This question tests understanding of radioactive decay kinetics and activity over time. Radioactive decay follows first-order kinetics, where the activity A(t) = A₀e^(-λt), showing exponential decrease over time. The decay probability per nucleus remains constant (λ), leading to exponential rather than linear decay behavior. For ⁹⁹ᵐTc with a 6.0 h half-life, the activity will decrease to half its initial value every 6.0 hours. Choice A correctly identifies the exponential decay pattern due to constant decay probability. Choices B and C incorrectly suggest linear decay or constant activity, while choice D impossibly suggests increasing activity. When evaluating decay patterns, remember that radioactive decay always follows exponential kinetics with constant decay probability per nucleus.

This question tests understanding of sequential nuclear decay processes. In ⁶⁰Co decay, β- emission occurs first, converting ⁶⁰Co to ⁶⁰Ni* (excited state) by changing a neutron to proton, thus changing the element. The excited ⁶⁰Ni* then undergoes gamma decay to ground state ⁶⁰Ni, releasing energy without changing nuclear composition. Choice A correctly describes beta emission changing the element followed by gamma emission relaxing the daughter nucleus. Choice B reverses the sequence impossibly, choice C gives incorrect mass changes, and choice D incorrectly relates emission order to particle speed. When analyzing decay chains, beta decay changes element identity while subsequent gamma decay only changes energy state.

This question tests understanding of nuclear decay and radioactivity, specifically beta-plus decay. In β+ decay, a proton in the nucleus converts to a neutron, emitting a positron (e+) and a neutrino (νe), which decreases the atomic number by 1 while the mass number remains constant. For ¹⁸F → ¹⁸O + e+ + νe, fluorine (Z=9) transforms to oxygen (Z=8), confirming the atomic number decreases by 1. Choice A correctly describes this process where a proton converts to a neutron with positron emission. Choice B incorrectly describes alpha decay, choice C describes β- decay (opposite process), and choice D describes gamma decay. When analyzing nuclear decay, verify that charge and mass numbers are conserved on both sides of the equation.

This question tests understanding of beta-minus decay and conservation laws. In β- decay of ¹³¹I to ¹³¹Xe, a neutron converts to a proton: n → p + e- + ν̄e. This increases atomic number from 53 (iodine) to 54 (xenon) while mass number remains 131, conserving baryon number. An electron is emitted to conserve charge. Choice A correctly describes mass number conservation at 131, atomic number increase by 1, and electron emission. Choice B incorrectly suggests neutron emission, choice C describes alpha decay, and choice D is physically impossible. When verifying nuclear reactions, check that mass number, charge, and baryon number are conserved on both sides of the equation.

This question tests understanding of alpha decay and its nuclear changes. Alpha particles are helium-4 nuclei (²He⁴) containing 2 protons and 2 neutrons. When emitted, the parent nucleus loses these 4 nucleons, decreasing atomic number by 2 and mass number by 4. This explains alpha particles' high mass and charge, leading to high linear energy transfer and short tissue range. Choice A correctly describes the loss of 2 protons and 2 neutrons with corresponding decreases in atomic and mass numbers. Choice B incorrectly suggests gaining nucleons, choice C describes a different process, and choice D incorrectly places alpha emission outside the nucleus. When analyzing alpha decay, remember the emitted particle is a complete helium nucleus.

This question tests understanding of electron capture (EC) as a mode of nuclear decay. In electron capture, an inner orbital electron is captured by a proton in the nucleus, converting it to a neutron: p + e- → n + νe, which decreases the atomic number by 1 while mass number stays constant. For ¹²⁵I + e- → ¹²⁵Te + νe, iodine (Z=53) captures an electron to become tellurium (Z=52), confirming the atomic number decreases by 1. Choice A correctly describes this process where a proton converts to a neutron. Choice B incorrectly describes β- decay, choice C describes alpha decay, and choice D incorrectly suggests mass number changes. When analyzing electron capture, remember it's functionally similar to β+ decay in terms of nuclear transformation but involves capturing an existing electron rather than emitting a positron.

This question tests understanding of radioactive decay over multiple half-lives. With t₁/₂ = 110 min and transport time = 220 min, exactly 2 half-lives have elapsed (220/110 = 2). After one half-life, 50% remains; after two half-lives, 25% remains. The formula is: fraction remaining = (1/2)^(t/t₁/₂) = (1/2)² = 1/4 = 25%. Choice A correctly identifies that 25% remains after two half-lives. Choice B incorrectly ignores time dependence, choice C miscalculates the fraction, and choice D incorrectly counts three half-lives. When calculating remaining activity, always determine the number of half-lives as t/t₁/₂ and apply $(1/2)^n$.

This question tests understanding of alpha decay and its effect on nuclear composition. In alpha decay, the nucleus emits an alpha particle (⁴He), which contains 2 protons and 2 neutrons, decreasing the mass number by 4 and atomic number by 2. For ²²²Rn → ²¹⁸Po + ⁴He, radon (Z=86) loses an alpha particle to become polonium (Z=84), with mass decreasing from 222 to 218. Choice A correctly describes the daughter having mass number 218 and atomic number decreased by 2. Choice B incorrectly suggests atomic number increase, choice C gives wrong mass change, and choice D incorrectly claims alpha particles carry no nucleons. When analyzing alpha decay, always subtract 4 from mass number and 2 from atomic number.

This question tests understanding of radioactive decay over multiple half-lives. After one half-life (4.0 h), N₀/2 nuclei remain; after two half-lives (8.0 h), N₀/4 remain; after three half-lives (12.0 h), N₀/8 remain. The general formula is N(t) = N₀(1/2)^(t/t₁/₂). With t = 12 h and t₁/₂ = 4.0 h, we have N(12) = N₀(1/2)³ = N₀/8. Choice A correctly identifies that three half-lives have elapsed, leaving N₀/8 nuclei. Choice B incorrectly assumes linear decay, choice C miscounts half-lives, and choice D overcounts to four half-lives. When calculating decay over time, always determine the number of half-lives elapsed as t/t₁/₂.

This question tests understanding of the relationship between half-life and decay constant. The decay constant λ = ln(2)/t₁/₂, so longer half-life means smaller decay constant. For the 30.0-y source: λ = 0.693/30.0 ≈ 0.0231 y⁻¹; for the 5.0-y source: λ = 0.693/5.0 ≈ 0.139 y⁻¹. The 30.0-y source has a smaller λ because it decays more slowly. Choice B correctly identifies that longer half-life corresponds to smaller decay constant. Choice A reverses the relationship, choice C incorrectly claims universal constants, and choice D incorrectly suggests λ=0 for long-lived isotopes. When comparing decay constants, remember that λ and t₁/₂ are inversely related.