IB Physics E.3: Radioactive decay worked solutions

The lower number is Z=27 protons; the upper number is the nucleon number A=60; neutrons N=A-Z=33. B swaps protons and neutrons; C and D treat 60 as the neutron or proton count and add the two numbers.

Isotopes have the same proton number, so the same electron arrangement and the same chemistry, but different neutron numbers, so different masses. A is wrong: isotopes of one element usually have very different half-lives (or are stable). C describes nuclei of different elements with equal A. D says nothing about the proton number.

The unified mass unit u, the \mathrm{MeV\,c^{-2}} and the kilogram are all mass units; these are three ways of writing the electron's rest mass. The MeV on its own is an energy: 0.511\,\mathrm{MeV} is the electron's rest energy, mc^2.

The binding-energy-per-nucleon curve peaks near A=60; iron-56 sits at the top at about 8.8\,\mathrm{MeV} per nucleon. Helium-4 (7.1\,\mathrm{MeV}) is a high point among the light nuclei but well below the peak; lead and uranium lie on the slow decline beyond it.

Total binding energy =A\times (binding energy per nucleon) =90\times8.71=784\,\mathrm{MeV}. A is the value for one nucleon; B multiplies by the proton number 40 and C by the neutron number 50.

Alpha particles ionize so strongly that they lose all their energy in a few centimetres of air. Gamma rays are the weakest ionizers (A); beta-minus particles are negative and alpha particles positive, so they bend opposite ways (B); gamma rays travel at c, faster than any beta particle (D).

Each nucleus has the same small chance of decaying per second, so more nuclei give more decays per second: B. A fails because decay is spontaneous; temperature, pressure and chemical bonding leave the half-life alone. C and D fail because decay is random: a nucleus has no memory, and only a probability can be given for any one nucleus.

In beta-plus decay a proton becomes a neutron: A stays 18, Z falls from 9 to 8 (oxygen), and the positron is accompanied by a neutrino. The antineutrino goes with beta-minus decay (B, D); C and D move Z the beta-minus way.

24 days is 3 half-lives, so (\tfrac12)^3=\tfrac18 remains and 1-\tfrac18=\tfrac78 has decayed. A is the fraction remaining; B and C wrongly treat the decay as linear.

At about 1.5\,\mathrm{fm} the strong force attracts any two nucleons, proton or neutron, and outweighs the electric repulsion of about 100\,\mathrm{N}. Gravity is some 10^{36} times too weak (A); the strong force acts between all nucleons (B); two positive charges always repel (C).

\lambda=\ln2/T_{1/2}=0.693/(15.0\times3600\,\mathrm{s})=1.28\times10^{-5}\,\mathrm{s^{-1}}. Since \lambda\Delta t\ll1, the probability is \lambda\Delta t=1.3\times10^{-5}. B leaves out \ln2; C uses hours for a one-second interval; D is 1/\lambda, a time, and no probability exceeds 1.

Carbon-11 lies below the band of stability (too few neutrons for its protons). Beta-plus decay turns a proton into a neutron, raising N/Z: Z falls to 5 (boron) with A=11 unchanged. A would make the neutron shortage worse; alpha decay is for heavy nuclei; D moves Z the wrong way for beta-plus.

(a) Alpha particles are stopped by a few centimetres of air or a sheet of paper, so none would reach the road surface (A1). (b) Sodium-24 (A1). It emits gamma rays, which penetrate 1.5\,\mathrm{m} of soil and can be detected at the surface (A1). Its half-life of 15 hours leaves about a third of the activity after one day, enough to trace the leak, while after a week (11 half-lives) less than 0.1\% remains; iodine-131 and cobalt-60 would stay active for weeks or years (A1). (c) 75 hours is 5 half-lives: (\tfrac12)^5=\tfrac1{32} (A1).

Both paths start at the parent and end at the daughter's ground state, so the photon carries the difference: 4.88-4.69=0.19\,\mathrm{MeV}. Fixed alpha and gamma energies mean the nucleus has discrete energy levels. Atomic levels are separated by electronvolts, not 0.19\,\mathrm{MeV} (B); C adds instead of subtracting; D gives the whole alpha energy to the photon.

\ln A=\ln A_0-\lambda t, so the gradient is -\lambda: \lambda=0.0866\,\mathrm{h^{-1}} and T_{1/2}=\ln2/\lambda=0.693/0.0866=8.0\,\mathrm{h}. A multiplies by \ln2; B takes the reciprocal of the gradient without \ln2; D doubles the half-life.

An artificial transmutation needs an incoming particle on the left: here an alpha particle strikes aluminium-27 and makes a new element, phosphorus-30. A, B and D have a single unstable particle on the left: they are spontaneous decays, even though A and B also change the element.

In 6.0\,\mathrm{h} X goes through 3 half-lives (factor \tfrac18) and Y through 2 (factor \tfrac14). Ratio =\tfrac18\div\tfrac14=\tfrac12. A and B are the ratio of the half-lives, not of the activities; C inverts the answer.

(a) R falls from 480\,\mathrm{s^{-1}} to 240\,\mathrm{s^{-1}} in 2.5\,\mathrm{min}, and from 240 to 120\,\mathrm{s^{-1}} in the next 2.5\,\mathrm{min}: two halvings read and checked (M1). T_{1/2}=2.5\times60=150\,\mathrm{s}, converted to seconds (A1). (b) The number of nuclei decaying per second is proportional to the number left, so the curve takes the same time to halve from any starting value (A1). (c) 12.5\,\mathrm{min} is 5 half-lives: 480/2^5=15\,\mathrm{s^{-1}} (A1).

(a) 925-25=900 counts per minute (A1). (b) 9.0\,\mathrm{h} is 3 half-lives, so the source contributes 900/8=112.5 counts per minute; halve the net rate, not the total (M1). The detector records 112.5+25\approx138 counts per minute (A1). Halving 925 three times would give 116, which is wrong. (c) The source's contribution falls towards zero, but the background remains, so the reading tends to about 25 counts per minute, not zero (A1).

(a) {}^{60}_{27}\mathrm{Co}\to{}^{60}_{28}\mathrm{Ni}+{}^{\;\;0}_{-1}\mathrm{e}+\bar{\nu}, with the antineutrino (A1). (b) The parent atom carries 27 electrons and the daughter atom 28; the extra electron counted in the nickel atomic mass stands in for the emitted beta particle, so the electron masses balance (A1). (c) \Delta m=59.933\,817-59.930\,786=0.003\,031\,\mathrm{u} (M1); E=0.003\,031\times931.5=2.82\,\mathrm{MeV} (A1). (d) Number of nuclei =\dfrac{1.0\times10^{-3}}{59.93}\times6.02\times10^{23}=1.0\times10^{19} (M1); energy =1.0\times10^{19}\times2.82\times1.60\times10^{-13}=4.5\times10^{6}\,\mathrm{J} (A1).

(a) \lambda=\ln2/(8.02\times86\,400)=1.00\times10^{-6}\,\mathrm{s^{-1}} (A1). (b) N=A/\lambda=3.7\times10^{8}/1.00\times10^{-6}=3.7\times10^{14} (A1). (c) m=\dfrac{3.7\times10^{14}}{6.02\times10^{23}}\times131\,\mathrm{g}=8.0\times10^{-8}\,\mathrm{g}=0.080\,\mathrm{\mu g} (A1): a tiny mass gives a large activity. (d) \lambda t=1.00\times10^{-6}\times2.5\times86\,400=0.216 (M1); A=3.7\times10^{8}\,e^{-0.216}=3.0\times10^{8}\,\mathrm{Bq} (A1).

Two beta-minus emissions: Z=82\to84, A=214. Then alpha: Z=82, A=210. Neutrons =210-82=128: recompute A-Z at the end. D is the starting neutron number; C keeps N fixed through the beta decays; A forgets that the alpha particle takes two protons.

(a) The total binding energy of thorium-230 plus the alpha particle is greater than that of uranium-234; the difference is released, mostly as kinetic energy (A1). (b) Totals: 230\times7.6310=1755.13\,\mathrm{MeV} and 4\times7.0739=28.30\,\mathrm{MeV} (M1); 234\times7.6005=1778.52\,\mathrm{MeV} (M1); energy released =1755.13+28.30-1778.52=4.91\,\mathrm{MeV} (A1). (c) Less: the products are more tightly bound, so their total mass is smaller (A1). (d) \Delta m=4.909/931.5=5.27\times10^{-3}\,\mathrm{u} (M1) =5.27\times10^{-3}\times1.661\times10^{-27}=8.75\times10^{-30}\,\mathrm{kg} (A1). (e) Binding energy per nucleon is an average; the energies to compare are the totals, binding energy per nucleon times nucleon number, for every nucleus (A1).

(a) Alpha particles would be absorbed completely by the foil, so no change of thickness could be detected (A1). Gamma rays would pass through almost unaffected; beta particles are partly absorbed, so the count rate changes noticeably with thickness (A1). (b) A higher count rate means less absorption, so the foil has become thinner (A1); the controller should increase the gap between the rollers, reducing the squeeze on the foil (A1). (c) Over days or months the source's own activity hardly changes, so any change in the reading is due to the foil and not to the source decaying (A1). (d) The source part of the reading is 1530-30=1500 counts per minute; after one half-life it is 750 (M1); the detector records 750+30=780 counts per minute (A1).

(a) A falls by 32, and only alpha decay changes A: 32/4=8 alpha decays (A1). Eight alpha decays lower Z by 16, to 76; Z must rise to 82, so there are 6 beta-minus decays (A1). (b) Number is proportional to mass divided by the mass of one nucleus: \dfrac{N_\mathrm{Pb}}{N_\mathrm{U}}=\dfrac{0.866/206}{1.00/238} (M1) =1.00 (A1). (c) Each lead-206 nucleus was once a uranium-238 nucleus, so the original number was twice the present uranium number (M1): one half-life has passed, 4.5\times10^{9} years (A1). (d) The crystal contained no lead-206 when it formed, or no uranium or lead has entered or left the crystal since (A1).

(a) \lambda=\ln2/110=6.30\times10^{-3}\,\mathrm{min^{-1}} (A1). (b) 0.40\,\mathrm{GBq}=400\,\mathrm{MBq}, the same unit top and bottom (M1); t=\dfrac{1}{\lambda}\ln\dfrac{400}{120}=\dfrac{1.204}{6.30\times10^{-3}}=191\,\mathrm{min} (M1); latest injection about 11:11 (A1). (c) t=90\,\mathrm{min} (M1); A=400\,e^{-6.30\times10^{-3}\times90}=227\,\mathrm{MBq}=0.23\,\mathrm{GBq} (A1). (d) \lambda\Delta t approximates the probability only when \lambda\Delta t\ll1; 110 minutes is one half-life, so the probability is exactly 0.5 (A1).

(a) Being doubly charged and massive, the alpha particle pulls strongly on the electrons of every atom it passes, so it gives up all its energy within a few centimetres (A1). A gamma photon has no charge and interacts only rarely, so most photons cross the chamber leaving few ion pairs (A1). (b) 5.49\times10^{6}/34=1.6\times10^{5} ion pairs (A1). (c) Electrons collected per second =2.6\times10^{-13}/1.60\times10^{-19}=1.6\times10^{6}\,\mathrm{s^{-1}} (M1); alpha particles per second =1.63\times10^{6}/1.61\times10^{5}=10, so the activity is about 10\,\mathrm{Bq} (A1). (d) 10/5.0\times10^{-4}=2.0\times10^{4}\,\mathrm{Bq\,m^{-3}} (A1).

(a) If only the beta particle and the nucleus were emitted, conservation of energy and momentum would give every beta particle the same energy, 1.71\,\mathrm{MeV} (A1). The spread shows that a third particle, the antineutrino, carries away the rest of the energy, in a share that varies from decay to decay (A1). (b) It has no electric charge, so it does not directly ionize material along a charged track (A1). It also has a very low probability of interacting with matter, so most pass through a detector without a signal (A1). Its small mass is inferred from the endpoint, rather than the reason it interacts rarely. (c) 1.71-0.60=1.11\,\mathrm{MeV} (A1). (d) In alpha decay only two bodies, the alpha particle and the daughter nucleus, share the energy; equal and opposite momenta then fix each share, so the energy is the same every time (A1).

(a) {}^{228}_{\;90}\mathrm{Th}\to{}^{224}_{\;88}\mathrm{Ra}+{}^{4}_{2}\mathrm{He} (A1). (b) The nucleus was at rest, so the two products have equal and opposite momenta p (M1). With E_\mathrm{k}=p^2/2m, the kinetic energy is inversely proportional to mass, so the light alpha particle takes the larger share (M1): E_\alpha=5.52\times\dfrac{224}{228}=5.42\,\mathrm{MeV} (A1). (c) v=\sqrt{2E_\mathrm{k}/m}=\sqrt{2\times5.42\times1.60\times10^{-13}/6.64\times10^{-27}} (M1) =1.6\times10^{7}\,\mathrm{m\,s^{-1}} (A1); about 0.05c, so the non-relativistic formula is fine. (d) qvB=mv^2/r gives r=\dfrac{mv}{qB} (M1) =\dfrac{6.64\times10^{-27}\times1.62\times10^{7}}{3.20\times10^{-19}\times0.80}=0.42\,\mathrm{m} (A1).

(a) P=A\times E=5.6\times10^{13}\times5.59\times1.60\times10^{-13} (M1) =50\,\mathrm{W} (A1). (b) Q=mc\Delta T=2.0\times900\times30=5.4\times10^{4}\,\mathrm{J} (M1); t=5.4\times10^{4}/50.1=1.1\times10^{3}\,\mathrm{s} (A1). (c) 176 years is 2 half-lives: 50/4=12.5\,\mathrm{W} (A1). (d) Alpha particles are stopped within a short distance in the heater itself, so almost all the decay energy is deposited there as thermal energy (A1); very little radiation escapes, so little shielding is needed to protect the instruments (A1). (e) P=\sigma AT^4 (M1): T=\left(\dfrac{50}{5.67\times10^{-8}\times0.050}\right)^{1/4}=365\,\mathrm{K} (A1).

(a) Nucleon numbers: 1+235=137+97+x, so x=2; proton numbers 92=52+40 check (A1). (b) Energy released = (total binding energy after) - (total binding energy before); free neutrons count zero: 137\times8.33+97\times8.60-235\times7.59 (M1) =192\,\mathrm{MeV} (A1). (c) Energy per fission =191.8\times1.60\times10^{-13}=3.07\times10^{-11}\,\mathrm{J} (M1); fissions per second =3.0\times10^{9}/3.07\times10^{-11}=9.8\times10^{19}\,\mathrm{s^{-1}} (A1). (d) Nuclei used per day =9.78\times10^{19}\times86\,400=8.4\times10^{24} (M1); mass =\dfrac{8.4\times10^{24}}{6.02\times10^{23}}\times0.235\,\mathrm{kg}=3.3\,\mathrm{kg} (A1). (e) Tellurium-137 and zirconium-97 lie nearer the peak of the curve than uranium-235, so they have a larger binding energy per nucleon; the total binding energy goes up and the difference is released (A1).

(a) 2.506-1.333=1.173\,\mathrm{MeV} and 1.333-0=1.333\,\mathrm{MeV} (A1). (b) \lambda=\dfrac{hc}{E}=\dfrac{1.99\times10^{-25}}{1.333\times10^{6}\times1.60\times10^{-19}} (M1) =9.3\times10^{-13}\,\mathrm{m} (A1). (c) 1.333\times10^{6}/1.89\approx7\times10^{5}: nuclear energy levels are separated by about a million times more energy than atomic levels, so gamma photons come from the nucleus, not the electrons (A1). (d) \lambda=\ln2/5.27=0.132\,\mathrm{y^{-1}} (M1); A=3.7\times10^{10}\,e^{-0.132\times2.0}=2.8\times10^{10}\,\mathrm{Bq} (A1). (e) Energy per decay =2.506\,\mathrm{MeV}=4.01\times10^{-13}\,\mathrm{J} (M1); P=2.84\times10^{10}\times4.01\times10^{-13}=0.011\,\mathrm{W} (A1).