15.7 Fission, Fusion, and Nuclear Decay

1. AThe strong force exerted by the nucleons, which here is attractive and larger than the repulsion Correct
   At nuclear scales the strong force dominates the interactions of nucleons. Between nucleons about 1 × 10⁻¹⁵ m apart it is attractive and larger than the electric repulsion between two protons, so each proton is held in the nucleus by the strong-force pull of the other nucleons.
2. BGravitational attraction between the nucleons, which is very large at such small separations
   A student who thinks gravity holds the nucleus together picks this. The gravitational force between two protons 1 × 10⁻¹⁵ m apart is Gm²/r² ≈ (6.67 × 10⁻¹¹)(1.67 × 10⁻²⁷)²/(10⁻¹⁵)² ≈ 2 × 10⁻³⁴ N, far too small to balance a repulsion of more than 200 N.
3. CThe neutrons, which cancel the charge of the protons so that the protons no longer repel
   A student who thinks neutrons neutralize the protons picks this. A neutron has no charge, so it cannot cancel a proton's charge; the two protons still repel with more than 200 N. Neutrons help hold a nucleus together only through the strong force.
4. DThe atom's electrons, whose attraction pulls inward on the protons from all sides
   A student who thinks the electrons hold the nucleus together picks this. The electrons are about 10⁻¹⁰ m away and spread around the nucleus, so their pull on each proton is tiny and does not press the protons together.

CED 15.7.A.1 · Read this in Fix

1. A88
   A student who leaves out the absorbed neutron picks this: 235 − 144 − 3 = 88. The neutron is a reactant, so the total on the left is 236.
2. B92
   A student who does not count the three free neutrons among the products picks this: 236 − 144 = 92. Each neutron has nucleon number 1, so the three neutrons account for 3 of the 236 nucleons.
3. C36
   A student who takes the lower number of the symbol as the nucleon number picks this. 36 is the proton number Z of krypton (92 = 56 + 36). The nucleon number is the upper number, found from 236 = 144 + A + 3.
4. D89 Correct
   Nucleon number is conserved. The reactants have 1 + 235 = 236 nucleons. The products have 144 + A + 3(1). So A = 236 − 144 − 3 = 89.

Working Conservation of nucleon number: 1 + 235 = 144 + A + 3 × 1, so A = 236 − 147 = 89. (Check charge: 0 + 92 = 56 + 36 + 0.) Errors: omit the absorbed neutron → 88; omit the 3 released neutrons → 92; lower number (Z) → 36.

CED 15.7.A.2 · Read this in Fix

1. A1.5 × 10⁷ m/s
   A student who thinks the two particles fly apart at equal speeds picks this. The momenta, not the speeds, are equal in magnitude, so the radon nucleus, with 55.5 times the mass, moves 55.5 times more slowly.
2. B2.7 × 10⁵ m/s Correct
   The nucleus was at rest, so the total momentum stays zero: m_α v_α = m_Rn v_Rn. v_Rn = (4.00 u/222 u)(1.5 × 10⁷ m/s) = 2.7 × 10⁵ m/s, in the direction opposite to the alpha particle.
3. C2.0 × 10⁶ m/s
   A student who gives the two particles equal kinetic energies picks this: v_Rn = v_α√(4.00/222) = 2.0 × 10⁶ m/s. With these speeds the momenta would not balance; momentum conservation gives v_Rn = (4.00/222)v_α.
4. D8.3 × 10⁸ m/s
   A student who inverts the mass ratio picks this: (222/4.00)(1.5 × 10⁷ m/s) = 8.3 × 10⁸ m/s, which is faster than light (3.00 × 10⁸ m/s). The heavier particle must be the slower one: v_Rn = (4.00/222)v_α.

Working Momentum before = 0, so m_α v_α = m_Rn v_Rn. v_Rn = (4.00/222)(1.5 × 10⁷) = 2.70 × 10⁵ m/s → 2.7 × 10⁵ m/s (2 s.f., from 1.5 × 10⁷). Errors: equal speeds → 1.5 × 10⁷; equal K: v_Rn = v_α√(4.00/222) = 2.01 × 10⁶; inverted ratio → 8.33 × 10⁸.

CED 15.7.A.3 · Read this in Fix

1. AFission and fusion change the total mass, but radioactive decay leaves the total mass unchanged.
   A student who links E = mc² with reactors, bombs and stars picks this. Mass–energy equivalence applies to every nuclear reaction: in an alpha decay, for instance, the new nucleus and the alpha particle together have less mass than the original nucleus.
2. BNone of the three processes change the total mass, since mass is conserved as in chemical reactions.
   A student who carries conservation of mass over from chemistry picks this. In nuclear processes the mass change is measurable: in deuterium–tritium fusion the products have 0.01889 u less mass than the reactants.
3. CEach of the three processes lowers the total mass by converting some of the nucleons into energy.
   A student who thinks nucleons are destroyed to make the energy picks this. The number of nucleons is the same before and after every nuclear reaction; the mass decreases because the products are more tightly bound, not because nucleons disappear.
4. DIn each of the three processes the total mass changes, while the total nucleon number stays the same. Correct
   For all nuclear reactions mass and energy may be exchanged. In each of these processes the products have less total mass than the reactants, and the decrease times c² is the energy released. The nucleon number is the same before and after.

CED 15.7.A.4 · Read this in Fix

1. AAll of it as gamma rays, since nuclear energy is given off as radiation
   A student who thinks nuclear energy is all 'radiation' picks this. Photons carry only a small part of the energy; most of it is the kinetic energy of the fragments that fly apart.
2. BAs heat produced inside the nucleus, which then flows outward and warms the fuel
   A student who thinks the nucleus produces heat directly picks this. The energy leaves as kinetic energy of particles and as photons; the fuel warms up afterward, as the fast fragments collide with atoms and are slowed down.
3. CMostly as kinetic energy of the flying fragments, with some carried by photons Correct
   Energy released in a nuclear process leaves as kinetic energy of the products or as photons. In fission, most of it (roughly 80%) is the kinetic energy of the two fragments as they fly apart; smaller amounts are carried by the neutrons' kinetic energy and by gamma-ray photons.
4. DAs extra mass of the two fragments, which end up heavier than the original nucleus
   A student who thinks more tightly bound fragments have more mass picks this. The fragments have less total mass than the uranium-236 nucleus they came from; that decrease in mass is what makes the energy available.

CED 15.7.A.5 · Read this in Fix

1. ARow 1
   A student who has exchanged the names fission and fusion picks this. In this reaction a uranium nucleus absorbs a neutron and splits into barium and krypton nuclei: that is fission, not fusion.
2. BRow 2 Correct
   Two hydrogen-2 nuclei (smaller nuclei) combine to form a larger nucleus, helium-3, and a neutron (a subatomic particle). That is fusion: smaller nuclei combining into a larger nucleus. Nucleon number (4 = 3 + 1) and charge (2 = 2 + 0) balance.
3. CRow 3
   A student who thinks any reaction producing helium is fusion picks this. Here a single uranium-238 nucleus breaks apart, emitting a helium-4 nucleus: that is alpha decay. In fusion, smaller nuclei combine into a larger one.
4. DRow 4
   A student who counts any joining of particles as fusion picks this. Hydrogen and oxygen molecules reacting to form water is a chemical reaction: the atoms rearrange, but no nucleus changes.

Working Fusion: smaller nuclei combine into a larger nucleus: ²₁H + ²₁H → ³₂He + ¹₀n (nucleon number 2 + 2 = 3 + 1; charge 1 + 1 = 2 + 0). The other rows: neutron-induced fission of uranium-235 (a nucleus splits); alpha decay of uranium-238 (one nucleus breaks up, emitting ⁴₂He); hydrogen burning in oxygen (chemical: no nucleus changes).

CED 15.7.A.6 · Read this in Fix

1. APlutonium-239, after taking in a neutron, breaks up into two medium-mass nuclei. Correct
   Fission is the splitting of a nucleus into two or more smaller nuclei, as well as subatomic particles. Here the plutonium nucleus splits into two smaller, medium-mass nuclei (and releases two or three neutrons).
2. BTwo hydrogen-2 nuclei join to form a helium-3 nucleus, releasing a neutron and energy.
   A student who has exchanged the names fission and fusion picks this. Two small nuclei combining into a larger one is fusion.
3. CA uranium-238 nucleus absorbs a neutron and becomes a heavier uranium-239 nucleus.
   A student who thinks any neutron absorption by uranium is fission picks this. The nucleus does not split; one larger nucleus, uranium-239, is formed. Fission requires splitting into smaller nuclei.
4. DA water molecule splits apart into two separate hydrogen atoms and a single oxygen atom.
   A student who counts any splitting as fission picks this. Splitting a molecule into atoms is a chemical process; every nucleus is unchanged. Fission splits a nucleus.

Working Fission: one nucleus splits into two or more smaller nuclei (plutonium-239 + n → two medium-mass nuclei + neutrons). ²H + ²H → ³He + n is fusion; ²³⁸U + n → ²³⁹U is neutron absorption with no split; H₂O → 2H + O is chemical, with no nucleus changed.

CED 15.7.A.7 · Read this in Fix

1. AFission of X would need an energy input, since X's fragments have a greater total binding energy than X.
   A student who thinks binding energy is energy stored in a nucleus picks this, reasoning that the fragments must be given the extra 180 MeV. More binding energy means the fragments are more tightly bound and have less mass than X, so fission of X releases about 180 MeV.
2. BNeither fission would need an energy input, since splitting a nucleus releases energy.
   A student who thinks every fission releases energy picks this. The chart shows Y's fragments with less total binding energy than Y (321 MeV compared with 342 MeV), so they have more mass than Y and energy must be supplied to form them.
3. CBoth fissions would need an energy input, since energy is needed to break a nucleus apart.
   A student who thinks splitting anything always costs energy picks this. Fission forms smaller nuclei, not separate nucleons. X's fragments have more total binding energy than X, so they have less mass and the fission of X releases energy.
4. DFission of Y would need an energy input, since Y's fragments have less total binding energy than Y. Correct
   Nucleon number is conserved, so fragments with less total binding energy (321 MeV compared with 342 MeV for Y) are less tightly bound and have more total mass than Y. Their extra rest energy, about 21 MeV, must be supplied, so the fission of Y needs an energy input. X's fragments have more binding energy than X (1970 MeV compared with 1790 MeV), so fission of X releases energy.

CED 15.7.A.8 · Read this in Fix

1. AA uranium-235 nucleus splits into two smaller nuclei after absorbing a neutron fired at it.
   A student who counts any nuclear change as decay picks this. This fission is triggered by the absorbed neutron, so it is not spontaneous; radioactive decay happens without an outside cause.
2. BAn excited nucleus emits a gamma-ray photon and drops to a lower energy level. Correct
   In radioactive decay, a nucleus spontaneously changes into one or more different nuclei, or drops to a lower energy level of the same nucleus. An excited nucleus emitting a gamma-ray photon without any outside cause is the second kind.
3. CAn excited atom emits a photon when one of its electrons drops to a lower energy level.
   A student who thinks any emission of radiation is radioactive decay picks this. The photon comes from a change in the atom's electron energy levels; the nucleus is unchanged, so this is not radioactive decay.
4. DA hydrogen peroxide molecule breaks down into molecules of water and of oxygen.
   A student who links radioactive decay with things breaking down picks this. The breakdown of hydrogen peroxide is a chemical reaction: the atoms rearrange, but no nucleus changes.

CED 15.7.B.1 · Read this in Fix

1. A(ln 2/t₁/₂)ln(N₀/N)
   A student who multiplies by λ instead of dividing when solving ln(N/N₀) = −λt picks this: it is λ ln(N₀/N). Its units are 1/time, not time; t = ln(N₀/N)/λ.
2. Bln(N₀/N)/(1/t₁/₂)
   A student who substitutes λ = 1/t₁/₂ into t = ln(N₀/N)/λ picks this. The relation is λ = (ln 2)/t₁/₂, so t = t₁/₂ ln(N₀/N)/ln 2. Check: this option gives t₁/₂ ln 2 = 0.69t₁/₂, not t₁/₂, when N = N₀/2.
3. Ct₁/₂ ln(N₀/N)/ln 2 Correct
   From ln(N/N₀) = −λt, t = ln(N₀/N)/λ. With λ = (ln 2)/t₁/₂, t = t₁/₂ ln(N₀/N)/ln 2. Check: when N = N₀/2, ln(N₀/N) = ln 2 and t = t₁/₂.
4. D2t₁/₂(N₀ − N)/N₀
   A student who thinks nuclei decay at a steady rate, N₀/2 in each half-life, picks this. The number decaying per unit time falls as N falls; for N = N₀/4, for example, this option gives 1.5t₁/₂, but two half-lives have passed.

Working ln(N/N₀) = −λt → t = ln(N₀/N)/λ; λ = ln 2/t₁/₂ → t = t₁/₂ ln(N₀/N)/ln 2. Checks at N = N₀/2: key → t₁/₂; λ ln(N₀/N) → (ln 2)²/t₁/₂ (units 1/time); λ = 1/t₁/₂ gives ln 2/(1/t₁/₂) = 0.69t₁/₂; linear → t₁/₂ (but at N₀/4 gives 1.5t₁/₂ instead of 2t₁/₂).

CED 15.7.B.2 · Read this in Fix

1. AIt is smaller by a factor of 10⁴, the same factor by which the electric repulsion falls
   A student who assumes the strong force follows an inverse-square law, like the electric force, picks this. If it did, the strong force would stay larger than the electric repulsion at every separation and would bind protons together in ordinary matter. It is exerted at nuclear scales only and falls to a negligible size.
2. BIt is negligible, far smaller than their electric repulsion at the larger separation Correct
   The strong force is exerted only at nuclear scales. At 1 × 10⁻¹³ m, about a hundred nuclear diameters, it is negligible, while the electric repulsion, which falls only as 1/r², is still about 2 × 10⁻² N. That is why protons in separate nuclei, even in the same molecule, do not stick together.
3. CIt is equal, since the strong force does not depend on how far apart they are
   A student who pictures the strong force as a fixed 'glue' belonging to nucleons picks this. The strong force depends strongly on separation: it is large at nuclear scales and negligible at 100 times that distance.
4. DIt is larger, as the pull of a rubber band grows when the band is stretched
   A student who applies the rubber-band picture of the force between quarks to nucleons picks this. Between two nucleons the strong force is exerted at nuclear scales and becomes negligible when they are pulled apart.

CED 15.7.A.1 · Read this in Fix

1. AIt is correct; the energy released is the rest energy of the nucleons that disappear.
   A student who thinks 'mass converted into energy' means particles vanish picks this. The nucleon numbers balance in every nuclear reaction; in ²₁H + ³₁H → ⁴₂He + ¹₀n there are 5 nucleons on each side. The mass lost (0.019 u) is a small fraction of one nucleon's mass.
2. BIt is incorrect; mass cannot change, so the energy comes from the nuclei's motion before fusing.
   A student who applies conservation of mass from chemistry picks this. In nuclear reactions the total mass does change measurably, and the energy released, far more than the kinetic energy the nuclei had before they fused, equals the decrease in mass times c².
3. CIt is incorrect; the number of nucleons is unchanged, and the products have less mass. Correct
   Nucleon number is conserved in every nuclear reaction, so no protons or neutrons disappear. The products of the fusion are more tightly bound and have slightly less total mass than the reactants; this decrease in mass, times c², is the energy released.
4. DIt is incorrect; the energy is binding energy that was stored in the nuclei before they fused.
   A student who thinks binding energy is energy stored in a nucleus picks this. Binding energy is the energy needed to pull a nucleus apart. The product of fusion has more binding energy than the reactants, which is why it has less mass and energy is released.

CED 15.7.A.2 · Read this in Fix

1. AThey are equal, since the energy released is shared out equally between them.
   A student who thinks the energy is shared equally picks this. Conservation of momentum fixes how the energy is divided: with equal momenta, K = p²/(2m) gives the lighter alpha particle about 50 times the kinetic energy of the new nucleus.
2. BThe new nucleus's is much greater, since the two move apart at equal speeds.
   A student who thinks the particles move apart at equal speeds picks this, then uses K = ½mv² to give the heavier nucleus more energy. Equal speeds would give the new nucleus 50 times the momentum of the alpha particle, so momentum would not be conserved.
3. CTheir total is zero, since the total kinetic energy before the decay was zero.
   A student who treats kinetic energy like momentum, as able to cancel, picks this. Kinetic energy is a scalar and cannot be negative; the products share the energy released, so their total kinetic energy is positive. It is the total momentum that stays zero.
4. DThe alpha particle's is much greater, as their momenta are equal in size. Correct
   The total momentum stays zero, so the two momenta have equal magnitudes p. Kinetic energy K = p²/(2m), so with equal p the particle with about 1/50 of the mass has about 50 times the kinetic energy.

CED 15.7.A.3 · Read this in Fix

1. A2.82 × 10⁻¹² J Correct
   Mass of reactants: 2.01410 u + 3.01605 u = 5.03015 u. Mass of products: 4.00260 u + 1.00866 u = 5.01126 u. Δm = 0.01889 u = 0.01889 × 1.66 × 10⁻²⁷ kg = 3.136 × 10⁻²⁹ kg. E = Δmc² = 3.136 × 10⁻²⁹ kg × (3.00 × 10⁸ m/s)² = 2.82 × 10⁻¹² J (about 17.6 MeV).
2. B9.41 × 10⁻²¹ J
   A student who multiplies the mass decrease by c instead of c² picks this: 3.136 × 10⁻²⁹ kg × 3.00 × 10⁸ m/s = 9.41 × 10⁻²¹. That product has units kg·m/s, not J; E = Δmc² gives 2.82 × 10⁻¹² J.
3. C7.52 × 10⁻¹⁰ J
   A student who thinks the whole mass of the reactants is converted picks this: 5.03015 u × 1.66 × 10⁻²⁷ kg/u × c² = 7.52 × 10⁻¹⁰ J. Only the decrease in mass, 0.01889 u, corresponds to the energy released.
4. D1.54 × 10⁻¹⁰ J
   A student who leaves the neutron out of the products picks this: Δm = 5.03015 u − 4.00260 u = 1.02755 u, giving 1.54 × 10⁻¹⁰ J. The neutron's mass, 1.00866 u, is part of the products' mass and must be subtracted too.

Working Δm = (2.01410 + 3.01605) − (4.00260 + 1.00866) = 0.01889 u = 3.136 × 10⁻²⁹ kg. E = Δmc² = 3.136 × 10⁻²⁹ × 9.00 × 10¹⁶ = 2.82 × 10⁻¹² J (≈ 17.6 MeV; 3 s.f. from 0.01889 u and the constants). Errors: Δm·c → 9.41 × 10⁻²¹; all reactant mass → 7.52 × 10⁻¹⁰; neutron omitted → 1.54 × 10⁻¹⁰.

CED 15.7.A.4 · Read this in Fix

1. A(M − m₁ − m₂)c
   A student who multiplies the mass decrease by c instead of c² picks this. The units would be kg·m/s, a momentum; an energy needs kg·m²/s², so the mass must be multiplied by c².
2. B(M − m₁ − m₂)/c²
   A student who divides by c², the step for converting an energy into a mass, picks this. Its units are kg·s²/m², not J. A mass is turned into its energy equivalent by multiplying by c².
3. C(M − m₁ − m₂)c² Correct
   Energy is conserved, counting rest energy. Before: Mc². After: m₁c² + m₂c² + K_total. So K_total = (M − m₁ − m₂)c², positive because the fragments have less total mass than the original nucleus.
4. D(m₁ + m₂)c² − Mc²
   A student who thinks the more tightly bound fragments have more mass picks this. For energy to be released, the fragments must have less total mass than the original nucleus, so this expression would be negative. The kinetic energy is (M − m₁ − m₂)c².

Working Conservation of energy (rest energy included), nucleus initially at rest: Mc² = m₁c² + m₂c² + K₁ + K₂, so K₁ + K₂ = (M − m₁ − m₂)c². Units kg·m²/s² = J. Errors: ×c → units kg·m/s; ÷c² → kg·s²/m²; products-minus-parent → negative.

CED 15.7.A.4 · Read this in Fix

1. AIt will decay exactly 10 minutes after it was formed, as every nucleus of the isotope does.
   A student who treats the half-life as the lifetime of each nucleus picks this. Nuclei of the same isotope decay at unpredictable times; the half-life only describes how long it takes half of a large number of them to decay.
2. BIt becomes more likely to decay in the next minute the longer it has already survived.
   A student who thinks nuclei age, or that a decay becomes 'due', picks this. A nucleus has the same probability of decaying per unit time however long it has existed.
3. CIt has a 50% chance of decaying in the next 10 minutes, but when it will decay cannot be predicted. Correct
   The time at which an individual nucleus decays is indeterminable, but its decay can be described by probability. Each undecayed nucleus has the same chance of decaying in any 10-minute interval, one-half, because that is what makes half of a large sample decay in one half-life.
4. DIt is certain to have decayed within 20 minutes, when the whole sample will have decayed.
   A student who thinks the sample is used up at a steady rate picks this. After 20 minutes, two half-lives, one-quarter of the nuclei remain undecayed, and any particular nucleus has a 25% chance of being one of them.

Working Each undecayed nucleus has the same probability of decaying per unit time, whatever its age. Probability of decaying within one half-life (10 min) = 1 − (1/2)¹ = 0.50; the moment of decay is not predictable. After 20 min (2 half-lives) a fraction (1/2)² = 0.25 remains, so a given nucleus is not certain to have decayed.

CED 15.7.B.1.i · Read this in Fix

1. AThe counts vary irregularly about a steady average, as expected when decays occur at random. Correct
   The counts range from 41 to 58 with no upward or downward trend. In a few minutes a sample with a 30-year half-life loses a negligible fraction of its nuclei, so the average rate is steady; the irregular variation is what random decay predicts, since the number of nuclei that happen to decay in each interval varies by chance.
2. BThe detector must be faulty, since every radioactive sample decays at one fixed, steady rate.
   A student who expects exactly equal counts in equal intervals picks this. Each decay is a random event, so counts of about 50 per interval are expected to vary by several counts either way even with a perfect detector.
3. CThe counts vary because the half-life of the sample changes as its number of nuclei decreases.
   A student who thinks the half-life depends on the amount of material picks this. The half-life is fixed for the isotope, and in a few minutes the number of nuclei in a sample with a 30-year half-life hardly changes.
4. DThe counts vary because the temperature of the room changed during the measurements.
   A student who thinks conditions such as temperature change the decay rate picks this. Everyday temperature changes affect atoms, not nuclei, and nothing in the data points to a temperature change; the variation is the random scatter of decay.

CED 15.7.B.1.i · Read this in Fix

1. A16 days, since twice as many nuclei have to decay before half are gone
   A student who thinks a larger sample takes longer to lose half its nuclei picks this. Twice as many nuclei give twice as many decays per second, so half of them are still gone after 8.0 days.
2. B8.0 days, since each nucleus has the same chance of decaying in a given time Correct
   The half-life depends only on the isotope. Each iodine-131 nucleus has the same probability of decaying per unit time, so in 8.0 days half of the nuclei decay whatever the number in the sample. The larger sample gives twice as many decays per second.
3. C4.0 days, since twice as many nuclei give twice as many decays per second
   A student who equates more decays per second with a shorter half-life picks this. The decays per second double because the number of nuclei doubles, not because each nucleus decays sooner; the half-life stays 8.0 days.
4. DIt cannot be predicted, since the decay of each individual nucleus is a random event
   A student who thinks randomness makes the whole sample unpredictable picks this. The time of each decay is random, but for a large number of nuclei the half-life is a fixed property of the isotope: 8.0 days.

Working Half-life is a property of the isotope, independent of N₀: the second sample's half-life is 8.0 days. Errors: proportional to amount → 2 × 8.0 = 16 days; faster because more decays per second → 8.0/2 = 4.0 days.

CED 15.7.B.1.ii · Read this in Fix

1. A0.067 min⁻¹
   A student who takes λ as 1/t₁/₂, by analogy with f = 1/T, picks this: 1/15 min = 0.067 min⁻¹. The relation is λ = (ln 2)/t₁/₂, which includes the factor ln 2 = 0.693.
2. B0.033 min⁻¹
   A student who treats the decay as steady picks this: if half the sample decays in 15 min at a steady rate, the fraction lost per minute is 0.5/15 = 0.033 min⁻¹. The graph is a curve, not a straight line, and λ = (ln 2)/t₁/₂ = 0.046 min⁻¹.
3. C0.020 min⁻¹
   A student who uses the common logarithm picks this: log 2/15 min = 0.301/15 = 0.020 min⁻¹. The relation uses the natural logarithm, ln 2 = 0.693.
4. D0.046 min⁻¹ Correct
   From the graph, N falls from 1600 to 800 in 15 min, from 800 to 400 in the next 15 min, and so on, so t₁/₂ = 15 min. λ = (ln 2)/t₁/₂ = 0.693/15 min = 0.046 min⁻¹.

Working Read t₁/₂: N halves every 15 min (1600 → 800 → 400 → 200 → 100 at 0, 15, 30, 45, 60 min). λ = ln 2/15 = 0.0462 min⁻¹ → 0.046 min⁻¹. Errors: 1/15 = 0.067; linear 0.5/15 = 0.033; log₁₀ 2/15 = 0.020.

CED 15.7.B.1.iii · Read this in Fix

1. ANe^λt Correct
   From N = N₀e^−λt, N₀ = N/e^−λt = Ne^λt. The exponent is positive, so N₀ is larger than N, as it must be.
2. BNe^−λt
   A student who applies the decay equation with N and N₀ exchanged picks this. It gives a value smaller than N, but the sample had more nuclei when it was prepared; solving N = N₀e^−λt for N₀ gives Ne^λt.
3. CN/(1 − λt)
   A student who uses a steady-rate model, N = N₀(1 − λt), picks this. Decay is exponential, not linear: the number decaying per unit time falls as the sample shrinks, so N = N₀e^−λt.
4. DN·2^t/λ
   A student who treats λ as if it were the half-life picks this, doubling N once for every 'λ' of elapsed time. λ is a probability per unit time (units s⁻¹), not a time; the half-life is (ln 2)/λ.

Working N = N₀e^−λt → N₀ = Ne^λt. Errors: roles swapped → Ne^−λt < N; linear model N = N₀(1 − λt) → N/(1 − λt); λ used as a half-life → N·2^t/λ.

CED 15.7.B.2 · Read this in Fix

1. A35 days
   A student who uses a steady-rate model, N = N₀(1 − λt), picks this: t = (1 − 0.30)/0.020 day⁻¹ = 35 days. Decay is exponential: use ln(N/N₀) = −λt, which gives 60 days.
2. B60 days Correct
   ln(0.30) = −1.20. The line reaches ln(N/N₀) = −1.20 at t = 60 days. (Equivalently, the slope gives λ = 2.0/100 days = 0.020 day⁻¹, and t = −ln(0.30)/λ = 1.20/0.020 day⁻¹ = 60 days.)
3. C26 days
   A student who uses the common logarithm picks this: log(0.30) = −0.52, which the line reaches at 26 days. The graph is of the natural logarithm; ln(0.30) = −1.20, reached at 60 days.
4. D15 days
   A student who reads the vertical axis as N/N₀ itself picks this, finding where the line reaches −0.30. The axis shows ln(N/N₀), so the ratio 0.30 must first be converted: ln(0.30) = −1.20, reached at 60 days.

Working Slope = −2.0/100 days → λ = 0.020 day⁻¹. ln(0.30) = −1.204 → t = 1.204/0.020 = 60.2 days ≈ 60 days (read directly from the line at −1.20). Errors: linear (1 − 0.30)/0.020 = 35; log₁₀: 0.523/0.020 = 26; reading −0.30 on the axis: 0.30/0.020 = 15.

CED 15.7.B.2 · Read this in Fix

1. AAfter 9.0 × 10⁹ years, no uranium-238 nuclei remain, just as no iodine-131 nuclei remain after 16 days.
   A student who thinks a sample is used up in two half-lives picks this. 9.0 × 10⁹ years is two half-lives of uranium-238 and 16 days is two half-lives of iodine-131; after two half-lives one-quarter of the nuclei remain undecayed.
2. BEvery iodine-131 nucleus lasts exactly 8.0 days, and every carbon-14 nucleus lasts exactly 5730 years.
   A student who takes the half-life to be the lifetime of each nucleus picks this. The time at which one nucleus decays cannot be predicted; in one half-life, half of a large number of nuclei decay, some soon after the start and some much later.
3. CAfter 1 year, under 10⁻¹² of iodine-131 nuclei remain, but over 99.9% of carbon-14 nuclei do. Correct
   One year is 365/8.0 ≈ 46 half-lives of iodine-131, so the fraction remaining is about (1/2)⁴⁶ ≈ 1 × 10⁻¹⁴, under 10⁻¹². One year is 1/5730 of carbon-14's half-life, so the fraction remaining is (1/2)^1/5730 ≈ 0.99988, over 99.9%. Half-lives of different isotopes differ enormously.
4. DA sample with half as many carbon-14 or iodine-131 nuclei would have half the half-life shown.
   A student who thinks the half-life depends on the amount of material picks this. The half-life is a property of the isotope: 5730 years for any sample of carbon-14 and 8.0 days for any sample of iodine-131. Half as many nuclei give half as many decays per second, not a shorter half-life.

Working Iodine-131: 365/8.0 = 45.6 half-lives → (1/2)^45.6 = 1.8 × 10⁻¹⁴ remaining (< 10⁻¹²). Carbon-14: 1/5730 of a half-life → 2^−1/5730 = 0.99988 remaining (> 99.9%). Uranium-238: 9.0 × 10⁹ y = 2 half-lives → 1/4 remains; iodine-131 after 16 days = 2 half-lives → 1/4 remains.

CED 15.7.B.3 · Read this in Fix

1. A0.25 Correct
   λ = (ln 2)/t₁/₂, so doubling λ halves the half-life: Y's half-life is half of X's. In a time equal to X's half-life, Y goes through two of its half-lives, so (1/2)² = 0.25 of the Y nuclei remain.
2. B0.75
   A student who gives the fraction that has decayed picks this. After two half-lives of Y, 0.75 of the Y nuclei have decayed and 0.25 remain.
3. C0.71
   A student who thinks a larger decay constant means a longer half-life picks this, making Y's half-life twice X's; Y would then have gone through half a half-life, leaving (1/2)^1/2 = 0.71. Doubling λ halves the half-life.
4. D0.00
   A student who treats decay as steady picks this: X loses half its nuclei in its half-life, so Y, decaying twice as fast, would lose all of them. Decay is exponential: Y loses half of its remaining nuclei in each of its half-lives, leaving 0.25.

Working t₁/₂Y = ln 2/λ_Y = ln 2/(2λ_X) = t₁/₂X/2. Time t = t₁/₂X = 2t₁/₂Y → N/N₀ = (1/2)² = 0.25. Errors: decayed fraction 0.75; half-life doubled → 2^−1/2 = 0.71; linear → 1 − 2 × 0.5 = 0.

CED 15.7.B.1.iii · Read this in Fix

1. A½mv²(M+m)/M Correct
   The total momentum is zero before the decay, so afterward MV = mv and the new nucleus recoils with speed V = mv/M. The energy released is the total kinetic energy: ½mv² + ½M(mv/M)² = ½mv²(1 + m/M) = ½mv²(M+m)/M, a little more than the alpha particle's own kinetic energy.
2. B½(M+m)v²
   A student who thinks the two particles fly apart with equal speeds picks this, giving the new nucleus speed v as well. Their momenta must be equal in magnitude, so the much more massive new nucleus moves far more slowly, at mv/M.
3. Cmv²
   A student who thinks the energy is shared equally picks this, giving the new nucleus the same kinetic energy as the alpha particle, ½mv². Because the momenta are equal in magnitude, the kinetic energies are in the inverse ratio of the masses, so the new nucleus has only m/M of the alpha particle's kinetic energy.
4. D½mv²
   A student who treats the new nucleus as staying at rest picks this, counting only the alpha particle's kinetic energy. The total momentum is zero after the decay as before, so the new nucleus recoils with momentum equal in magnitude to the alpha particle's and carries some of the energy released.

Working Momentum: 0 = mv − MV, so V = mv/M. Energy released Q = ½mv² + ½MV² = ½mv² + ½m²v²/M = ½mv²(M+m)/M. Check (m = 4.00 u, M = 222 u, v = 1.52 × 10⁷ m/s): Q = 7.81 × 10⁻¹³ J, and momentum and energy both balance. Errors: equal speeds → ½(M+m)v²; equal kinetic energies (½mv² each) → mv²; new nucleus at rest → ½mv².

CED 15.7.A.3 · Read this in Fix

1. A(1/λ) ln(1 + r) Correct
   If P parent nuclei remain, there are D = rP daughter nuclei, and each came from one parent nucleus that decayed. So the rock started with N₀ = P + rP = P(1 + r) parent nuclei and has N = P now. From ln(N/N₀) = −λt, t = (1/λ)ln(N₀/N) = (1/λ) ln(1 + r).
2. B(1/λ)ln(1 + r)/ln 2
   A student who counts half-lives, n = ln(1 + r)/ln 2 (since N₀/N = 2ⁿ = 1 + r), and then takes the half-life to be 1/λ picks this. The half-life is t₁/₂ = (ln 2)/λ, so t = n t₁/₂ = (1/λ) ln(1 + r).
3. C(1/λ) ln(1 + 1/r)
   A student who uses the fraction that has decayed, D/(P + D) = r/(1 + r), as the fraction remaining, N/N₀, picks this. The fraction remaining is P/(P + D) = 1/(1 + r), so N₀/N = 1 + r.
4. D2r ln 2/(λ(1 + r))
   A student who thinks nuclei decay at a steady rate, half in one half-life t₁/₂ = (ln 2)/λ and the rest in the next, sets the fraction decayed, r/(1 + r), equal to t/(2t₁/₂) and picks this. The number of decays per unit time falls as the number of parent nuclei falls, so N = N₀e^−λt.

Working If P parent nuclei remain and D = rP daughter nuclei are present, each daughter came from one decayed parent, so N₀ = P + D = P(1 + r) and N = P. ln(N/N₀) = −λt ⇒ t = (1/λ)ln(N₀/N) = (1/λ) ln(1 + r). Errors: counting half-lives n = ln(1 + r)/ln 2 and taking t₁/₂ = 1/λ → (1/λ)ln(1 + r)/ln 2; fraction decayed D/(P + D) = r/(1 + r) used as N/N₀ → (1/λ) ln(1 + 1/r); steady decay rate (straight-line fall, half gone in t₁/₂ = (ln 2)/λ, all gone in 2t₁/₂): D/(P + D) = t/(2t₁/₂) ⇒ t = 2r ln 2/(λ(1 + r)).

CED 15.7.B.2 · Read this in Fix

1. A1.7 g
   A student who takes the fraction found, e^−λt = 0.21, to be the fraction that has decayed picks this. e^−λt = N/N₀ is the fraction remaining, so 1 − 0.21 = 0.79 of the sample has decayed: 6.3 g.
2. B6.3 g Correct
   λ = (ln 2)/t₁/₂ = 0.693/(5.3 yr) = 0.131 yr⁻¹, so the fraction remaining is e^{−(0.131 yr⁻¹)(12 yr)} = 0.21. The mass remaining is 8.0 g × 0.21 = 1.7 g, so 8.0 g − 1.7 g = 6.3 g has decayed.
3. C7.2 g
   A student who uses λ = 1/t₁/₂ = 0.189 yr⁻¹ picks this: e^−12/5.3 = 0.10 remains. The decay constant is λ = (ln 2)/t₁/₂ = 0.131 yr⁻¹, which leaves 0.21 of the sample.
4. D4.0 g
   A student who uses the log key, log 2 = 0.301, in place of ln 2 = 0.693 picks this: λ = 0.057 yr⁻¹ leaves 0.51 of the sample. The decay equations use natural logarithms: λ = (ln 2)/t₁/₂.

Working λ = (ln 2)/t₁/₂ = 0.693/(5.3 yr) = 0.131 yr⁻¹. Fraction remaining: e^−λt = e^{−(0.131)(12)} = e^−1.57 = 0.208. Mass remaining 8.0 g × 0.208 = 1.67 g, so mass decayed = 8.0 g − 1.67 g = 6.33 g ≈ 6.3 g. (0.208 taken as the fraction decayed: 1.7 g; λ = 1/t₁/₂: 7.2 g; log instead of ln: 4.0 g.)

CED 15.7.B.2 · Read this in Fix