A radioactive sample has a half-life of \(8.0\) days. Calculate the fraction of the original activity remaining after \(24\) days.
Number of half-lives \(= 24/8 = 3\).
Fraction remaining \(= (1/2)^{3} = 1/8\).
Nuclear Physics
10 exam-style questions on radioactivity, half-life, binding energy, fission and fusion. Tap Show answer for full working.
3 Easy 4 Medium 3 Hard
Write the nuclear equation for the alpha decay of \(^{238}_{92}\mathrm{U}\) and state the mass number and atomic number of the daughter nucleus.
\(^{238}_{92}\mathrm{U} \to ^{234}_{90}\mathrm{Th} + ^{4}_{2}\mathrm{He}\).
Daughter: mass number \(234\), atomic number \(90\).
Define binding energy and binding energy per nucleon. State which nuclei have the highest binding energy per nucleon.
Binding energy: the energy required to separate a nucleus into its individual protons and neutrons (or the energy released when the nucleus is formed).
Binding energy per nucleon: binding energy divided by the mass number \(A\).
Nuclei around iron-56 have the highest binding energy per nucleon.
The activity of a radioactive source falls from \(8.0\times10^{4}\,\mathrm{Bq}\) to \(1.0\times10^{4}\,\mathrm{Bq}\) in \(36\) hours. Calculate the half-life of the source and the decay constant.
\(A = A_{0}e^{-\lambda t} \Rightarrow 1/8 = e^{-\lambda\times36} \Rightarrow \ln8 = \lambda\times36 \Rightarrow \lambda = 0.0578\,\mathrm{h}^{-1}\).
\(T_{1/2} = \ln2/\lambda = 0.693/0.0578 = 12.0\,\mathrm{h}\).
A nucleus of \(^{235}\mathrm{U}\) absorbs a neutron and fissions into \(^{141}\mathrm{Ba}\) and \(^{92}\mathrm{Kr}\) plus neutrons. Write a balanced equation for the reaction and calculate the number of neutrons released.
\(^{235}_{92}\mathrm{U} + ^{1}_{0}\mathrm{n} \to ^{141}_{56}\mathrm{Ba} + ^{92}_{36}\mathrm{Kr} + 3^{1}_{0}\mathrm{n}\).
Mass number: \(235+1 = 141+92 + n \Rightarrow n = 3\).
The mass of a \(^{4}\mathrm{He}\) nucleus is \(4.0015\,\mathrm{u}\) and the mass of a proton is \(1.0073\,\mathrm{u}\), of a neutron \(1.0087\,\mathrm{u}\). Calculate the binding energy of the helium nucleus in MeV. (\(1\,\mathrm{u} = 931.5\,\mathrm{MeV}/c^{2}\).)
Mass of \(2p + 2n = 2\times1.0073 + 2\times1.0087 = 4.0320\,\mathrm{u}\).
Mass defect \(= 4.0320 - 4.0015 = 0.0305\,\mathrm{u}\).
Binding energy \(= 0.0305\times931.5 = 28.4\,\mathrm{MeV}\).
Explain the difference between nuclear fission and nuclear fusion. State which process releases energy for nuclei lighter than iron and which for nuclei heavier than iron.
Fission: a heavy nucleus splits into two lighter nuclei (plus neutrons). Fusion: two light nuclei combine to form a heavier nucleus.
Fusion of nuclei lighter than iron increases binding energy per nucleon and releases energy. Fission of nuclei heavier than iron also increases binding energy per nucleon and releases energy.
A sample contains \(2.5\times10^{16}\) radioactive nuclei with a decay constant of \(1.2\times10^{-6}\,\mathrm{s}^{-1}\). Calculate the activity of the sample and the time taken for the activity to fall to \(1.0\times10^{9}\,\mathrm{Bq}\).
\(A_{0} = \lambda N = 1.2\times10^{-6}\times2.5\times10^{16} = 3.0\times10^{10}\,\mathrm{Bq}\).
\(A = A_{0}e^{-\lambda t} \Rightarrow 1.0\times10^{9}/3.0\times10^{10} = e^{-\lambda t} \Rightarrow \ln(30) = \lambda t\).
\(t = 3.40/(1.2\times10^{-6}) = 2.83\times10^{6}\,\mathrm{s}\) (\(\approx 32.8\) days).
In a fusion reaction \(^{2}_{1}\mathrm{H} + ^{3}_{1}\mathrm{H} \to ^{4}_{2}\mathrm{He} + ^{1}_{0}\mathrm{n}\). The masses are: \(^{2}\mathrm{H} = 2.0141\,\mathrm{u}\), \(^{3}\mathrm{H} = 3.0160\,\mathrm{u}\), \(^{4}\mathrm{He} = 4.0026\,\mathrm{u}\), neutron \(= 1.0087\,\mathrm{u}\). Calculate the energy released in MeV.
Mass of reactants \(= 2.0141 + 3.0160 = 5.0301\,\mathrm{u}\).
Mass of products \(= 4.0026 + 1.0087 = 5.0113\,\mathrm{u}\).
Mass defect \(= 0.0188\,\mathrm{u}\); energy \(= 0.0188\times931.5 = 17.5\,\mathrm{MeV}\).
A radioactive isotope is used as a tracer. Its half-life is \(6.0\) hours. A patient is injected with a sample of activity \(2.4\times10^{6}\,\mathrm{Bq}\). Calculate the activity after \(18\) hours, ignoring biological removal. Explain why a short half-life is desirable for a medical tracer.
Three half-lives: activity \(= 2.4\times10^{6}/8 = 3.0\times10^{5}\,\mathrm{Bq}\).
Short half-life means the activity (and therefore the radiation dose to the patient) falls quickly after the diagnostic procedure is complete, reducing the long-term radiation exposure.