College Physics Quiz: Types Of Radioactive Decay
17 questions · exam conditions
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Types Of Radioactive DecayQuestion 1 of 17

A positron emission tomography (PET) scan uses fluorine-18, which decays by emitting a positron. In this decay process, what happens to the nuclear composition, and what additional particle must be emitted to conserve lepton number?

Proton converts to neutron; electron neutrino is emitted
Neutron converts to proton; electron antineutrino is emitted
Proton converts to neutron; electron antineutrino is emitted
Proton converts to neutron; electron neutrino and photon are emitted
Neutron converts to proton; electron neutrino is emitted
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College Physics Quiz

College Physics Quiz: Types Of Radioactive Decay

Practice Types Of Radioactive Decay in College 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 Types Of Radioactive Decay, giving you a quick way to practice the rules, question types, and explanations that matter most for College Physics.

How to use this quiz

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

All questions

Question 1

A positron emission tomography (PET) scan uses fluorine-18, which decays by emitting a positron. In this decay process, what happens to the nuclear composition, and what additional particle must be emitted to conserve lepton number?

  1. Proton converts to neutron; electron neutrino is emitted (correct answer)
  2. Neutron converts to proton; electron antineutrino is emitted
  3. Proton converts to neutron; electron antineutrino is emitted
  4. Proton converts to neutron; electron neutrino and photon are emitted
  5. Neutron converts to proton; electron neutrino is emitted
Explanation: When you encounter questions about radioactive decay, focus on the conservation laws that must be satisfied: charge, mass-energy, and lepton number. Positron emission (β⁺ decay) is a specific type of decay where an unstable nucleus has too many protons relative to neutrons. In positron emission, a proton in the nucleus converts to a neutron, emitting a positron (e⁺) in the process. This reduces the atomic number by 1 while keeping the mass number constant. However, this creates a lepton number problem: the positron carries a lepton number of -1, but we started with zero leptons. To conserve lepton number, an electron neutrino (νₑ) must also be emitted, which has a lepton number of +1, balancing the equation. Looking at the wrong answers: Choice B describes β⁻ decay (neutron to proton), not positron emission, and incorrectly suggests an antineutrino is emitted. Choice C correctly identifies the proton-to-neutron conversion but wrongly states an electron antineutrino is emitted—this would make lepton number conservation impossible since both the positron and antineutrino would carry negative lepton numbers. Choice D correctly identifies the nuclear change and neutrino type but unnecessarily adds a photon, which isn't required for lepton number conservation. The correct answer is A: the proton converts to a neutron, and an electron neutrino is emitted alongside the positron. Study tip: Remember that in positron emission, you always get a neutrino (not antineutrino) to balance the negative lepton number of the positron. The pattern is β⁺ + νₑ for lepton number conservation.

Question 2

A radioactive nucleus undergoes decay and emits a particle with charge +2e and mass approximately 4 u. After this decay, the daughter nucleus has an atomic number that is 2 less than the parent and a mass number that is 4 less than the parent. However, the daughter nucleus is found to be in an excited state and subsequently emits electromagnetic radiation to reach its ground state. What is the correct sequence of decay processes?

  1. Alpha decay followed by gamma emission (correct answer)
  2. Beta-plus decay followed by gamma emission
  3. Beta-minus decay followed by alpha decay
  4. Gamma emission followed by alpha decay
  5. Alpha decay followed by beta-minus decay
Explanation: When analyzing nuclear decay problems, you need to track both the particles emitted and the changes in atomic number (Z) and mass number (A) to identify the decay type. The question describes a particle with charge +2e and mass 4 u being emitted, which is an alpha particle (helium nucleus, 4He2+^4He^{2+}). When a nucleus undergoes alpha decay, it loses 2 protons and 2 neutrons, so the daughter nucleus has Z decreased by 2 and A decreased by 4 - exactly what's described. The subsequent electromagnetic radiation emission is gamma decay, which occurs when an excited nucleus releases energy to reach its ground state without changing its composition. Looking at the wrong answers: B) Beta-plus decay involves emitting a positron (charge +1e, negligible mass) and increases Z by 1 while keeping A constant - this doesn't match the described particle or nuclear changes. C) Beta-minus decay emits an electron (charge -1e) and decreases Z by 1, followed by alpha decay - but the question describes only one charged particle emission before gamma rays. D) Gamma emission followed by alpha decay reverses the actual sequence; gamma rays don't change nuclear composition, so they couldn't produce the initial particle described. Answer A correctly identifies both processes: alpha decay produces the +2e, 4 u particle and the specified nuclear changes, followed by gamma emission from the excited daughter nucleus. Study tip: Always match the described particle properties (charge, mass) to known decay particles first, then verify that the nuclear changes (ΔZ, ΔA) are consistent with that decay type.

Question 3

An isotope of carbon-14 undergoes radioactive decay. The decay process converts a neutron in the nucleus into a proton, an electron, and an antineutrino. The electron is ejected from the nucleus. What type of decay is this, and what is the resulting nucleus?

  1. Alpha decay producing boron-10
  2. Beta-minus decay producing nitrogen-14 (correct answer)
  3. Beta-plus decay producing boron-14
  4. Gamma decay producing carbon-14
  5. Beta-minus decay producing nitrogen-13
Explanation: When you encounter radioactive decay problems, focus on what's happening to the particles in the nucleus and what gets emitted. The key details here are that a neutron converts to a proton plus an ejected electron (and antineutrino). This process describes beta-minus decay, where a neutron transforms into a proton while emitting an electron (called a beta particle) and an antineutrino. Since the atomic number increases by 1 (gaining a proton) while the mass number stays the same, carbon-14 (6 protons) becomes nitrogen-14 (7 protons). The equation is: 14C→14N+e−+νˉe^{14}C \rightarrow ^{14}N + e^- + \bar{\nu}_e Looking at the wrong answers: Choice A describes alpha decay, which involves emitting a helium nucleus (2 protons + 2 neutrons), not converting neutrons to protons. Choice C represents beta-plus decay, where a proton converts to a neutron (opposite of what's described), and boron-14 would require losing a proton, not gaining one. Choice D suggests gamma decay, which only involves energy emission without changing the nuclear composition—no particle conversion occurs. Remember this pattern: in beta-minus decay, the atomic number increases by 1 while mass number stays constant, and you'll see neutron-to-proton conversion with electron emission. In beta-plus decay, it's the reverse. For college physics exams, always track what happens to proton and neutron counts to identify the decay type and predict the resulting nucleus.

Question 4

A nucleus undergoes electron capture, where an inner orbital electron is captured by a proton in the nucleus. This process competes with positron emission in certain isotopes. What fundamental difference distinguishes when electron capture is energetically favorable compared to positron emission?

  1. Electron capture requires less energy because no positron mass-energy is created (correct answer)
  2. Electron capture occurs only in heavy nuclei with high binding energies
  3. Positron emission conserves lepton number while electron capture does not
  4. Electron capture produces gamma rays while positron emission does not
  5. Positron emission requires neutron-rich nuclei while electron capture requires proton-rich nuclei
Explanation: When analyzing nuclear decay processes, the key factor is energy balance—specifically, how much mass-energy must be invested to create new particles versus how much binding energy is released. Both electron capture and positron emission achieve the same nuclear transformation (converting a proton to a neutron), but they differ crucially in their energy requirements. In electron capture, an orbital electron combines directly with a nuclear proton: p+e−→n+νep + e^- \rightarrow n + \nu_e. The electron already exists, so no new particle mass must be created. In positron emission, the nucleus must create a positron from pure energy: p→n+e++νep \rightarrow n + e^+ + \nu_e, requiring the rest mass energy of the positron (0.511 MeV). This makes answer A correct—electron capture requires less energy because no positron mass-energy needs to be created. When the available decay energy (Q-value) is less than 1.022 MeV (twice the electron rest mass), only electron capture is energetically possible. B is wrong because electron capture occurs across all nuclear masses, not just heavy nuclei. C contains a fundamental error—both processes conserve lepton number perfectly (electron capture: lepton number changes from 0 to +1, positron emission: changes from 0 to +1). D is incorrect because both processes typically produce gamma rays from nuclear de-excitation, and positron emission also creates characteristic 0.511 MeV annihilation photons. Remember: when comparing competing decay modes, always consider the mass-energy cost of particle creation—existing particles (like orbital electrons) are "free" while creating new particles requires significant energy investment.

Question 5

A radioactive nucleus emits a 2.3 MeV gamma ray immediately after undergoing alpha decay. The gamma ray has the same energy regardless of the direction of the recoiling daughter nucleus. What does this observation indicate about the gamma emission process?

  1. The gamma ray is emitted simultaneously with the alpha particle
  2. The daughter nucleus has negligible recoil energy compared to the gamma ray energy (correct answer)
  3. The alpha decay and gamma emission violate conservation of momentum
  4. The gamma ray energy comes from the kinetic energy of the alpha particle
  5. The process involves internal conversion instead of gamma emission
Explanation: When analyzing nuclear decay processes, you need to consider energy and momentum conservation separately for each stage of a multi-step decay. The key insight here is understanding what affects the energy of emitted radiation. The correct answer is B because the observation tells us something crucial about energy scales. If the gamma ray energy remains constant at 2.3 MeV regardless of the daughter nucleus's recoil direction, this means the daughter's recoil kinetic energy is negligible compared to the gamma ray energy. In alpha decay, daughter nuclei typically recoil with energies of tens to hundreds of keV, which is indeed much smaller than 2.3 MeV. This small recoil energy doesn't significantly affect the available energy for gamma emission, so the gamma ray energy stays constant. Option A is incorrect because simultaneous emission would actually cause the gamma ray energy to vary with recoil direction due to momentum conservation constraints. Option C misses the point entirely—both processes individually conserve momentum; the constant energy simply reflects the energy scale difference. Option D is wrong because gamma rays originate from nuclear energy level transitions in the daughter nucleus, not from the alpha particle's kinetic energy. The gamma ray comes from the daughter nucleus transitioning from an excited state to a lower energy state after the alpha decay, and since the recoil energy is small relative to nuclear binding energies, it doesn't significantly affect this transition energy. Study tip: In nuclear physics problems, always compare energy scales. Nuclear transition energies (MeV) typically dwarf recoil energies (keV), which often simplifies the analysis.

Question 6

In beta-minus decay, the energy spectrum of emitted electrons shows a continuous distribution from zero up to a maximum energy EmaxE_{max}, rather than a discrete energy. This observation led to the prediction of which particle, and why was this particle necessary?

  1. Electron neutrino; to conserve electric charge in the decay process
  2. Electron antineutrino; to conserve energy and momentum in three-body decay (correct answer)
  3. Positron; to conserve lepton number in the decay process
  4. Electron antineutrino; to conserve angular momentum only
  5. Electron neutrino; to conserve mass number in nuclear decay
Explanation: When you encounter questions about beta decay and continuous energy spectra, you're dealing with one of the most important discoveries in particle physics that revealed the need for conservation laws to explain experimental observations. In beta-minus decay, a neutron converts to a proton, electron, and another particle: n→p+e−+νˉen → p + e^- + \bar{\nu}_e. The puzzle was that if only two particles (proton and electron) were produced, the electron should have a fixed energy based on the mass difference. Instead, electrons showed a continuous energy spectrum up to EmaxE_{max}, suggesting a third particle was carrying away the "missing" energy and momentum. Wolfgang Pauli proposed the neutrino (specifically the electron antineutrino νˉe\bar{\nu}_e) to solve this problem. In three-body decay, the available energy can be shared among all three particles in various combinations, creating the observed continuous spectrum. This preserves both energy and momentum conservation, making answer B correct. Let's examine why the other options fail: A is wrong because electric charge was already conserved with just the proton and electron (neutron is neutral, proton +1, electron -1). C suggests a positron, but positrons appear in beta-plus decay, not beta-minus, and wouldn't solve the energy distribution problem. D incorrectly identifies angular momentum as the primary concern—while the antineutrino does help conserve angular momentum, the main issue was energy and momentum conservation in the continuous spectrum. Remember: continuous energy spectra in decay processes typically indicate missing particles needed to satisfy conservation laws, especially energy and momentum conservation.

Question 7

Internal conversion is an alternative to gamma emission where the nuclear excitation energy is transferred to an inner orbital electron, ejecting it from the atom. For a particular nuclear transition, the conversion coefficient (ratio of conversion electrons to gamma rays) is higher for K-shell electrons than L-shell electrons. What property of the orbital electrons explains this difference?

  1. K-shell electrons have higher binding energies than L-shell electrons
  2. K-shell electrons have larger orbital radii than L-shell electrons
  3. K-shell electrons have higher probability density at the nucleus than L-shell electrons (correct answer)
  4. K-shell electrons have lower kinetic energies than L-shell electrons
  5. K-shell electrons have opposite spin compared to L-shell electrons
Explanation: Internal conversion is a nuclear decay process where excited nuclear energy is directly transferred to orbital electrons rather than emitted as gamma rays. The key to understanding conversion coefficients lies in recognizing that this process depends on the overlap between the nuclear wave function and the electron wave function. The correct answer is C because K-shell electrons have the highest probability density at the nucleus. Since internal conversion requires direct interaction between the nucleus and orbital electrons, electrons that spend more time near the nucleus have a much higher probability of receiving the nuclear excitation energy. K-shell electrons, being in the innermost orbital (n=1), have wave functions that peak closest to the nucleus, making them most likely to undergo internal conversion. Option A is incorrect because while K-shell electrons do have higher binding energies, this actually makes conversion less favorable energetically, not more. Option B is wrong because K-shell electrons actually have smaller orbital radii than L-shell electrons - they're closer to the nucleus, not farther away. Option D is incorrect because the kinetic energy difference between shells doesn't determine the conversion probability; it's the spatial overlap with the nucleus that matters. When studying nuclear physics, remember that processes involving direct nuclear-electron interactions (like internal conversion and electron capture) favor inner shell electrons because of their proximity to the nucleus. The wave function overlap principle governs these interactions - the closer the electron orbital to the nucleus, the stronger the coupling.

Question 8

A radioactive isotope undergoes beta-plus decay with a Q-value (total energy release) of 2.5 MeV. In this decay, what is the minimum kinetic energy that the emitted positron can have, and what determines this minimum value?

  1. 0 MeV; energy is shared continuously with neutrino and recoil nucleus (correct answer)
  2. 0.511 MeV; minimum energy equals the positron rest mass energy
  3. 1.022 MeV; minimum energy equals the pair creation threshold energy
  4. 2.5 MeV; minimum energy equals the total Q-value of the decay
  5. 1.489 MeV; energy sharing follows three-body conservation laws
Explanation: When analyzing beta-plus decay, you need to understand that this is a three-body decay process where energy and momentum are shared among three particles: the positron, neutrino, and recoiling daughter nucleus. In beta-plus decay, a proton converts to a neutron, positron, and neutrino. The Q-value represents the total energy available for distribution among these three particles. Since we have three particles sharing this energy, the positron's kinetic energy can range from zero (when the neutrino carries maximum energy) to nearly the full Q-value (when the neutrino carries minimal energy). The continuous energy spectrum of emitted positrons is a hallmark of beta decay. The minimum kinetic energy occurs when the positron is emitted with just enough energy to satisfy momentum conservation while the neutrino carries away the maximum possible energy. This minimum can indeed be zero. Answer B incorrectly confuses the positron's rest mass energy (0.511 MeV) with its minimum kinetic energy. The rest mass energy is already accounted for in the Q-value calculation and doesn't determine the kinetic energy distribution. Answer C mistakenly applies pair creation physics to beta decay. The 1.022 MeV pair creation threshold is irrelevant here since the positron is created through nuclear transformation, not pair production. Answer D assumes the positron must carry all the decay energy, ignoring the three-body nature of the process and the neutrino's role in energy sharing. Study tip: Remember that beta decay spectra are continuous because energy is shared among three particles, unlike alpha decay where only two particles share the energy, creating discrete energies.

Question 9

A radioactive sample contains nuclei that can undergo three different decay modes: alpha decay with 60% probability, beta-minus decay with 30% probability, and electron capture with 10% probability. If 1000 nuclei decay simultaneously, approximately how many of the resulting daughter nuclei will have the same mass number as the original parent nuclei?

  1. 100 nuclei from electron capture only
  2. 300 nuclei from beta-minus decay only
  3. 400 nuclei from beta-minus and electron capture (correct answer)
  4. 600 nuclei from alpha decay only
  5. 900 nuclei from all three decay modes
Explanation: When you encounter radioactive decay problems, focus on how each decay type affects the mass number (A) and atomic number (Z). The key insight is that only certain decay modes preserve the mass number. Let's examine each decay type: Alpha decay reduces both mass number and atomic number (A decreases by 4, Z decreases by 2), so the daughter nucleus has a different mass number. Beta-minus decay converts a neutron to a proton, keeping the mass number unchanged while increasing the atomic number by 1. Electron capture converts a proton to a neutron, also preserving the mass number while decreasing the atomic number by 1. With 1000 decaying nuclei, we get: 600 alpha decays (60%), 300 beta-minus decays (30%), and 100 electron captures (10%). Since only beta-minus decay and electron capture preserve the mass number, we have 300 + 100 = 400 daughter nuclei with the same mass number as the parent. Answer A is wrong because it only counts electron capture (100 nuclei) and ignores beta-minus decay. Answer B incorrectly suggests only beta-minus decay preserves mass number, missing the electron capture contribution. Answer D is incorrect because alpha decay always changes the mass number by reducing it by 4. For nuclear decay problems, remember this pattern: alpha and beta-plus decay change mass number, while beta-minus decay and electron capture preserve it. Always check which decay modes maintain the specific nuclear property the question asks about.

Question 10

A sample of technetium-99m (the 'm' indicates metastable) is used in medical imaging. This isotope decays by emitting a 140 keV gamma ray with a half-life of 6 hours. After gamma emission, the resulting nucleus is technetium-99, which is also radioactive. What type of decay process produces technetium-99m, and what happens to the nuclear composition during gamma decay?

  1. Beta-minus decay produces Tc-99m; gamma decay decreases mass number by 1
  2. Alpha decay produces Tc-99m; gamma decay changes atomic number by 1
  3. Beta-minus decay produces Tc-99m; gamma decay leaves nuclear composition unchanged (correct answer)
  4. Electron capture produces Tc-99m; gamma decay decreases atomic number by 1
  5. Alpha decay produces Tc-99m; gamma decay decreases mass number by 1
Explanation: When you encounter nuclear decay problems, you need to distinguish between different types of radioactive processes and understand what happens to nuclear composition in each. Technetium-99m is produced when molybdenum-99 undergoes beta-minus decay. In beta-minus decay, a neutron converts to a proton plus an electron (beta particle), increasing the atomic number by 1 while keeping the mass number constant. This transforms Mo-99 (Z=42) into Tc-99m (Z=43). The "m" indicates the technetium nucleus is left in a metastable (excited) state. During gamma decay, the excited Tc-99m nucleus simply releases energy as a 140 keV photon to reach its ground state, becoming stable Tc-99. Crucially, gamma emission involves no particle emission that would change the number of protons or neutrons—only energy is released. The nuclear composition remains identical. Choice A correctly identifies beta-minus decay as the production mechanism but incorrectly claims gamma decay decreases mass number—gamma rays are massless photons. Choice B suggests alpha decay (which would decrease both mass and atomic numbers significantly) and wrongly states gamma decay changes atomic number. Choice D proposes electron capture (which decreases atomic number) as the production method and incorrectly claims gamma decay affects atomic number. Remember this key distinction: particle emission (alpha, beta, positron) changes nuclear composition by altering proton/neutron numbers, while gamma emission only releases excess energy without changing the nucleus's identity. This makes gamma decay useful for medical imaging since it provides detectable radiation without creating different elements.

Question 11

Consider the decay chain: 238U→234Th→234Pa→234U^{238}U \rightarrow ^{234}Th \rightarrow ^{234}Pa \rightarrow ^{234}U. Based on the changes in mass number and atomic number in this sequence, what types of decay occur at each step?

  1. Alpha, beta-minus, beta-minus (correct answer)
  2. Beta-minus, alpha, beta-plus
  3. Alpha, beta-plus, beta-minus
  4. Beta-plus, beta-minus, alpha
  5. Alpha, beta-minus, beta-plus
Explanation: Nuclear decay questions require you to track how mass number (A) and atomic number (Z) change between parent and daughter nuclei. Each decay type has a characteristic signature: alpha decay reduces both A by 4 and Z by 2, beta-minus decay increases Z by 1 while A stays constant, and beta-plus decay decreases Z by 1 while A stays constant. Let's trace each step in the decay chain. First step: 238U^{238}U (A=238, Z=92) → 234Th^{234}Th (A=234, Z=90). The mass number dropped by 4 and atomic number dropped by 2, which is the signature of alpha decay (4He2+^4He^{2+} emission). Second step: 234Th^{234}Th (A=234, Z=90) → 234Pa^{234}Pa (A=234, Z=91). Mass number unchanged, atomic number increased by 1. This indicates beta-minus decay, where a neutron converts to a proton plus electron. Third step: 234Pa^{234}Pa (A=234, Z=91) → 234U^{234}U (A=234, Z=92). Again, mass number unchanged but atomic number increased by 1, indicating another beta-minus decay. The sequence is alpha, beta-minus, beta-minus, making A correct. Option B incorrectly suggests alpha decay in the middle step where mass number doesn't change. Option C places beta-plus decay in the second step, but Z increases rather than decreases. Option D misplaces all decay types and puts alpha decay last when both A and Z remain constant. Study tip: Always calculate ΔA and ΔZ for each step first. Alpha changes both significantly, while beta decays only change Z by ±1.

Question 12

During alpha decay of 238^{238}U, the daughter nucleus recoils with kinetic energy KdK_d. If the alpha particle has kinetic energy KαK_α, which relationship correctly describes the momentum and energy conservation in this decay process?

  1. Kα=KdK_α = K_d and the alpha particle and daughter nucleus have equal magnitude momenta in opposite directions
  2. Kα>KdK_α > K_d and the alpha particle and daughter nucleus have equal magnitude momenta in opposite directions (correct answer)
  3. Kα<KdK_α < K_d and the alpha particle has greater momentum magnitude than the daughter nucleus
  4. Kα=KdK_α = K_d and the alpha particle has greater momentum magnitude than the daughter nucleus
Explanation: In alpha decay, momentum conservation requires the alpha particle and daughter nucleus to have equal magnitude momenta in opposite directions: pα=pdp_α = p_d. However, since kinetic energy K=p2/(2m)K = p^2/(2m) and the alpha particle (mass ≈ 4u) is much lighter than the thorium daughter nucleus (mass ≈ 234u), the alpha particle has much greater kinetic energy: Kα=p2/(2×4u)>>Kd=p2/(2×234u)K_α = p^2/(2×4u) >> K_d = p^2/(2×234u). Choice A incorrectly equates kinetic energies. Choice C reverses the energy relationship and violates momentum conservation. Choice D violates momentum conservation.

Question 13

A nucleus with mass number 60 and atomic number 27 undergoes electron capture. During this process, an inner orbital electron combines with a proton in the nucleus. What happens to the atomic structure and what additional particle is necessarily emitted?

  1. The atomic number becomes 26, a characteristic X-ray photon is emitted when the electron vacancy is filled, and a neutrino carries away excess energy (correct answer)
  2. The atomic number becomes 28, an Auger electron is ejected from an outer shell, and an antineutrino is emitted to conserve lepton number
  3. The atomic number becomes 26, the mass number decreases to 59, and a positron is emitted to balance the nuclear charge
  4. The atomic number becomes 28, the nuclear binding energy increases significantly, and gamma radiation is emitted to release excess energy
Explanation: In electron capture, an inner orbital electron combines with a proton to form a neutron and neutrino: p+e−→n+νep + e^- → n + ν_e. This decreases the atomic number by 1 (27→26) while the mass number stays 60. The electron vacancy in the inner shell is filled by electrons from higher shells, emitting characteristic X-rays. A neutrino is emitted to conserve energy and lepton number. Choice B incorrectly increases atomic number and mentions antineutrino. Choice C incorrectly decreases mass number and mentions positron emission. Choice D increases atomic number and doesn't account for the neutrino.

Question 14

A research team is studying the decay products from a sample of 90^{90}Sr, which undergoes beta-minus decay with a half-life of 28.8 years. The daughter nucleus produced is also radioactive and undergoes further beta-minus decay with a half-life of 64 hours.

Based on the passage above, what is the final stable nucleus produced in this decay chain, and what is the total change in atomic number from the original 90^{90}Sr nucleus?

  1. 86^{86}Kr with atomic number 36, representing a decrease of 2 in atomic number through alpha decay processes
  2. 90^{90}Mo with atomic number 42, representing an increase of 4 in atomic number due to multiple decay processes
  3. 90^{90}Zr with atomic number 40, representing an increase of 2 in atomic number through two successive beta-minus decays (correct answer)
  4. 90^{90}Y with atomic number 39, representing an increase of 1 in atomic number through a single effective decay process
Explanation: When you encounter radioactive decay chains, focus on tracking how each type of decay affects the atomic number and mass number. Beta-minus decay is key here: it converts a neutron to a proton, increasing the atomic number by 1 while keeping the mass number constant. Let's trace this decay chain step by step. Strontium-90 (90^{90}Sr) has atomic number 38. When it undergoes beta-minus decay, the atomic number increases by 1, producing 90^{90}Y (yttrium-90) with atomic number 39. This daughter nucleus is also radioactive and undergoes another beta-minus decay, again increasing the atomic number by 1, yielding 90^{90}Zr (zirconium-90) with atomic number 40. Zirconium-90 is stable, ending the decay chain. The total change in atomic number is +2 (from 38 to 40). Answer A incorrectly suggests alpha decay and a decrease in atomic number. Alpha decay would change the mass number from 90 to 86, but the passage specifies beta-minus decay. Answer B shows the wrong final nucleus (molybdenum) and an impossible +4 change from just two beta-minus decays. Answer D stops at the intermediate product (yttrium-90) rather than the final stable nucleus, showing only one decay instead of the complete chain. Remember: in beta-minus decay, atomic number always increases by 1 while mass number stays the same. Count each decay step carefully in multi-step chains, and make sure you identify the final stable product, not an intermediate one.

Question 15

A nucleus undergoes internal conversion instead of gamma emission, where a 0.5 MeV nuclear transition energy is transferred to a K-shell electron (binding energy 80 keV). What is the kinetic energy of the ejected conversion electron, and what subsequent atomic process must occur?

  1. 420 keV kinetic energy, followed by characteristic X-ray emission with energy approximately 80 keV as L-shell electrons fill the K-shell vacancy
  2. 580 keV kinetic energy, followed by Auger electron emission as the only mechanism to fill the inner shell vacancy
  3. 500 keV kinetic energy, followed by both X-ray emission and possible Auger electron emission to fill the electron vacancy
  4. 420 keV kinetic energy, followed by either characteristic X-ray emission or Auger electron emission as competing processes to fill the K-shell vacancy (correct answer)
Explanation: In internal conversion, the conversion electron receives kinetic energy equal to the transition energy minus the binding energy: Ke=Etransition−BK=500K_e = E_{transition} - B_K = 500 keV −80- 80 keV =420= 420 keV. The K-shell vacancy can be filled by either: (1) X-ray emission when an L-shell electron drops down, emitting a photon with energy ≈ 80 keV, or (2) Auger process where the transition energy ejects another electron instead. Both processes compete. Choice A correctly gives kinetic energy but incorrectly states only X-ray emission occurs. Choice B gives wrong kinetic energy and wrong filling mechanism. Choice C gives wrong kinetic energy.

Question 16

In gamma decay following beta-minus decay, the daughter nucleus transitions from an excited state to its ground state. If the gamma ray has energy 1.17 MeV and the daughter nucleus has mass number 60, what is the approximate recoil kinetic energy of the daughter nucleus, and why is this energy typically negligible in most applications?

  1. Approximately 11 eV, negligible compared to chemical and nuclear energy scales
  2. Approximately 110 eV, negligible because photons carry most transition energy
  3. Approximately 1100 eV, small compared to gamma energy but chemically significant
  4. Approximately 12 eV, negligible because massive nucleus requires little kinetic energy for momentum conservation (correct answer)
Explanation: Using momentum conservation: p_nucleus = p_photon = E_γ/c. The recoil kinetic energy is K = p²/(2m) = E_γ²/(2mc²). With E_γ = 1.17 MeV and m ≈ 60u ≈ 55,860 MeV/c², we get K = (1.17)²/(2 × 55,860) ≈ 12 eV. This is negligible because the massive nucleus requires very little kinetic energy to satisfy momentum conservation with the emitted photon.

Question 17

A heavy nucleus undergoes spontaneous fission, splitting into two fragments with mass numbers 95 and 138, plus several neutrons. If the original nucleus was ²³⁶U (92 protons) and the lighter fragment has 38 protons, how many neutrons are released, and what determines the kinetic energy distribution between the two fragments?

  1. 3 neutrons released; kinetic energies proportional to masses due to binding energy differences
  2. 2 neutrons released; kinetic energies equal due to identical Coulomb repulsion forces
  3. 3 neutrons released; kinetic energies inversely proportional to fragment masses due to momentum conservation (correct answer)
  4. 4 neutrons released; kinetic energies depend on specific nuclear shell configurations
Explanation: When analyzing nuclear fission problems, you need to apply two fundamental conservation laws: conservation of mass number (nucleons) and conservation of momentum. Let's track the nucleons first. The original ²³⁶U has 236 total nucleons. After fission, you have fragments with mass numbers 95 and 138. The difference is 236−95−138=3236 - 95 - 138 = 3 neutrons released. This immediately eliminates options B and D. For the kinetic energy distribution, momentum conservation is key. Initially, the uranium nucleus is at rest, so total momentum is zero. After fission, the two fragments must have equal and opposite momenta: m1v1=m2v2m_1v_1 = m_2v_2. Since kinetic energy is KE=12mv2=p22mKE = \frac{1}{2}mv^2 = \frac{p^2}{2m}, and both fragments have the same momentum magnitude, their kinetic energies are inversely proportional to their masses. The lighter fragment gets more kinetic energy than the heavier one. Option A correctly identifies 3 neutrons but incorrectly states kinetic energies are proportional to masses due to binding energy differences. While binding energy provides the total energy released, momentum conservation determines how that energy is distributed between fragments. Option B is wrong on both counts: only 2 neutrons and equal kinetic energies. Equal Coulomb forces don't produce equal kinetic energies when masses differ. Option D gives the wrong neutron count and incorrectly emphasizes shell configurations over fundamental conservation laws. Remember: In fission problems, always check nucleon conservation first, then apply momentum conservation to determine energy sharing between fragments.