PROBLEM \(\PageIndex{1}\) Write the following isotopes in nuclide notation (e.g., " \(\ce{^{14}_6C}\) ") oxygen-14 copper-70 tantalum-175 francium-217 Answer a \(\ce{^{14}_8O}\) Answer b \(\ce{^{70}_29Cu}\) Answer c \(\ce{^{175}_73Ta}\) Answer d \(\ce{^{217}_87Fr}\) PROBLEM \(\PageIndex{2}\) For the following isotopes that have missing information, fill in the missing information to complete the notation \(\ce{^{34}_{14}X}\) \(\ce{^{36}_P}\) \(\ce{^{57}_{X}Mn}\) \(\ce{^{121}_{56}X}\) Answer a \(\ce{^{34}_{14}Si}\) Answer b \(\ce{^{36}_{15}P}\) Answer c \(\ce{^{57}_{25}Mn}\) Answer d \(\ce{^{121}_{56}Ba}\) PROBLEM \(\PageIndex{3}\) Write the nuclide notation, including charge if applicable, for atoms with the following characteristics: 25 protons, 20 neutrons, 24 electrons 45 protons, 24 neutrons, 43 electrons 53 protons, 89 neutrons, 54 electrons 97 protons, 146 neutrons, 97 electrons Answer a \(\ce{^{45}_{25}Mn^{+1}}\) Answer b \(\ce{^{69}_{45}Rh^{+2}}\) Answer c \(\ce{^{142}_{53}I^{−1}}\) Answer d \(\ce{^{247}_{97}Bk}\) PROBLEM \(\PageIndex{4}\) Which of the following nuclei lie within the band of stability? chlorine-37 calcium-40 204 Bi 56 Fe 206 Pb 211 Pb 222 Rn carbon-14 Answer (a), (b), (c), (d), and (e) PROBLEM \(\PageIndex{5}\) Which of the following nuclei lie within the band of stability? argon-40 oxygen-16 122 Ba 58 Ni 205 Tl 210 Tl 226 Ra magnesium-24 Answer (b), (e - very close), and (h) Click here for a video of the solution PROBLEM \(\PageIndex{6}\) Write a brief description or definition of each of the following: nucleon α particle β particle positron γ ray nuclide mass number atomic number Answer a collective term for protons and neutrons in a nucleus Answer b (α or \(\ce{^4_2He}\) or \(\ce{^4_2α}\)) high-energy helium nucleus; a helium atom that has lost two electrons and contains two protons and two neutrons Answer c ( \(β\) or \(\ce{^0_{-1}e}\) or \(\ce{^0_{-1}β}\) ) high-energy electron Answer d antiparticle to the electron; it has identical properties to an electron, except for having the opposite (positive) charge Answer e (γ or \(\ce{^0_0γ}\) ) short wavelength, high-energy electromagnetic radiation that exhibits wave-particle duality Answer f nucleus of a particular isotope Answer g sum of the numbers of neutrons and protons in the nucleus of an atom Answer h number of protons in the nucleus of an atom PROBLEM \(\PageIndex{7}\) Complete each of the following equations by adding the missing species: \(\ce{^{27}_{13}Al + ^4_2He⟶\:? + ^1_0n}\) \(\ce{^{239}_{94}Pu +\, ? ⟶ ^{242}_{96}Cm + ^1_0n}\) \(\ce{^{14}_7N + ^4_2He⟶\:? + ^1_1H}\) \(\ce{^{235}_{92}U⟶\:? + ^{135}_{55}Cs + 4^1_0n}\) Answer a \(\ce{^{27}_{13}Al + ^4_2He⟶ ^{30}_{15}P + ^1_0n}\) Answer b \(\ce{Pu + He^2 ⟶ ^{242}_{96}Cm + ^1_0n}\) Answer c \(\ce{^{14}_7N + ^4_2He⟶ ^{17}_8O + ^1_1H}\) Answer d \(\ce{^{235}_{92}U⟶ ^{96}_{37}Rb + ^{135}_{55}Cs + 4^1_0n}\) PROBLEM \(\PageIndex{8}\) Complete each of the following equations: \(\ce{^7_3Li +\, ?⟶2^4_2He}\) \(\ce{^{14}_6C⟶ ^{14}_7N +\, ?}\) \(\ce{^{27}_{13}Al + ^4_2He⟶\,? + ^1_0n}\) \(\ce{^{250}_{96}Cm ⟶\, ? + ^{98}_{38}Sr + 4^1_0n}\) Answer a \(\ce{^7_3Li +^1_1H ⟶2^4_2He}\) Answer b \(\ce{^{14}_6C⟶ ^{14}_7N +^0_{-1}e}\) Answer c \(\ce{^{27}_{13}Al + ^4_2He⟶^30_15P + ^1_0n}\) Answer d \(\ce{^{250}_{96}Cm ⟶^148_{58}Ce + ^{98}_{38}Sr + 4^1_0n}\) Click here to see a video of the solution PROBLEM \(\PageIndex{9}\) Write a balanced equation for each of the following nuclear reactions: the production of 17 O from 14 N by α particle bombardment the production of 14 C from 14 N by neutron bombardment the production of 233 Th from 232 Th by neutron bombardment the production of 239 U from 238 U by \(\ce{^2_1H}\) bombardment Answer a \(\ce{^{14}_7N + He^2 ⟶ ^{17}_8O + ^1_1H}\) Answer b \(\ce{^{14}_7N + ^1_0n⟶ ^{14}_6N + ^1_1H}\) Answer c \(\ce{^{232}_{90}Th + ^1_0n⟶ ^{233}_{90}Th}\) Answer d \(\ce{^{238}_{92}U + ^2_1H⟶ ^{239}_{92}U + ^1_1H}\) PROBLEM \(\PageIndex{10}\) Technetium-99 is prepared from 98 Mo. Molybdenum-98 combines with a neutron to give molybdenum-99, an unstable isotope that emits a β particle to yield an excited form of technetium-99, represented as 99 Tc * . This excited nucleus relaxes to the ground state, represented as 99 Tc, by emitting a γ ray. The ground state of 99 Tc then emits a β particle. Write the equations for each of these nuclear reactions. Answer \(\ce{^{98}_{42}Mo + ^1_0n \rightarrow ^{99}_{42}Mo}\) \(\ce{^{99}_{42}Mo} \rightarrow ^0_{-1}e + ^{99}_{43}Tc^*\) \(\ce{^{99}_{43}Tc}^* \rightarrow ^0_{0}\gamma + ^{99}_{43}Tc\) \(\ce{^{99}_{43}Tc \rightarrow ^0_{-1}e + ^{99}_{44}Ru}\) Click here to see a video of the solution PROBLEM \(\PageIndex{11}\) What changes occur to the atomic number and mass of a nucleus during each of the following decay scenarios? an α particle is emitted a β particle is emitted γ radiation is emitted a positron is emitted an electron is captured Answer a Since an α particle is the same as a \(\ce{^4_2He}\) nucleus, the mass number will decrease by 4 and the atomic number will decrease by 2. Answer b Since a β particle is the same as \(\ce{^0_{-1}e}\), the mass number will not change but the atomic number will increase by 1. Answer c Since a γ ray has no mass (it is energy) the mass number and atomic number do not change. Answer d A positron is the opposite of a β particle, it is \(\ce{^0_{+1}e}\), the mass number will not change but the atomic number will decrease by 1 Answer e Electron capture has the same effect on the nucleus as positron emission: The atomic number is decreased by one and the mass number does not change. PROBLEM \(\PageIndex{12}\) What is the change in the nucleus that results from the following decay scenarios? emission of a β particle emission of a β + particle capture of an electron Answer a conversion of a neutron to a proton: \(\ce{^1_0n ⟶ ^1_1p + ^0_{+1}e}\) Answer b conversion of a proton to a neutron; the positron has the same mass as an electron and the same magnitude of positive charge as the electron has negative charge when the n:p ratio of a nucleus is too low, a proton is converted into a neutron with the emission of a positron: \(\ce{^1_1p ⟶ ^1_0n + ^0_{+1}e}\) Answer c In a proton-rich nucleus, an inner atomic electron can be absorbed. In simplest form, this changes a proton into a neutron: \(\ce{^1_1p + ^0_{-1}e ⟶ ^1_0p}\) PROBLEM \(\PageIndex{13}\) Explain how unstable heavy nuclides (atomic number > 83) may decompose to form nuclides of greater stability if (a) they are below the band of stability and (b) they are above the band of stability Answer Nuclei below the band of stability will undergo positron decay, while those above the band of stability will undergo beta decay. Heavy nuclei past the band of stability will undergo alpha decay PROBLEM \(\PageIndex{14}\) Which of the following nuclei is most likely to decay by positron emission? Explain your choice. chromium-53 manganese-51 iron-59 Answer Manganese-51 is most likely to decay by positron emission. The n:p ratio for Cr-53 is \(\dfrac{29}{24}\) = 1.21; for Mn-51, it is \(\dfrac{26}{25}\) = 1.04; for Fe-59, it is \(\dfrac{33}{26}\) = 1.27. Positron decay occurs when the n:p ratio is low. Mn-51 has the lowest n:p ratio and therefore is most likely to decay by positron emission. Besides, \(\ce{^{53}_{24}Cr}\) is a stable isotope, and \(\ce{^{59}_{26}Fe}\) decays by beta emission. PROBLEM \(\PageIndex{15}\) The following nuclei do not lie in the band of stability. How would they be expected to decay? Explain your answer. \(\ce{^{34}_{15}P}\) \(\ce{^{239}_{92}U}\) \(\ce{^{38}_{20}Ca}\) \(\ce{^3_1H}\) \(\ce{^{245}_{94}Pu}\) Answer a Above the band of stability, beta decay is expected Answer b Beyond the band of stability, heavy nuclei undergo alpha decay Answer c Below the band of stability, positron decay is expected Answer d Above the band of stability, beta decay is expected Answer e Beyond the band of stability, heavy nuclei undergo alpha decay PROBLEM \(\PageIndex{16}\) Write a nuclear reaction for each step in the formation of \(\ce{^{218}_{84}Po}\) from \(\ce{^{238}_{92}U}\), which proceeds by a series of decay reactions involving the step-wise emission of α, β, β, α, α, α, α particles, in that order. Answer \(\ce{^{238}_{92}U⟶ ^{234}_{90}Th + ^4_2He}\) \(\ce{^{234}_{90}Th⟶ ^{234}_{91}Pa + ^0_{-1}e}\) \(\ce{^{234}_{91}Pa⟶^{234}_{92}U + ^0_{-1}e}\) \(\ce{^{234}_{92}U⟶ ^{230}_{90}Th + ^4_2He}\) \(\ce{^{230}_{90}Th⟶ ^{226}_{88}Ra + ^4_2He } \) \( \ce {^{226}_{88}Ra⟶^{222}_{86}Rn + ^4_2He}\) \(\ce{^{222}_{86}Rn⟶^{218}_{84}Po + ^4_2He}\) PROBLEM \(\PageIndex{17}\) Write a nuclear reaction for each step in the formation of \(\ce{^{208}_{82}Pb}\) from \(\ce{^{228}_{90}Th}\), which proceeds by a series of decay reactions involving the step-wise emission of α, α, α, α, β, β, α particles, in that order. Click here for a video solution. Answer \(\ce{^{228}_{90}Th \rightarrow ^4_2He + ^{224}_{88}Ra}\) \(\ce{^{224}_{88}Ra \rightarrow ^4_2He + ^{220}_{86}Rn}\) \(\ce{^{220}_{86}Th \rightarrow ^4_2He + ^{216}_{84}Po}\) \(\ce{^{216}_{84}Po \rightarrow ^4_2He + ^{212}_{82}Pb}\) \(\ce{^{212}_{82}Pb \rightarrow ^0_{-1}e + ^{212}_{83}Bi}\) \(\ce{^{212}_{83}Bi \rightarrow ^0_{-1}e + ^{212}_{84}Po}\) \(\ce{^{212}_{84}Po \rightarrow ^4_2He + ^{208}_{82}Pb}\) Click here to see a video of the solution Have a video solution request? Let your professors know here. ***Please know that you are helping future students - videos will be made in time for next term's class. Feedback Think one of the answers above is wrong? Let us know here .