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Chapter 19: Problem 5

Write balanced nuclear equations for (a) the alpha emission resulting in the formation of \(\mathrm{Pa}-233\). (b) the loss of a positron by \(\mathrm{Y}-85\). (c) the fusion of two C-12 nuclei to give sodium-23 and another particle. (d) the fission of Pu-239 to give tin-130, another nucleus, and an excess of two neutrons.

Short Answer

Expert verified
Question: Write balanced nuclear equations for the following processes: a) Alpha emission resulting in the formation of Pa-233 b) Loss of a positron by Y-85 c) Fusion of two C-12 nuclei to give sodium-23 and another particle d) Fission of Pu-239 to give Sn-130, another nucleus, and an excess of two neutrons Answer: a) ^{237}_{93}Np → ^{233}_{91}Pa + ^4_2He b) ^{85}_{39}Y → ^{85}_{38}Sr + ^0_{-1}β c) ^{12}_6C + ^{12}_6C → ^{23}_{11}Na + ^1_1p d) ^{239}_{94}Pu → ^{130}_{50}Sn + ^{107}_{44}Ru + 2×^1_0n

Step by step solution

01

Identify the reactant nucleus and the emission particle

The original reactant nucleus is the one which undergoes alpha emission to produce Pa-233. An alpha particle consists of 2 protons and 2 neutrons, represented as ^4_2He.
02

Determine the reactant nucleus and write the balanced nuclear equation

To find the reactant nucleus, we need to consider the conservation of mass number and atomic number. Let's denote Reactant nucleus as X. The mass number and atomic number of Pa (Protactinium) are 91 and 233 respectively. So, the equation can be written as: \[^A_ZX \rightarrow ^{233}_{91}\mathrm{Pa} + ^4_2\mathrm{He}\] By applying conservation of mass number: \[A = 233 + 4 \Rightarrow A = 237\] By applying conservation of atomic number: \[Z = 91 + 2 \Rightarrow Z = 93\] So, the reactant nucleus is ^237_93X, which is an isotope of Neptunium (Np). The balanced nuclear equation is: \[^{237}_{93}\mathrm{Np} \rightarrow ^{233}_{91}\mathrm{Pa} + ^4_2\mathrm{He}\] #b) Loss of a positron by Y-85#
03

Identify the original nucleus and the emission particle

The original nucleus is Y-85 (Yttrium), and the particle emitted is a positron, denoted as ^0_1e or ^0_{-1}\beta.
04

Determine the product nucleus and write the balanced nuclear equation

To find the product nucleus, we need to consider the conservation of mass number and atomic number. The mass number and the atomic number of Y are 39 and 85 respectively. So, the equation can be written as: \[^{85}_{39}\mathrm{Y} \rightarrow ^A_ZX + ^0_{-1}\beta\] By applying conservation of mass number: \[85 = A\] By applying conservation of atomic number: \[39 = Z - (-1) \Rightarrow Z = 38\] So, the product nucleus is ^{85}_{38}X, which is an isotope of Sr (Strontium). The balanced nuclear equation is: \[^{85}_{39}\mathrm{Y} \rightarrow ^{85}_{38}\mathrm{Sr} + ^0_{-1}\beta\] #c) Fusion of two C-12 nuclei to give sodium-23 and another particle#
05

Write the reactants and known product in the nuclear equation

The reactants are two C-12 nuclei, and the known product is Na-23 (Sodium). The equation can be written as: \[^{12}_6\mathrm{C} + ^{12}_6\mathrm{C} \rightarrow ^{23}_{11}\mathrm{Na} + ^A_ZX\]
06

Determine the missing particle and write the balanced nuclear equation

To find the missing particle, we need to consider the conservation of mass number and atomic number. By applying conservation of mass number: \[12 + 12 = 23 + A \Rightarrow A = 1\] By applying conservation of atomic number: \[6 + 6 = 11 + Z \Rightarrow Z = 1\] So, the missing particle is ^1_1X, which is a proton (p). The balanced nuclear equation is: \[^{12}_6\mathrm{C} + ^{12}_6\mathrm{C} \rightarrow ^{23}_{11}\mathrm{Na} + ^1_1\mathrm{p}\] #d) Fission of Pu-239 to give Sn-130, another nucleus, and an excess of two neutrons#
07

Write the reactants and known products in the nuclear equation

The reactant is Pu-239 (Plutonium), and the known products are Sn-130 (Tin) and two neutrons. The equation can be written as: \[^{239}_{94}\mathrm{Pu} \rightarrow ^{130}_{50}\mathrm{Sn} + ^A_ZX + 2\times^1_0\mathrm{n}\]
08

Determine the other nucleus and write the balanced nuclear equation

To find the other nucleus, we need to consider the conservation of mass number and atomic number. By applying conservation of mass number: \[239 = 130 + A + 2 \Rightarrow A = 107\] By applying conservation of atomic number: \[94 = 50 + Z \Rightarrow Z = 44\] So, the other nucleus is ^{107}_{44}X, which is an isotope of Ru (Ruthenium). The balanced nuclear equation is: \[^{239}_{94}\mathrm{Pu} \rightarrow ^{130}_{50}\mathrm{Sn} + ^{107}_{44}\mathrm{Ru} + 2\times^1_0\mathrm{n}\]

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Most popular questions from this chapter

Classify the following statements as true or false. If false, correct the statement to make it true. (a) The mass number increases in beta emission. (b) A radioactive species with a large rate constant, \(k\), decays very slowly. (c) Fusion gives off less energy per gram of fuel than fission.

Consider the fission reaction $${ }_{0}^{1} n+{ }_{92}^{235} \mathrm{U} \longrightarrow{ }_{3}{ }_{3}^{89} \mathrm{Rb}+{ }_{55}^{144} \mathrm{Ce}+3_{-1}^{0} e+3{ }_{0}^{1} n$$ How many liters of octane, \(\mathrm{C}_{8} \mathrm{H}_{18}\), the primary component of gasoline, must be burned to \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(g)\) to produce as much energy as the fission of one gram of U-235 fuel? Octane has a density of \(0.703 \mathrm{~g} / \mathrm{mL} ;\) its heat of formation is \(-249.9 \mathrm{~kJ} / \mathrm{mol}\).

Consider the fusion of boron- 10 with an alpha particle. The products of the fusion are carbon-13 and a proton. (a) Write a nuclear reaction for this process. (b) How much energy is released when \(1.00 \mathrm{~g}\) of \(\mathrm{B}-10\) is fused with an \(\alpha\) -particle?

Balance the following equations by filling in the blanks. (a) \({ }_{92}^{235} \mathrm{U}+{ }_{0} n \longrightarrow{ }_{54}^{137}=2{ }_{0}^{1} n+\) (b) \({ }_{90}^{232} \mathrm{Th}+{ }_{6}^{12}\) \(\longrightarrow\) \(1 . n+{ }_{96}^{240} \mathrm{Cm}\) (c) \({ }_{2}^{4} \mathrm{He}+{ }_{42}^{96} \mathrm{Mo} \longrightarrow{ }_{43}^{100}\) (d) \(+{ }_{1}^{2} \mathrm{H} \longrightarrow{ }_{84}^{210}+{ }_{0}^{1} n\)

Smoke detectors contain a small amount of americium-241. Its decay product is neptunium-237. Identify the emission from americium-241.
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