The Workbook

1. The \(\alpha\) mode removes \(\lambda_1 N\) nuclei per second and the \(\beta\) mode removes \(\lambda_2 N\). Total: \(\dfrac{dN}{dt} = -\lambda_1 N - \lambda_2 N = -(\lambda_1 + \lambda_2)N\).

2. This is the ordinary decay equation with \(\lambda = \lambda_1 + \lambda_2\). Its solution is \(N = N_0 e^{-(\lambda_1 + \lambda_2)t}\). Decay constants add because the two chances of decay add.

\(\lambda = \dfrac{\ln 2}{T}\), so \(\dfrac{1}{T} = \dfrac{1}{T_1} + \dfrac{1}{T_2} = \dfrac{1}{20} + \dfrac{1}{30} = \dfrac{3 + 2}{60} = \dfrac{1}{12}\). \(T = 12\) min.

Half-lives combine like resistors in parallel; the effective half-life is shorter than either.

Fraction by \(\alpha\): \(\dfrac{\lambda_1}{\lambda_1 + \lambda_2} = \dfrac{1/20}{1/12} = \dfrac{12}{20} = 0.6\). Fraction by \(\beta\): \(\dfrac{1/30}{1/12} = 0.4\).

The faster route takes the larger share. Every nucleus decays exactly once, so the fractions add to \(1\).

1. \(24\) min is two effective half-lives: \(N = \dfrac{10^{20}}{4} = 2.5 \times 10^{19}\).

2. Decayed: \(7.5 \times 10^{19}\). By \(\alpha\): \(0.6 \times 7.5 \times 10^{19} = 4.5 \times 10^{19}\). By \(\beta\): \(0.4 \times 7.5 \times 10^{19} = 3 \times 10^{19}\).

1. \(\dfrac{1}{T} = \dfrac{1}{1} + \dfrac{1}{3} = \dfrac{4}{3}\) h\(^{-1}\), so \(T = \dfrac34\) h \(= 45\) min.

2. Fraction to Y \(= \dfrac{\lambda_1}{\lambda_1 + \lambda_2} = \dfrac{1}{4/3} = \dfrac34\). Three quarters of the X nuclei become Y.

3. \(45\) min is one effective half-life, so \(\dfrac12\) remains.

\(\lambda = \dfrac{1}{\tau}\), so \(\dfrac{1}{\tau} = \dfrac{1}{\tau_1} + \dfrac{1}{\tau_2} = \dfrac{\tau_1 + \tau_2}{\tau_1\tau_2}\) and \(\tau = \dfrac{\tau_1\tau_2}{\tau_1 + \tau_2}\).

\(\tau = \dfrac{2 \times 6}{2 + 6} = 1.5\) h.

1. \(\dfrac{dN_B}{dt} = \lambda_A N_A - \lambda_B N_B\). The first term is the rate at which A decays into B (B is created). The second is the rate at which B itself decays into C (B is lost).

2. \(N_B\) is constant when \(\dfrac{dN_B}{dt} = 0\), that is, when \(\lambda_A N_A = \lambda_B N_B\): B is destroyed as fast as it is made.

1. \(\lambda = \dfrac{\ln 2}{T}\). So \(\dfrac{N_B}{N_A} = \dfrac{\lambda_A}{\lambda_B} = \dfrac{\ln 2/T_A}{\ln 2/T_B} = \dfrac{T_B}{T_A}\).

2. \(\dfrac{N_{Rn}}{N_{Ra}} = \dfrac{3.8}{1600 \times 365} = \dfrac{3.8}{5.84 \times 10^{5}} \approx 6.5 \times 10^{-6}\). About \(6\) radon atoms for every million radium atoms.

1. The curve levels off at the dashed line, \(N_{eq} = 1000\).

2. \(N_B = 500\) at \(t = 2\) h. From \(1 - e^{-\lambda_B t} = \dfrac12\), \(e^{-\lambda_B t} = \dfrac12\), which is the half-life of B: \(T_B = 2\) h.

3. At \(6\) h, three half-lives, \(e^{-\lambda_B t} = \dfrac18\). \(N_B = 1000\left(1 - \dfrac18\right) = 875\), as the dotted line shows.

1. A month is many radon half-lives, so equilibrium has been reached. The radon activity equals the radium activity, \(3.7 \times 10^{10}\) Bq.

2. \(A = \lambda N\), so \(N = \dfrac{A}{\lambda} = \dfrac{A\,T}{\ln 2} = \dfrac{3.7 \times 10^{10} \times 3.8 \times 86400}{0.693} \approx 1.75 \times 10^{16}\) atoms.

3. Mass \(= \dfrac{1.75 \times 10^{16}}{6.02 \times 10^{23}} \times 222 \approx 6.5 \times 10^{-6}\) g \(= 6.5\ \mu\)g. A tiny mass, yet as active as the whole gram of radium.

1. \(\dfrac{N_B}{N_A} = \dfrac{T_B}{T_A} = \dfrac{1 \text{ day}}{1000 \times 365 \text{ days}} \approx 2.7 \times 10^{-6}\).

2. In equilibrium \(\lambda_A N_A = \lambda_B N_B\): the activities are equal. Every B that decays is replaced by a fresh one from A.

3. No. C is stable and is created at the rate \(\lambda_B N_B\), so \(N_C\) grows steadily.

Activity is \(\lambda N\), the number decaying per second. The daughter has a huge \(\lambda\) (short half-life) and a tiny \(N\); the parent has a tiny \(\lambda\) and a huge \(N\). The products are equal.

The daughter is like a small tank with a wide outlet, fed by a slow tap. The level stays low, but the outflow equals the inflow.

1. Initial mass: \(14.00307 + 4.00260 = 18.00567\) u. Final mass: \(16.99913 + 1.00783 = 18.00696\) u.

   \(Q = (18.00567 - 18.00696) \times 931.5 = -0.00129 \times 931.5 \approx -1.20\) MeV.

2. \(Q < 0\): endothermic. The products are heavier than the reactants, so \(1.20\) MeV of kinetic energy must be supplied and turned into mass. A positive \(Q\) would mean energy released.

Initial: \(2 \times 2.01410 = 4.02820\) u. Final: \(3.01603 + 1.00866 = 4.02469\) u.

\(Q = 0.00351 \times 931.5 \approx 3.27\) MeV, released as kinetic energy of the products.

Initial: \(7.01600 + 1.00783 = 8.02383\) u. Final: \(2 \times 4.00260 = 8.00520\) u.

\(Q = 0.01863 \times 931.5 \approx 17.35\) MeV. This large release is why the reaction was one of the first studied with accelerated protons.

Energy conservation: \(K_{initial} + Q = K_{final}\), so \(Q = K_{final} - K_{initial} = 4 - 1 = 3\) MeV.

The products carry more kinetic energy than came in; the extra \(3\) MeV came from the mass lost in the reaction.

1. The \(\alpha\) particle brings momentum. After the collision the products must carry the same momentum, so they cannot be at rest. Some kinetic energy must remain as motion, and is not available to supply \(|Q|\).

2. \(K_{th} = 1.20 \times \dfrac{4 + 14}{14} = 1.20 \times \dfrac{18}{14} \approx 1.54\) MeV.

Before: momentum \(p = \sqrt{2mK}\). At threshold all products move together with total mass \(m + M\) and the same \(p\). Their kinetic energy is \(K' = \dfrac{p^2}{2(m + M)} = \dfrac{m}{m + M}K\).

The energy available for the reaction is \(K - K' = \dfrac{M}{m + M}K\). Setting this equal to \(|Q|\): \(K_{th} = |Q|\,\dfrac{m + M}{M}\).

1. No. With \(Q > 0\) the products can be made even from reactants at rest; energy is released, not needed.

2. The proton and the lithium nucleus are both positive. They repel, and the proton must have enough kinetic energy to come within reach of the short-range nuclear force. This Coulomb barrier, not \(Q\), sets the practical minimum.

1. The photon must supply at least \(|Q| = 2.22\) MeV, the binding energy of the deuteron.

2. Correction \(= |Q| \times \dfrac{|Q|}{2Mc^2} = 2.22 \times \dfrac{2.22}{2 \times 1876} = 2.22 \times 5.9 \times 10^{-4} \approx 1.3 \times 10^{-3}\) MeV \(= 1.3\) keV. The recoil correction is tiny because the deuteron is so heavy compared with the photon energy.

1. \(K_{final} = K_{initial} + Q = 1 + 3.27 = 4.27\) MeV.

2. No. The incoming deuteron had momentum and the target had none, so the total momentum was not zero. It is conserved, so the products together carry the same momentum along the original direction.

1. The neutron and the boron nucleus are both practically at rest, so the total momentum is zero. The two products must have equal and opposite momenta.

2. With equal momenta \(p\), \(K = \dfrac{p^2}{2m} \propto \dfrac{1}{m}\). \(\dfrac{K_\alpha}{K_{Li}} = \dfrac{m_{Li}}{m_\alpha} = \dfrac{7}{4}\).

   \(K_\alpha + K_{Li} = 2.79\) MeV and \(K_\alpha = \dfrac{7}{11} \times 2.79 \approx 1.78\) MeV, \(K_{Li} = \dfrac{4}{11} \times 2.79 \approx 1.01\) MeV.

Both \(A\) (top) and \(Z\) (bottom) add up on the two sides.

1. \(A = 238 - 4 = 234\), \(Z = 92 - 2 = 90\): \(^{234}_{90}\)Th.

2. The electron has \(A = 0\), \(Z = -1\). So \(A = 14\), \(Z = 6 + 1 = 7\): \(^{14}_{7}\)N.

3. The positron has \(Z = +1\). So \(A = 22\), \(Z = 11 - 1 = 10\): \(^{22}_{10}\)Ne.

\(\alpha\): \(A\) falls by \(4\), \(Z\) falls by \(2\). \(\beta^-\): \(A\) unchanged, \(Z\) rises by \(1\). \(\beta^+\): \(A\) unchanged, \(Z\) falls by \(1\). \(\gamma\): \(A\) and \(Z\) unchanged; the nucleus only loses energy.

In \(\beta^-\) decay a neutron inside the nucleus turns into a proton: \(n \to p + e^- + \bar{\nu}\). The electron and the antineutrino are created at that moment and leave; they were not in the nucleus before.

\(\Delta m = 238.05079 - (234.04363 + 4.00260) = 238.05079 - 238.04623 = 0.00456\) u.

\(Q = 0.00456 \times 931.5 \approx 4.25\) MeV. The electron masses cancel because atomic masses are used on both sides.

1. \(\Delta m = 14.003242 - 14.003074 = 0.000168\) u. \(Q = 0.000168 \times 931.5 \approx 0.156\) MeV. (The atomic mass of nitrogen already includes the extra electron.)

2. Three bodies share the energy: the daughter nucleus, the electron and the antineutrino. The energy can be divided between the electron and the antineutrino in any proportion. So the electron energy is spread from \(0\) to \(Q\).

1. The curve reaches zero at \(K_{\max} = 0.156\) MeV, the \(Q\)-value of the decay.

2. The antineutrino carries the rest, about \(0.156 - 0.06 = 0.096\) MeV (apart from the tiny recoil of the nucleus).

3. If only the electron and the nucleus were produced, every electron would have the same energy, as in \(\alpha\) decay. The spread suggested that energy was lost. Pauli proposed the neutrino, an unseen third particle, to carry the missing energy.

1. The nucleus was at rest, so the total momentum was zero. After the decay the two products must still have zero total momentum: equal magnitudes, opposite directions.

2. With equal momenta \(p\), \(K = \dfrac{p^2}{2m} \propto \dfrac{1}{m}\). So \(\dfrac{K_\alpha}{K_{Th}} = \dfrac{234}{4}\) and \(K_\alpha = Q\,\dfrac{234}{238}\).

   \(K_\alpha = 4.25 \times \dfrac{234}{238} \approx 4.18\) MeV. Thorium gets \(4.25 - 4.18 = 0.07\) MeV. The light particle takes almost all the energy.

1. \(\lambda = \dfrac{\ln 2}{T_{1/2}} = \dfrac{0.693}{6.93} = 0.1\) h\(^{-1}\). \(\tau = \dfrac{1}{\lambda} = 10\) h.

2. Each nucleus has a probability of \(0.1\) of decaying in any one hour, no matter how old it is. Which nucleus decays is pure chance; only the average rate is fixed.

After \(n\) half-lives the fraction remaining is \(\dfrac{1}{2^n}\) and the fraction decayed is \(1 - \dfrac{1}{2^n}\).

1. \(\dfrac12\) remains, \(\dfrac12\) has decayed.

2. \(\dfrac18\) remains, \(\dfrac78\) has decayed.

3. \(\dfrac{1}{32}\) remains, \(\dfrac{31}{32}\) has decayed.

1. \(15\) days is \(3\) half-lives. \(N = \dfrac{8 \times 10^{20}}{8} = 10^{20}\) remain and \(7 \times 10^{20}\) have decayed.

2. \(\lambda = \dfrac{0.693}{5 \times 86400} = 1.6 \times 10^{-6}\) s\(^{-1}\).

   \(A_0 = \lambda N_0 = 1.6 \times 10^{-6} \times 8 \times 10^{20} \approx 1.3 \times 10^{15}\) Bq.

1. \(N_0 = 8000\) at \(t = 0\). \(N\) falls to \(4000\) at \(t = 2\) h, so \(T_{1/2} = 2\) h.

2. At \(6\) h the curve reads \(1000\). Check: \(6\) h is \(3\) half-lives, \(8000/2^3 = 1000\).

3. \(\lambda = \dfrac{0.693}{2} \approx 0.35\) h\(^{-1}\).

1. \(N = \dfrac{1}{226} \times 6.02 \times 10^{23} \approx 2.66 \times 10^{21}\).

2. \(\lambda = \dfrac{0.693}{1600 \times 3.156 \times 10^{7}} = 1.37 \times 10^{-11}\) s\(^{-1}\).

   \(A = \lambda N = 1.37 \times 10^{-11} \times 2.66 \times 10^{21} \approx 3.7 \times 10^{10}\) Bq. One curie is defined as \(3.7 \times 10^{10}\) decays per second.

1. Take the natural log of both sides: \(\ln N = \ln N_0 + \ln e^{-\lambda t} = \ln N_0 - \lambda t\). This is \(y = c + mx\) with slope \(-\lambda\).

2. Slope \(= \dfrac{18 - 20}{10 - 0} = -0.2\) h\(^{-1}\), so \(\lambda = 0.2\) h\(^{-1}\). \(T_{1/2} = \dfrac{0.693}{0.2} \approx 3.5\) h.

3. The intercept is \(\ln N_0 = 20\), so \(N_0 = e^{20} \approx 4.9 \times 10^{8}\).

\(A = A_0 e^{-\lambda t}\) with \(\dfrac{A}{A_0} = \dfrac{1}{10}\): \(e^{\lambda t} = 10\), so \(t = \dfrac{\ln 10}{\lambda} = \dfrac{2.303}{0.693}T_{1/2} = 3.32\,T_{1/2}\).

\(t = 3.32 \times 3 \approx 10\) days.

1. \(\lambda = \dfrac{\ln 2}{T_{1/2}}\), so \(\tau = \dfrac{1}{\lambda} = \dfrac{T_{1/2}}{\ln 2} = \dfrac{T_{1/2}}{0.693} \approx 1.44\,T_{1/2}\).

2. At \(t = \tau\): \(N = N_0 e^{-\lambda\tau} = N_0 e^{-1} = 0.37\,N_0\). About \(37\%\) remain, \(63\%\) have decayed.

1. \(A = \lambda N\) and \(\lambda \propto \dfrac{1}{T_{1/2}}\). Same \(N\): \(\dfrac{A_P}{A_Q} = \dfrac{\lambda_P}{\lambda_Q} = \dfrac{2}{1}\).

2. After \(2\) h, P (\(2\) half-lives) has \(\dfrac{N_0}{4}\) and Q (\(1\) half-life) has \(\dfrac{N_0}{2}\). \(\dfrac{A_P}{A_Q} = \dfrac{\lambda_P N_0/4}{\lambda_Q N_0/2} = 2 \times \dfrac{1}{2} = 1\). The activities are equal.

1. Cosmic rays strike nitrogen in the upper air and make \(^{14}\)C at a steady rate. It mixes into the CO\(_2\) of the air, which plants take in. Intake balances decay, so a living plant keeps a fixed ratio of \(^{14}\)C to ordinary carbon.

2. A dead plant takes in no new carbon. Its \(^{14}\)C decays with no replacement, so the ratio falls with time. The fall measures the time since death.

One quarter means two half-lives: \(t = 2 \times 5730 = 11460\) years.

\(12.5\% = \dfrac18 = \dfrac{1}{2^3}\): three half-lives. \(t = 3 \times 5730 = 17190\) years.

1. \(50\%\) at \(5.7\) ky (one half-life), \(25\%\) at \(11.5\) ky, \(12.5\%\) at \(17.2\) ky.

2. \(6\%\) is just under \(6.25\% = 1/16\), four half-lives, so the age is a little over \(23\) ky. Beyond about \(50000\) years (\(9\) half-lives) less than \(0.2\%\) remains. The activity is too weak to measure reliably.

\(\dfrac{A_0}{A} = \dfrac{15}{3} = 5 = e^{\lambda t}\), so \(t = \dfrac{\ln 5}{\lambda} = \dfrac{T_{1/2}}{\ln 2}\ln 5\).

\(t = \dfrac{5730}{0.693} \times 1.61 \approx 8270 \times 1.61 \approx 13300\) years.

1. Every lead atom was once a uranium atom. Equal numbers mean half the uranium has decayed: one half-life, \(4.5 \times 10^{9}\) years.

2. That the rock had no \(^{206}\)Pb when it formed, and that no uranium or lead has leaked in or out since.

Of every \(4\) original uranium atoms, \(3\) remain: \(\dfrac{N}{N_0} = \dfrac34\), so \(e^{\lambda t} = \dfrac43\).

\(t = \dfrac{\ln(4/3)}{\lambda} = \dfrac{T_{1/2}}{\ln 2} \times 0.288 = \dfrac{4.5 \times 10^{9}}{0.693} \times 0.288 \approx 1.9 \times 10^{9}\) years.

1. A\(\to\)B moves \(2\) left and \(2\) down: \(Z\) and \(N\) each fall by \(2\), the loss of an \(\alpha\) particle. B\(\to\)C and C\(\to\)D move \(1\) right and \(1\) down: \(Z\) rises by \(1\) and \(N\) falls by \(1\). A neutron has turned into a proton, which is \(\beta^-\) decay.

2. \(A = Z + N\). A: \((92, 146)\), \(^{238}_{92}\)U. B: \((90, 144)\), \(^{234}_{90}\)Th. C: \((91, 143)\), \(^{234}_{91}\)Pa. D: \((92, 142)\), \(^{234}_{92}\)U.

\(\dfrac{6400}{100} = 64 = 2^6\): the activity must fall through six half-lives.

\(t = 6 \times 8 = 48\) days.