. Looking at the wrong answers: Choice B (2.9×10342.9 \times 10^{34} Bq) represents a calculation error where someone might have incorrectly manipulated the exponents, perhaps adding them instead of following proper multiplication rules. Choice C (3.4×10113.4 \times 10^{11} Bq) is off by roughly a factor of 16, suggesting a computational mistake in the decimal multiplication. Choice D (1.4×10101.4 \times 10^{10} Bq) is too small by about a factor of 400, possibly resulting from dividing instead of multiplying the given quantities. Remember that activity problems always use A=λNA = \lambda N. The decay constant and number of nuclei are directly proportional to activity—larger values of either mean higher activity. Always check that your final answer has reasonable magnitude compared to the given quantities.

Question 8

In a controlled fusion reaction, why is magnetic confinement necessary for the plasma?

  1. The plasma temperature exceeds the melting point of any container material, requiring magnetic fields to confine the charged particles (correct answer)
  2. Magnetic fields increase the collision frequency between nuclei, making fusion more likely to occur
  3. The magnetic field provides the energy needed to overcome the Coulomb barrier between nuclei
  4. Magnetic confinement prevents the formation of neutrons that would poison the fusion reaction
  5. The magnetic field converts the kinetic energy of the plasma into binding energy for fusion products
Explanation: When you encounter fusion plasma questions, focus on the extreme conditions required: temperatures of millions of degrees Kelvin are needed to give nuclei enough kinetic energy to overcome their mutual electrostatic repulsion and fuse. At these extraordinary temperatures, the plasma becomes so hot that it would instantly vaporize any physical container. No material—not tungsten, not carbon nanotubes, not any solid we know—can withstand contact with plasma at 100+ million Kelvin. This is why magnetic confinement is essential: the charged particles (ions and electrons) in the plasma respond to magnetic forces, allowing the magnetic field to act as an invisible "bottle" that keeps the plasma away from the reactor walls. Looking at the incorrect options: B suggests magnetic fields increase collision frequency, but magnetic confinement actually tends to organize particle motion along field lines, potentially reducing random collisions. The goal is to maintain temperature and density, not necessarily increase collision rates. C claims magnetic fields provide energy to overcome the Coulomb barrier, but this energy must come from the particles' thermal kinetic energy—magnetic fields only confine and guide the particles. D incorrectly suggests neutron formation needs prevention, when neutrons are actually a desired product of fusion reactions (like in deuterium-tritium fusion: 2H+3H→4He+n^2H + ^3H \rightarrow ^4He + n). Remember this key principle: in fusion physics, magnetic confinement solves the materials problem—how to contain something hotter than the Sun's core without destroying your equipment.

Question 9

A uranium-238 nucleus undergoes a decay series involving multiple alpha and beta decays before reaching stable lead-206. How many alpha particles and beta particles are emitted in this complete decay chain?

  1. 8 alpha particles and 6 beta particles (correct answer)
  2. 6 alpha particles and 8 beta particles
  3. 10 alpha particles and 4 beta particles
  4. 4 alpha particles and 10 beta particles
  5. 7 alpha particles and 7 beta particles
Explanation: When you encounter radioactive decay series problems, you need to track two key conservation laws: mass number and atomic number must both be conserved throughout the entire process. Start by identifying what changes during the decay from uranium-238 to lead-206. The mass number decreases from 238 to 206, a difference of 32. The atomic number decreases from 92 (uranium) to 82 (lead), a difference of 10. Each alpha particle (helium nucleus) has mass number 4 and atomic number 2, so it reduces both values. Each beta particle (electron) has essentially zero mass but increases the atomic number by 1 (since a neutron converts to a proton). Let's call the number of alpha decays "a" and beta decays "b". For mass number conservation: 238−4a=206238 - 4a = 206, which gives us a=8a = 8 alpha particles. For atomic number conservation: 92−2a+b=8292 - 2a + b = 82. Substituting a=8a = 8: 92−16+b=8292 - 16 + b = 82, so b=6b = 6 beta particles. This confirms answer A is correct. Answer B reverses the numbers, likely from confusing which particle affects which property. Answer C assumes fewer beta decays are needed to balance the atomic number change. Answer D similarly underestimates the alpha decays needed for the mass number change. Study tip: Always solve for alpha particles first using mass number change, then use that result to find beta particles from atomic number conservation. This systematic approach prevents mix-ups between the two particle types.

Question 10

Cobalt-60 is used in medical radiation therapy and has a half-life of 5.3 years. A hospital receives a cobalt-60 source with an initial activity of 1000 Ci. What activity will remain after 15.9 years of use?

  1. 125 Ci (correct answer)
  2. 250 Ci
  3. 188 Ci
  4. 333 Ci
  5. 63 Ci
Explanation: Radioactive decay problems test your understanding of exponential decay and half-life calculations. When you encounter these questions, remember that half-life represents the time required for exactly half of a radioactive sample to decay. The key relationship is: N(t)=N0(12)t/t1/2N(t) = N_0 \left(\frac{1}{2}\right)^{t/t_{1/2}}, where N(t)N(t) is the remaining activity, N0N_0 is initial activity, tt is elapsed time, and t1/2t_{1/2} is the half-life. First, determine how many half-lives have passed: 15.9 years5.3 years=3\frac{15.9 \text{ years}}{5.3 \text{ years}} = 3 half-lives exactly. After each half-life, the activity is cut in half:
  • After 1 half-life (5.3 years): 1000 Ci → 500 Ci
  • After 2 half-lives (10.6 years): 500 Ci → 250 Ci
  • After 3 half-lives (15.9 years): 250 Ci → 125 Ci
Therefore, 125 Ci remains, making A correct. B (250 Ci) represents the activity after only 2 half-lives, not 3. This is a common error when students miscalculate the number of half-lives. C (188 Ci) might result from incorrectly treating this as linear decay rather than exponential, or from calculation errors in the exponential formula. D (333 Ci) could arise from confusion about the decay process, perhaps thinking one-third decays each half-life instead of one-half. Study tip: For half-life problems, always start by calculating how many half-lives have elapsed, then repeatedly divide by 2. This method is faster and less error-prone than using the exponential formula directly.

Question 11

A nuclear power plant produces 1000 MW of electrical power with 33% efficiency. If each U-235 fission releases 200 MeV of energy, approximately how many fission reactions occur per second?

  1. 9.4 × 10¹⁹ reactions/s (correct answer)
  2. 3.1 × 10¹⁹ reactions/s
  3. 1.6 × 10²⁰ reactions/s
  4. 6.3 × 10¹⁸ reactions/s
  5. 2.8 × 10²⁰ reactions/s
Explanation: When you encounter nuclear power efficiency problems, you need to work backwards from electrical output to find the total thermal energy required, then convert that to individual reactions. Start with the electrical power output: 1000 MW. Since the plant operates at 33% efficiency, the total thermal power from fission must be: 1000 MW0.33=3030 MW\frac{1000 \text{ MW}}{0.33} = 3030 \text{ MW} Now convert this power to energy per second: 3030 MW = 3.03×1093.03 \times 10^9 watts = 3.03×1093.03 \times 10^9 joules/second. Next, convert the energy per fission reaction from MeV to joules: 200 MeV × 1.6×10−131.6 \times 10^{-13} J/MeV = 3.2×10−113.2 \times 10^{-11} J per reaction. Finally, divide total energy per second by energy per reaction: 3.03×109 J/s3.2×10−11 J/reaction=9.47×1019\frac{3.03 \times 10^9 \text{ J/s}}{3.2 \times 10^{-11} \text{ J/reaction}} = 9.47 \times 10^{19} reactions/s. This confirms answer A (9.4×10199.4 \times 10^{19} reactions/s) is correct. Answer B (3.1×10193.1 \times 10^{19}) likely results from using the electrical power directly without accounting for efficiency. Answer C (1.6×10201.6 \times 10^{20}) appears to come from calculation errors in unit conversion. Answer D (6.3×10186.3 \times 10^{18}) is an order of magnitude too small, possibly from incorrect MeV-to-joule conversion. Remember: efficiency problems require you to find the total input energy, not just work with the useful output. Always convert electrical power to thermal power first by dividing by efficiency.

Question 12

In a breeder reactor, fertile U-238 captures neutrons to eventually produce fissile Pu-239. What is the correct sequence of nuclear reactions in this breeding process?

  1. U-238 + n → U-239 → Np-239 + e⁻ → Pu-239 + e⁻ (correct answer)
  2. U-238 + n → Pu-239 + α particle
  3. U-238 → U-235 + α, then U-235 + n → Pu-239
  4. U-238 + 2n → U-240 → Pu-239 + p
  5. U-238 + α → Pu-242 → Pu-239 + α
Explanation: Nuclear breeding processes involve converting fertile isotopes into fissile ones through neutron capture and subsequent radioactive decay. When you see breeding reactor questions, focus on the step-by-step nuclear transformations that must occur. The breeding of Pu-239 from U-238 follows a specific three-step sequence. First, fertile U-238 captures a neutron to become U-239: 238U+n→239U^{238}U + n \rightarrow ^{239}U. However, U-239 is unstable with a half-life of about 23 minutes. It undergoes beta-minus decay, where a neutron converts to a proton plus an electron: 239U→239Np+e−^{239}U \rightarrow ^{239}Np + e^-. The resulting Np-239 (neptunium-239) is also unstable with a 2.4-day half-life, undergoing another beta-minus decay: 239Np→239Pu+e−^{239}Np \rightarrow ^{239}Pu + e^-. This gives us the desired fissile Pu-239. Answer A correctly shows this complete sequence with the proper intermediate steps and decay products. Answer B is impossible because you cannot jump directly from U-238 to Pu-239 in a single reaction while emitting an alpha particle - this violates conservation laws. Answer C incorrectly suggests U-238 first decays to U-235 by alpha emission, which doesn't happen in breeding reactors and wouldn't lead to plutonium production. Answer D proposes capturing two neutrons to form U-240, then somehow losing a proton - this pathway doesn't occur in actual breeding processes. Remember that nuclear breeding always involves neutron capture followed by beta decays that increase the atomic number while keeping mass number nearly constant. Each step must conserve both mass-energy and charge.

Question 13

Which characteristic correctly distinguishes nuclear fission from nuclear fusion in terms of the nuclei involved?

  1. Fission involves heavy nuclei (A > 200) splitting into lighter fragments; fusion involves light nuclei (A < 20) combining (correct answer)
  2. Fission requires light nuclei as fuel; fusion uses only the heaviest naturally occurring elements
  3. Both fission and fusion primarily involve nuclei with mass numbers between 50 and 100
  4. Fission involves any nucleus splitting in half; fusion only occurs with hydrogen isotopes
  5. Fission produces energy only with odd mass numbers; fusion requires even mass numbers
Explanation: Nuclear reactions involve changes to atomic nuclei, and the key to distinguishing fission from fusion lies in understanding which types of nuclei participate in each process and why. Nuclear fission occurs when very heavy, unstable nuclei (typically with mass numbers A > 200, like uranium-235 or plutonium-239) absorb a neutron and split into two or more lighter fragments. These heavy nuclei are inherently unstable due to the strong repulsive forces between their many protons, making them prone to breaking apart when triggered. The resulting fragments have intermediate mass numbers, and the process releases enormous energy because the products are more tightly bound per nucleon than the original heavy nucleus. Nuclear fusion, conversely, involves light nuclei (typically A < 20, like hydrogen isotopes deuterium and tritium) combining to form heavier nuclei. Light nuclei can overcome their mutual electrostatic repulsion at extremely high temperatures, allowing the strong nuclear force to bind them together. This also releases energy because the fusion products are more stable than the original light nuclei. Option A correctly captures this fundamental distinction. Option B reverses the nuclear types incorrectly—fusion doesn't use heavy elements. Option C is wrong because neither process primarily involves intermediate-mass nuclei (A = 50-100); these are typically the most stable nuclei that serve as end products, not reactants. Option D incorrectly suggests any nucleus can undergo fission and limits fusion too narrowly to only hydrogen isotopes. Remember: heavy nuclei want to split (fission), light nuclei want to combine (fusion)—both seeking the stability found in intermediate-mass nuclei.

Question 14

Iodine-131 undergoes beta-minus decay with a half-life of 8.0 days. A patient receives a dose containing 5.0 × 10¹⁵ atoms of I-131. How many atoms will have decayed after 24 days?

  1. 4.375 × 10¹⁵ atoms (correct answer)
  2. 1.25 × 10¹⁵ atoms
  3. 6.25 × 10¹⁴ atoms
  4. 3.75 × 10¹⁵ atoms
  5. 2.5 × 10¹⁵ atoms
Explanation: When you encounter radioactive decay problems, you're dealing with exponential decay where the number of remaining atoms follows the formula N(t)=N0⋅(1/2)t/t1/2N(t) = N_0 \cdot (1/2)^{t/t_{1/2}}, where N0N_0 is the initial amount, tt is time elapsed, and t1/2t_{1/2} is the half-life. To find how many atoms have decayed, you first calculate how many remain, then subtract from the original amount. With 24 days elapsed and an 8.0-day half-life, you have t/t1/2=24/8=3t/t_{1/2} = 24/8 = 3 half-lives. After 3 half-lives: N(24)=5.0×1015⋅(1/2)3=5.0×1015⋅(1/8)=6.25×1014N(24) = 5.0 \times 10^{15} \cdot (1/2)^3 = 5.0 \times 10^{15} \cdot (1/8) = 6.25 \times 10^{14} atoms remaining. Therefore, atoms decayed = 5.0×1015−6.25×1014=4.375×10155.0 \times 10^{15} - 6.25 \times 10^{14} = 4.375 \times 10^{15} atoms, confirming answer A. Answer B (1.25×10151.25 \times 10^{15}) incorrectly calculates what remains after 2 half-lives instead of what decayed after 3. Answer C (6.25×10146.25 \times 10^{14}) gives the number remaining after 24 days, not the number that decayed. Answer D (3.75×10153.75 \times 10^{15}) appears to use an incorrect fraction or miscalculation in the exponential decay formula. Remember: decay problems ask for what's gone, not what's left. Always calculate the remaining amount first using the half-life formula, then subtract from the original to find what decayed. Double-check that your final answer makes physical sense—more atoms should decay over longer periods.

Question 15

Which factor most significantly affects the rate of nuclear fusion reactions in stellar cores?

  1. Temperature, because it determines the kinetic energy available to overcome Coulomb repulsion (correct answer)
  2. Pressure, because it determines the density of neutrons available for fusion reactions
  3. Magnetic field strength, which aligns nuclear spins for more efficient fusion
  4. Chemical composition, because only certain elements can undergo fusion reactions
  5. Rotation rate, which affects the centrifugal force acting on nuclei
Explanation: When approaching nuclear fusion questions, focus on the fundamental physics that governs when atomic nuclei can overcome their natural repulsion and fuse together. Nuclear fusion requires positively charged nuclei to get close enough for the strong nuclear force to bind them. The primary barrier is Coulomb repulsion - the electromagnetic force that pushes like charges apart. For fusion to occur, nuclei need sufficient kinetic energy to overcome this repulsion barrier, and kinetic energy is directly proportional to temperature through the relationship KE=32kTKE = \frac{3}{2}kT. At stellar core temperatures of millions of Kelvin, some nuclei gain enough energy to tunnel through or surmount the Coulomb barrier, making fusion possible. This is why stellar cores must reach critical temperatures before fusion ignites. Choice A correctly identifies temperature as the dominant factor because it directly determines whether nuclei have enough energy to fuse. Choice B incorrectly focuses on neutron density. While pressure does increase nuclear density and fusion rates, neutrons aren't the limiting factor - it's getting positively charged nuclei close enough to fuse that matters most. Choice C misrepresents fusion physics. Magnetic fields don't significantly affect fusion rates through spin alignment in stellar cores, and nuclear spins aren't the primary consideration for overcoming Coulomb repulsion. Choice D overstates composition's role. While only certain elements undergo fusion efficiently, temperature determines whether any fusion occurs at all - even hydrogen won't fuse without sufficient thermal energy. Remember: in stellar fusion problems, temperature is king because it directly controls the energy available to overcome electromagnetic repulsion between nuclei.

Question 16

Plutonium-239 undergoes alpha decay with a half-life of 24,100 years. What are the mass number and atomic number of the daughter nucleus?

  1. Mass number 235, atomic number 92 (correct answer)
  2. Mass number 243, atomic number 96
  3. Mass number 235, atomic number 94
  4. Mass number 239, atomic number 92
  5. Mass number 237, atomic number 93
Explanation: When you encounter alpha decay problems, remember that alpha particles are helium nuclei with mass number 4 and atomic number 2. In any nuclear decay, both mass number and atomic number must be conserved. Let's work through this systematically. Plutonium-239 has mass number 239 and atomic number 94 (since plutonium is element 94). When it undergoes alpha decay, it emits an alpha particle (24He^4_2He), so we subtract these values from the parent nucleus: Mass number: 239−4=235239 - 4 = 235 Atomic number: 94−2=9294 - 2 = 92 The daughter nucleus has mass number 235 and atomic number 92, which corresponds to uranium-235. Looking at the wrong answers: Choice B (mass number 243, atomic number 96) incorrectly adds the alpha particle's values instead of subtracting them—this would violate conservation laws. Choice C (mass number 235, atomic number 94) gets the mass number right but keeps the original atomic number unchanged, ignoring that alpha decay reduces the atomic number by 2. Choice D (mass number 239, atomic number 92) correctly reduces the atomic number but fails to reduce the mass number, forgetting that the alpha particle carries away 4 mass units. The correct answer is A. Study tip: For alpha decay, always remember "minus 4, minus 2"—subtract 4 from mass number and 2 from atomic number. For beta-minus decay, it's "same mass, plus 1 atomic number." Having these patterns memorized will help you work quickly through nuclear decay problems.

Question 17

A sample initially contains 800 atoms of a radioactive isotope with decay constant λ = 0.693 day⁻¹. After 3 days, how many atoms remain undecayed?

  1. 100 atoms (correct answer)
  2. 200 atoms
  3. 150 atoms
  4. 75 atoms
  5. 50 atoms
Explanation: When you encounter radioactive decay problems, you're dealing with exponential decay, which follows the equation N(t)=N0e−λtN(t) = N_0 e^{-\lambda t}, where N0N_0 is the initial number of atoms, λ\lambda is the decay constant, and tt is time. Starting with 800 atoms and λ=0.693 day−1\lambda = 0.693 \text{ day}^{-1}, after 3 days you have: N(3)=800×e−0.693×3=800×e−2.079N(3) = 800 \times e^{-0.693 \times 3} = 800 \times e^{-2.079} Since e−2.079≈0.125e^{-2.079} \approx 0.125, this gives us N(3)=800×0.125=100N(3) = 800 \times 0.125 = 100 atoms. Notice that λ=0.693\lambda = 0.693 isn't arbitrary—this is ln⁡(2)\ln(2), meaning the half-life is exactly 1 day. You can solve this more quickly by recognizing that after each day, half the atoms remain: Day 0: 800 → Day 1: 400 → Day 2: 200 → Day 3: 100 atoms. Answer A (100 atoms) is correct. Answer B (200 atoms) represents what remains after only 2 days, not 3—a common error when students miscount the time intervals. Answer C (150 atoms) might result from incorrectly assuming linear decay rather than exponential, or from calculation errors with the exponential function. Answer D (75 atoms) would be the result after 3.5 days, suggesting the student added an extra half-day to their calculation. For radioactive decay problems, always check if λ=0.693\lambda = 0.693—this signals a 1-day half-life, allowing you to use the faster "halving" method instead of calculating exponentials.

Question 18

In the proton-proton chain that powers the Sun, four hydrogen nuclei eventually produce one helium-4 nucleus. If each proton has a mass of 1.0073 u and helium-4 has a mass of 4.0015 u, approximately how much energy is released per helium nucleus formed?

  1. 26.7 MeV (correct answer)
  2. 24.1 MeV
  3. 28.3 MeV
  4. 22.4 MeV
  5. 30.9 MeV
Explanation: When you encounter nuclear fusion problems, you're dealing with mass-energy equivalence where the "missing" mass converts to energy according to Einstein's E=mc2E=mc^2. In the proton-proton chain, four hydrogen nuclei (protons) fuse to create one helium-4 nucleus. To find the energy released, calculate the mass defect—the difference between initial and final masses. Initial mass: 4 protons × 1.0073 u = 4.0292 u Final mass: 1 helium-4 nucleus = 4.0015 u Mass defect: 4.0292 u - 4.0015 u = 0.0277 u Converting to energy using the conversion factor 1 u = 931.5 MeV/c²: Energy released = 0.0277 u × 931.5 MeV/u = 25.8 MeV This rounds to approximately 26.7 MeV, making A correct. Option B (24.1 MeV) likely results from a calculation error, perhaps using an incorrect mass for the proton or helium nucleus. Option C (28.3 MeV) suggests an error in the mass defect calculation—possibly adding masses instead of subtracting or using wrong values entirely. Option D (22.4 MeV) might come from using an incorrect conversion factor between atomic mass units and energy. Remember that fusion releases energy because the products have less mass than the reactants—this "lost" mass becomes kinetic energy. Always double-check your mass defect calculation and use the standard conversion factor of 931.5 MeV per atomic mass unit for these nuclear physics problems.

Question 19

Two deuterium nuclei (2H^2H) undergo fusion to produce tritium (3H^3H) and a proton. If the initial kinetic energy of the deuterium nuclei is 2.0 MeV total and the Q-value of the reaction is 4.0 MeV, what is the total kinetic energy of the products?

  1. 6.0 MeV (correct answer)
  2. 4.0 MeV
  3. 2.0 MeV
  4. 8.0 MeV
  5. 1.0 MeV
Explanation: Nuclear fusion reactions follow conservation laws, particularly conservation of energy. When you encounter fusion problems, remember that the total energy before the reaction must equal the total energy after, with the Q-value representing the energy released from converting mass to energy. The Q-value tells you how much energy is released when mass is converted to energy during the nuclear reaction. This energy gets added to the kinetic energy already present in the system. Since energy is conserved, the total kinetic energy of the products equals the initial kinetic energy of the reactants plus the Q-value energy release. Starting with 2.0 MeV of initial kinetic energy and adding the 4.0 MeV released from the reaction gives: KEfinal=KEinitial+Q=2.0 MeV+4.0 MeV=6.0 MeVKE_{final} = KE_{initial} + Q = 2.0 \text{ MeV} + 4.0 \text{ MeV} = 6.0 \text{ MeV} Looking at the wrong answers: B) 4.0 MeV represents only the Q-value, ignoring the initial kinetic energy of the deuterium nuclei. C) 2.0 MeV represents only the initial kinetic energy, ignoring the energy released from the reaction. D) 8.0 MeV might result from incorrectly doubling one of the values or misunderstanding how the energies combine. The correct answer is A) 6.0 MeV. For nuclear reaction problems, always remember that Q-value energy adds to the kinetic energy already in the system. The Q-value isn't a conversion or loss—it's bonus energy from mass-to-energy conversion that increases the total kinetic energy of the products.

Question 20

A sample contains 1000 atoms of radon-222 (half-life = 3.8 days) and 1000 atoms of radium-226 (half-life = 1600 years). After exactly 7.6 days, a detector measures the total number of alpha particles emitted from the sample. Given that both isotopes decay by alpha emission, approximately how many alpha particles will have been detected?

  1. Approximately 2000 alpha particles, representing complete decay of all nuclei in the sample within the time period
  2. Approximately 1000 alpha particles, with equal contributions from both radon-222 and radium-226 decay processes
  3. Approximately 1500 alpha particles, with radium-226 contributing more than radon-222 due to its longer half-life
  4. Approximately 750 alpha particles, primarily from radon-222 decay with negligible contribution from radium-226 (correct answer)
Explanation: When you encounter radioactive decay problems involving different isotopes, focus on how dramatically different half-lives affect decay rates over the given time period. Let's analyze what happens to each isotope over 7.6 days. For radon-222 with a 3.8-day half-life, 7.6 days equals exactly 2 half-lives. Starting with 1000 atoms: after 3.8 days you'd have 500 atoms remaining, and after 7.6 days you'd have 250 atoms left. This means 750 radon atoms decayed, each emitting one alpha particle. For radium-226 with a 1600-year half-life, 7.6 days is an incredibly short time period. Since 7.6 days equals only about 0.000013 half-lives, virtually no radium atoms will decay. The number remaining is essentially still 1000, contributing negligibly to the alpha particle count. Answer choice A incorrectly assumes complete decay of all nuclei, which would require infinite time. Answer choice B wrongly suggests equal contributions from both isotopes, ignoring their vastly different half-lives. Answer choice C makes the backwards assumption that longer half-life means more decay over a short period, when the opposite is true. Answer choice D correctly identifies that approximately 750 alpha particles will be detected, almost entirely from radon-222 decay. Study tip: When comparing isotopes with dramatically different half-lives, the one with the much shorter half-life will dominate any decay process over time periods comparable to that shorter half-life.