Week 13: Radioactive Decay & Nuclear Reactions

\(\Delta\) ← if you see the word "Delta" instead of a triangle, please refresh the page! Thanks :)

[1] Overview⚓︎

I recommend reviewing the following concepts, they will probably come up in recitation:

• Radioactive decay and half-life

  • \(N(t)=N_0e^{-\lambda t}\)
  • \(A(t)=A_0e^{-\lambda t}\)
  • \(\lambda=\dfrac{\ln 2}{T_{1/2}}\)
  • after one half-life, the activity is cut in half

• Activity units

  • \(1\ \mathrm{Bq}=1\ \text{decay/s}\)
  • \(1\ \mathrm{Ci}=3.70\times10^{10}\ \mathrm{Bq}\)
  • \(1\ \mathrm{mCi}=3.70\times10^7\ \mathrm{Bq}\)
  • activity is proportional to the number of undecayed nuclei

• Balancing nuclear reactions

  • mass number \(A\) must be conserved
  • atomic number \(Z\) must be conserved
  • the missing particle or nuclide can usually be found by balancing \(A\) and \(Z\) separately

• Common decay modes

  • alpha decay: emits \(^4_2\mathrm{He}\)
  • beta-minus decay: emits \(^0_{-1}\mathrm{e}\) and increases \(Z\) by 1
  • beta-plus decay: emits \(^0_{+1}\mathrm{e}\) and decreases \(Z\) by 1
  • electron capture: absorbs an inner electron and decreases \(Z\) by 1

• Reaction energy / Q-values

  • \(Q=(m_{\mathrm{reactants}}-m_{\mathrm{products}})c^2\)
  • \(1\ \mathrm{u}=931.5\ \mathrm{MeV}/c^2\)
  • positive \(Q\) means energy is released
  • negative \(Q\) means energy is absorbed / required

• Using atomic masses carefully

  • for most balanced nuclear reactions, neutral atomic masses can be used directly if the same total number of electrons appears on both sides
  • for \(\beta^-\) decay, atomic masses can be used as \(Q=(M_{\mathrm{parent}}-M_{\mathrm{daughter}})c^2\)
  • for \(\beta^+\) decay, atomic masses require subtracting \(2m_e\): \(Q=(M_{\mathrm{parent}}-M_{\mathrm{daughter}}-2m_e)c^2\)

Practice Problems⚓︎

• If you're comfortable with the "★★★" problems, you should do great during recitation.
• To print these questions, simply press Ctrl + P while on this page, and it should come out formatted nicely -- just make sure to refresh first so all the math renders properly.
  • For a copy without solutions, click here.

Difficulty key:

• ★☆☆ = beginner
• ★★☆ = standard
• ★★★ = challenging / multi-step

[2.1] Radioactive Decay, Half-Life, and Activity⚓︎

1. ★☆☆
   A radioactive isotope has a half-life of \(12.0\ \mathrm{min}\) and an initial activity of \(4.80\times10^5\ \mathrm{Bq}\).

   • (a) Find the decay constant in \(\mathrm{min^{-1}}\).
   • (b) Find the decay constant in \(\mathrm{s^{-1}}\).
   • (c) What is the activity after \(12.0\ \mathrm{min}\)?
   • (d) What is the activity after \(24.0\ \mathrm{min}\)?
   Solution

   The decay constant is related to the half-life by

   \[\lambda=\frac{\ln 2}{T_{1/2}}.\]

   Since \(T_{1/2}=12.0\ \mathrm{min}\),

   \[\lambda=\frac{0.693}{12.0\ \mathrm{min}}=0.0578\ \mathrm{min^{-1}}.\]

   To convert to seconds,

   \[12.0\ \mathrm{min}=720\ \mathrm{s},\]

   so

   \[\lambda=\frac{0.693}{720\ \mathrm{s}}=9.63\times10^{-4}\ \mathrm{s^{-1}}.\]

   After one half-life, the activity is cut in half:

   \[A(12.0\ \mathrm{min})=\frac{4.80\times10^5}{2}=2.40\times10^5\ \mathrm{Bq}.\]

   After two half-lives,

   \[A(24.0\ \mathrm{min})=\frac{4.80\times10^5}{4}=1.20\times10^5\ \mathrm{Bq}.\]

   
   Answer: (a) \(0.0578\ \mathrm{min^{-1}}\); (b) \(9.63\times10^{-4}\ \mathrm{s^{-1}}\); (c) \(2.40\times10^5\ \mathrm{Bq}\); (d) \(1.20\times10^5\ \mathrm{Bq}\).

2. ★☆☆
   A sample of a radioactive tracer has an activity of \(18.0\ \mathrm{mCi}\) at noon. Its half-life is \(6.00\ \mathrm{h}\).

   • (a) What is the activity at 6:00 p.m.?
   • (b) What is the activity at midnight?
   • (c) What percentage of the original activity remains at midnight?
   Solution

   From noon to 6:00 p.m. is one half-life, so

   \[A=\frac{18.0\ \mathrm{mCi}}{2}=9.00\ \mathrm{mCi}.\]

   From noon to midnight is two half-lives, so

   \[A=\frac{18.0\ \mathrm{mCi}}{2^2}=4.50\ \mathrm{mCi}.\]

   The fraction remaining after two half-lives is

   \[\frac{1}{2^2}=\frac{1}{4}=0.250.\]

   So the percentage remaining is

   \[25.0\%.\]

   
   Answer: (a) \(9.00\ \mathrm{mCi}\); (b) \(4.50\ \mathrm{mCi}\); (c) \(25.0\%\).

3. ★★☆
   A radioisotope has a half-life of \(4.20\ \mathrm{h}\) and an initial activity of \(2.50\ \mathrm{mCi}\).

   • (a) Find the activity after \(10.5\ \mathrm{h}\).
   • (b) Convert your answer to becquerels.
   Solution

   The activity follows

   \[A=A_0\left(\frac{1}{2}\right)^{t/T_{1/2}}.\]

   Here,

   \[\frac{t}{T_{1/2}}=\frac{10.5}{4.20}=2.50.\]

   Therefore,

   \[A=(2.50\ \mathrm{mCi})\left(\frac{1}{2}\right)^{2.50}=0.442\ \mathrm{mCi}.\]

   Since

   \[1\ \mathrm{mCi}=3.70\times10^7\ \mathrm{Bq},\]

   we get

   \[A=(0.442)(3.70\times10^7)=1.64\times10^7\ \mathrm{Bq}.\]

   
   Answer: (a) \(0.442\ \mathrm{mCi}\); (b) \(1.64\times10^7\ \mathrm{Bq}\).

4. ★★☆
   A radioactive sample has a half-life of \(2.50\ \mathrm{days}\). Its activity is initially \(1600\ \mathrm{Bq}\).

   • (a) How long will it take for the activity to fall to \(100\ \mathrm{Bq}\)?
   • (b) How many half-lives have passed during this time?
   Solution

   Use

   \[A=A_0\left(\frac{1}{2}\right)^n,\]

   where \(n\) is the number of half-lives.

   The ratio is

   \[\frac{A}{A_0}=\frac{100}{1600}=\frac{1}{16}.\]

   Since

   \[\frac{1}{16}=\left(\frac{1}{2}\right)^4,\]

   four half-lives have passed.

   The time is therefore

   \[t=4(2.50\ \mathrm{days})=10.0\ \mathrm{days}.\]

   
   Answer: (a) \(10.0\ \mathrm{days}\); (b) 4 half-lives.

5. ★★★
   A sample of fluorine-18 has a half-life of \(110\ \mathrm{min}\). At 7:30 a.m., its activity is \(12.0\ \mathrm{mCi}\). A patient is scheduled to receive the sample at 9:40 a.m.

   • (a) What will the activity be at 9:40 a.m.?
   • (b) If the patient needs at least \(5.00\ \mathrm{mCi}\) at injection time, is this sample sufficient?
   • (c) What activity would the sample need to have at 7:30 a.m. in order to have exactly \(5.00\ \mathrm{mCi}\) at 9:40 a.m.?
   Solution

   The elapsed time from 7:30 a.m. to 9:40 a.m. is

   \[t=130\ \mathrm{min}.\]

   The activity is

   \[A=A_0\left(\frac{1}{2}\right)^{t/T_{1/2}}.\]

   So

   \[A=(12.0\ \mathrm{mCi})\left(\frac{1}{2}\right)^{130/110}.\]

   This gives

   \[A=5.29\ \mathrm{mCi}.\]

   Since \(5.29\ \mathrm{mCi}>5.00\ \mathrm{mCi}\), the sample is sufficient.

   To find the required initial activity, solve

   \[5.00\ \mathrm{mCi}=A_0\left(\frac{1}{2}\right)^{130/110}.\]

   Thus,

   \[A_0=\frac{5.00}{(1/2)^{130/110}}=11.3\ \mathrm{mCi}.\]

   
   Answer: (a) \(5.29\ \mathrm{mCi}\); (b) yes, barely; (c) \(11.3\ \mathrm{mCi}\) at 7:30 a.m.

6. ★★★
   A sample initially contains \(6.40\times10^6\) undecayed nuclei of a radioactive isotope with half-life \(3.00\ \mathrm{h}\).

   • (a) Find the decay constant in \(\mathrm{s^{-1}}\).
   • (b) Find the initial activity of the sample.
   • (c) How many undecayed nuclei remain after \(7.50\ \mathrm{h}\)?
   • (d) What is the activity after \(7.50\ \mathrm{h}\)?
   Solution

   First convert the half-life to seconds:

   \[3.00\ \mathrm{h}=3.00(3600\ \mathrm{s})=1.08\times10^4\ \mathrm{s}.\]

   The decay constant is

   \[\lambda=\frac{0.693}{1.08\times10^4\ \mathrm{s}}=6.42\times10^{-5}\ \mathrm{s^{-1}}.\]

   The activity is related to the number of undecayed nuclei by

   \[A=\lambda N.\]

   Initially,

   \[A_0=(6.42\times10^{-5})(6.40\times10^6)=4.11\times10^2\ \mathrm{Bq}.\]

   The number of half-lives after \(7.50\ \mathrm{h}\) is

   \[\frac{7.50}{3.00}=2.50.\]

   Therefore,

   \[N=(6.40\times10^6)\left(\frac{1}{2}\right)^{2.50}=1.13\times10^6.\]

   The activity at that time is

   \[A=\lambda N=(6.42\times10^{-5})(1.13\times10^6)=72.6\ \mathrm{Bq}.\]

   
   Answer: (a) \(6.42\times10^{-5}\ \mathrm{s^{-1}}\); (b) \(4.11\times10^2\ \mathrm{Bq}\); (c) \(1.13\times10^6\) nuclei; (d) \(72.6\ \mathrm{Bq}\).

[2.2] Balancing Nuclear Reactions and Identifying Decay Modes⚓︎

1. ★☆☆
   Complete the following nuclear reaction and identify the decay mode:

   \[^{226}_{88}\mathrm{Ra}\rightarrow X+{}^4_2\mathrm{He}\]
   Solution

   Conserve mass number \(A\):

   \[226=A_X+4,\]

   so

   \[A_X=222.\]

   Conserve atomic number \(Z\):

   \[88=Z_X+2,\]

   so

   \[Z_X=86.\]

   The element with \(Z=86\) is radon, so

   \[X=^{222}_{86}\mathrm{Rn}.\]

   Since the emitted particle is \(^4_2\mathrm{He}\), this is alpha decay.

   
   Answer: \(X=^{222}_{86}\mathrm{Rn}\); alpha decay.

2. ★☆☆
   Complete the following nuclear reaction and identify the decay mode:

   \[^{14}_{6}\mathrm{C}\rightarrow X+{}^0_{-1}\mathrm{e}\]
   Solution

   Conserve mass number:

   \[14=A_X+0,\]

   so

   \[A_X=14.\]

   Conserve atomic number:

   \[6=Z_X+(-1),\]

   so

   \[Z_X=7.\]

   The element with \(Z=7\) is nitrogen, so

   \[X=^{14}_{7}\mathrm{N}.\]

   Since an electron is emitted from the nucleus, this is beta-minus decay.

   
   Answer: \(X=^{14}_{7}\mathrm{N}\); beta-minus decay.

3. ★★☆
   Complete the following reaction:

   \[^{27}_{13}\mathrm{Al}+{}^4_2\mathrm{He}\rightarrow X+{}^1_0\mathrm{n}\]
   • (a) Find the mass number of \(X\).
   • (b) Find the atomic number of \(X\).
   • (c) Identify the nuclide \(X\).
   Solution

   Conserve mass number:

   \[27+4=A_X+1.\]

   So

   \[31=A_X+1,\]

   which gives

   \[A_X=30.\]

   Conserve atomic number:

   \[13+2=Z_X+0.\]

   So

   \[Z_X=15.\]

   The element with \(Z=15\) is phosphorus, so

   \[X=^{30}_{15}\mathrm{P}.\]

   
   Answer: (a) \(A=30\); (b) \(Z=15\); (c) \(X=^{30}_{15}\mathrm{P}\).

4. ★★☆
   Complete the following decay reaction and classify the decay:

   \[^{64}_{29}\mathrm{Cu}\rightarrow{}^{64}_{28}\mathrm{Ni}+X\]
   Solution

   Conserve mass number:

   \[64=64+A_X,\]

   so

   \[A_X=0.\]

   Conserve atomic number:

   \[29=28+Z_X,\]

   so

   \[Z_X=+1.\]

   A particle with \(A=0\) and \(Z=+1\) is a positron:

   \[X=^0_{+1}\mathrm{e}.\]

   Since a positron is emitted, this is beta-plus decay.

   
   Answer: \(X=^0_{+1}\mathrm{e}\); beta-plus decay.

5. ★★★
   Complete the following reaction and classify the missing particle:

   \[^{23}_{11}\mathrm{Na}+{}^2_1\mathrm{H}\rightarrow{}^{24}_{12}\mathrm{Mg}+X\]
   • (a) Find \(A_X\).
   • (b) Find \(Z_X\).
   • (c) What particle is \(X\)?
   • (d) Explain why this is not an alpha particle or a proton.
   Solution

   Conserve mass number:

   \[23+2=24+A_X.\]

   Therefore,

   \[25=24+A_X,\]

   so

   \[A_X=1.\]

   Conserve atomic number:

   \[11+1=12+Z_X.\]

   Therefore,

   \[12=12+Z_X,\]

   so

   \[Z_X=0.\]

   A particle with \(A=1\) and \(Z=0\) is a neutron:

   \[X=^1_0\mathrm{n}.\]

   This is not an alpha particle because an alpha particle has \(A=4\) and \(Z=2\). It is not a proton because a proton has \(A=1\) and \(Z=1\).

   
   Answer: (a) \(A_X=1\); (b) \(Z_X=0\); (c) \(X=^1_0\mathrm{n}\); (d) it has the mass number of a nucleon but no positive charge.

6. ★★★
   A uranium nucleus \(^{238}_{92}\mathrm{U}\) undergoes two alpha decays followed by one beta-minus decay.

   • (a) What is the final mass number?
   • (b) What is the final atomic number?
   • (c) What is the final nuclide?
   • (d) Would the final answer change if the beta-minus decay occurred between the two alpha decays instead?
   Solution

   Each alpha decay decreases the mass number by 4 and the atomic number by 2.

   After two alpha decays,

   \[A=238-2(4)=230,\]

   and

   \[Z=92-2(2)=88.\]

   A beta-minus decay leaves the mass number unchanged but increases the atomic number by 1.

   Thus,

   \[A=230,\]

   and

   \[Z=88+1=89.\]

   The element with \(Z=89\) is actinium, so the final nuclide is

   \[^{230}_{89}\mathrm{Ac}.\]

   The final answer would not change if the beta-minus decay happened between the two alpha decays, because the total changes in \(A\) and \(Z\) are the same.

   
   Answer: (a) \(A=230\); (b) \(Z=89\); (c) \(^{230}_{89}\mathrm{Ac}\); (d) no, the order does not matter for the final \(A\) and \(Z\).

[2.3] Nuclear Reaction Energetics and Q-Values⚓︎

1. ★☆☆
   Consider the fusion reaction

   \[^2_1\mathrm{H}+{}^3_1\mathrm{H}\rightarrow{}^4_2\mathrm{He}+{}^1_0\mathrm{n}.\]

   Use the following atomic masses:

   \[m(^2_1\mathrm{H})=2.014102\ \mathrm{u},\quad m(^3_1\mathrm{H})=3.016049\ \mathrm{u},\]
   \[m(^4_2\mathrm{He})=4.002603\ \mathrm{u},\quad m_n=1.008665\ \mathrm{u}.\]
   • (a) Find the total mass of the reactants.
   • (b) Find the total mass of the products.
   • (c) Find the reaction energy \(Q\) in MeV.
   • (d) Is energy released or absorbed?
   Solution

   The reactant mass is

   \[m_{\mathrm{reactants}}=2.014102+3.016049=5.030151\ \mathrm{u}.\]

   The product mass is

   \[m_{\mathrm{products}}=4.002603+1.008665=5.011268\ \mathrm{u}.\]

   The mass difference is

   \[\Delta m=m_{\mathrm{reactants}}-m_{\mathrm{products}}\]

   \[\Delta m=5.030151-5.011268=0.018883\ \mathrm{u}.\]

   Therefore,

   \[Q=(0.018883)(931.5)=17.6\ \mathrm{MeV}.\]

   Since \(Q\) is positive, energy is released.

   
   Answer: (a) \(5.030151\ \mathrm{u}\); (b) \(5.011268\ \mathrm{u}\); (c) \(Q=+17.6\ \mathrm{MeV}\); (d) energy is released.

2. ★☆☆
   Tritium can beta-minus decay according to

   \[^3_1\mathrm{H}\rightarrow{}^3_2\mathrm{He}+{}^0_{-1}\mathrm{e}+\bar\nu.\]

   The atomic mass of \(^3_1\mathrm{H}\) is \(3.016049\ \mathrm{u}\), and the atomic mass of \(^3_2\mathrm{He}\) is \(3.016029\ \mathrm{u}\).

   • (a) Using atomic masses, find the mass difference.
   • (b) Find the energy released in MeV.
   Solution

   For beta-minus decay, atomic masses can be used as

   \[Q=(M_{\mathrm{parent}}-M_{\mathrm{daughter}})c^2.\]

   The mass difference is

   \[\Delta m=3.016049-3.016029=0.000020\ \mathrm{u}.\]

   Therefore,

   \[Q=(0.000020)(931.5)=0.0186\ \mathrm{MeV}.\]

   This is a small positive \(Q\), so energy is released.

   
   Answer: (a) \(2.0\times10^{-5}\ \mathrm{u}\); (b) \(Q=0.0186\ \mathrm{MeV}\) released.

3. ★★☆
   Consider the reaction

   \[^7_3\mathrm{Li}+{}^1_1\mathrm{H}\rightarrow{}^4_2\mathrm{He}+{}^4_2\mathrm{He}.\]

   Use the following atomic masses:

   \[m(^7_3\mathrm{Li})=7.016004\ \mathrm{u},\quad m(^1_1\mathrm{H})=1.007825\ \mathrm{u},\]
   \[m(^4_2\mathrm{He})=4.002603\ \mathrm{u}.\]
   • (a) Find the total reactant mass.
   • (b) Find the total product mass.
   • (c) Find \(Q\) in MeV.
   • (d) Is this reaction exothermic or endothermic?
   Solution

   The reactant mass is

   \[m_{\mathrm{reactants}}=7.016004+1.007825=8.023829\ \mathrm{u}.\]

   The product mass is

   \[m_{\mathrm{products}}=2(4.002603)=8.005206\ \mathrm{u}.\]

   The mass difference is

   \[\Delta m=8.023829-8.005206=0.018623\ \mathrm{u}.\]

   Therefore,

   \[Q=(0.018623)(931.5)=17.3\ \mathrm{MeV}.\]

   Since \(Q>0\), this reaction is exothermic.

   
   Answer: (a) \(8.023829\ \mathrm{u}\); (b) \(8.005206\ \mathrm{u}\); (c) \(Q=+17.3\ \mathrm{MeV}\); (d) exothermic.

4. ★★☆
   Consider the reaction

   \[^{14}_{7}\mathrm{N}+{}^4_2\mathrm{He}\rightarrow{}^{17}_{8}\mathrm{O}+{}^1_1\mathrm{H}.\]

   Use the following atomic masses:

   \[m(^{14}_{7}\mathrm{N})=14.003074\ \mathrm{u},\quad m(^4_2\mathrm{He})=4.002603\ \mathrm{u},\]
   \[m(^{17}_{8}\mathrm{O})=16.999132\ \mathrm{u},\quad m(^1_1\mathrm{H})=1.007825\ \mathrm{u}.\]
   • (a) Find the mass of the reactants.
   • (b) Find the mass of the products.
   • (c) Find \(Q\) in MeV.
   • (d) Is energy released or absorbed?
   Solution

   The reactant mass is

   \[m_{\mathrm{reactants}}=14.003074+4.002603=18.005677\ \mathrm{u}.\]

   The product mass is

   \[m_{\mathrm{products}}=16.999132+1.007825=18.006957\ \mathrm{u}.\]

   The mass difference is

   \[\Delta m=18.005677-18.006957=-0.001280\ \mathrm{u}.\]

   So

   \[Q=(-0.001280)(931.5)=-1.19\ \mathrm{MeV}.\]

   Since \(Q\) is negative, energy is absorbed. Equivalently, \(1.19\ \mathrm{MeV}\) must be supplied.

   
   Answer: (a) \(18.005677\ \mathrm{u}\); (b) \(18.006957\ \mathrm{u}\); (c) \(Q=-1.19\ \mathrm{MeV}\); (d) energy is absorbed.

5. ★★★
   Sodium-22 can decay by positron emission:

   \[^{22}_{11}\mathrm{Na}\rightarrow{}^{22}_{10}\mathrm{Ne}+{}^0_{+1}\mathrm{e}+\nu.\]

   The atomic mass of \(^{22}_{11}\mathrm{Na}\) is \(21.994436\ \mathrm{u}\), and the atomic mass of \(^{22}_{10}\mathrm{Ne}\) is \(21.991385\ \mathrm{u}\).

   • (a) Why is this classified as beta-plus decay?
   • (b) When using atomic masses, what mass difference should be used for a beta-plus decay?
   • (c) Find the energy released in MeV.
   Solution

   This is beta-plus decay because the emitted particle is a positron,

   \[^0_{+1}\mathrm{e},\]

   and the atomic number decreases from \(Z=11\) to \(Z=10\).

   For beta-plus decay using atomic masses,

   \[Q=(M_{\mathrm{parent}}-M_{\mathrm{daughter}}-2m_e)c^2.\]

   The extra \(2m_e\) appears because neutral atomic masses already include different numbers of atomic electrons.

   The mass difference is

   \[\Delta m=21.994436-21.991385-2(0.00054858).\]

   Thus,

   \[\Delta m=0.00195384\ \mathrm{u}.\]

   The energy released is

   \[Q=(0.00195384)(931.5)=1.82\ \mathrm{MeV}.\]

   
   Answer: (a) positron emission; (b) subtract \(2m_e\) when using atomic masses; (c) \(Q=1.82\ \mathrm{MeV}\) released.

6. ★★★
   Consider the reaction

   \[^{11}_{5}\mathrm{B}+{}^1_1\mathrm{H}\rightarrow3\left({}^4_2\mathrm{He}\right).\]

   Use the following atomic masses:

   \[m(^{11}_{5}\mathrm{B})=11.009305\ \mathrm{u},\quad m(^1_1\mathrm{H})=1.007825\ \mathrm{u},\]
   \[m(^4_2\mathrm{He})=4.002603\ \mathrm{u}.\]
   • (a) Verify that mass number and atomic number are conserved.
   • (b) Find the total mass of the reactants.
   • (c) Find the total mass of the products.
   • (d) Find \(Q\) in MeV and state whether energy is released or absorbed.
   Solution

   First check conservation.

   For mass number,

   \[11+1=12,\]

   and

   \[3(4)=12.\]

   So mass number is conserved.

   For atomic number,

   \[5+1=6,\]

   and

   \[3(2)=6.\]

   So atomic number is also conserved.

   The reactant mass is

   \[m_{\mathrm{reactants}}=11.009305+1.007825=12.017130\ \mathrm{u}.\]

   The product mass is

   \[m_{\mathrm{products}}=3(4.002603)=12.007809\ \mathrm{u}.\]

   The mass difference is

   \[\Delta m=12.017130-12.007809=0.009321\ \mathrm{u}.\]

   Therefore,

   \[Q=(0.009321)(931.5)=8.68\ \mathrm{MeV}.\]

   Since \(Q\) is positive, energy is released.

   
   Answer: (a) both \(A\) and \(Z\) are conserved; (b) \(12.017130\ \mathrm{u}\); (c) \(12.007809\ \mathrm{u}\); (d) \(Q=+8.68\ \mathrm{MeV}\) released.