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Complete and balance the nuclear equations for the following fission reactions: (a) \({ }_{94}^{239} \mathrm{Pu}+{ }_{0}^{1} \mathrm{n} \longrightarrow{ }_{52}^{137} \mathrm{Te}+{ }_{42}^{100} \mathrm{Mo}+\) (b) \({ }_{100}^{256} \mathrm{Fm}+{ }_{0}^{1} \mathrm{n} \longrightarrow{ }_{46}^{113} \mathrm{Pd}+{ }_{-}+4{ }_{0}^{1} \mathrm{n}\)

Short Answer

Expert verified
(a) Add three neutrons (b) Add xenon (Xe) with four neutrons.

Step by step solution

01

Determine Missing Information in Equation (a)

First, analyze the given reaction:\[{ }_{94}^{239} \mathrm{Pu}+{ }_{0}^{1} \mathrm{n} \longrightarrow{ }_{52}^{137} \mathrm{Te}+{ }_{42}^{100} \mathrm{Mo}+?\]Calculate the total atomic and mass numbers on both sides. On the left side, the atomic number is 94 (Pu) and the mass number is 240 (239 + 1). On the right, the atomic numbers add up to 94 (52 from Te and 42 from Mo), and the mass numbers add up to 237 (137 + 100). The missing particle needs 3 mass units and 0 atomic units.
02

Identify Missing Particle for Equation (a)

The missing particle must have mass number 3 and atomic number 0 to balance the equation. This corresponds to three neutrons:\[{ }_{0}^{1} \mathrm{n}+{ }_{0}^{1} \mathrm{n}+{ }_{0}^{1} \mathrm{n} \text{ or simply } 3{ }_{0}^{1} \mathrm{n}\].Thus, the balanced equation is:\[{ }_{94}^{239} \mathrm{Pu}+{ }_{0}^{1} \mathrm{n} \longrightarrow { }_{52}^{137} \mathrm{Te}+{ }_{42}^{100} \mathrm{Mo}+3{ }_{0}^{1} \mathrm{n}\].
03

Equate the Atomic and Mass Numbers for Equation (b)

Examine the second reaction given:\[{ }_{100}^{256} \mathrm{Fm}+{ }_{0}^{1} \mathrm{n} \longrightarrow { }_{46}^{113} \mathrm{Pd}+{ ? }+4{ }_{0}^{1} \mathrm{n}\]Address the ambiguity marked as { ? }. The total atomic number on the left is 100 and the mass number is 257 (256 + 1). On the right side, including the four neutrons, the atomic number is 46 and the mass number is 117 (113 from Pd plus 4 neutrons). The missing particle must have 54 atomic units and 140 mass units.
04

Identify and Complete Equation (b)

The missing particle must have atomic number 54 and mass number 140, which corresponds to the element xenon (Xe):\[{ }_{54}^{140} \mathrm{Xe}\].Therefore, the balanced equation is:\[{ }_{100}^{256} \mathrm{Fm}+{ }_{0}^{1} \mathrm{n} \longrightarrow { }_{46}^{113} \mathrm{Pd}+{ }_{54}^{140} \mathrm{Xe}+4{ }_{0}^{1} \mathrm{n}\].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Nuclear Equations
Nuclear equations are similar to chemical equations but involve nuclear reactions rather than chemical ones. They depict the process of nuclear fission, where an atomic nucleus splits into two or more smaller nuclei, along with other particles. In nuclear equations, both atomic number and mass number must be balanced to represent the conservation of charge and mass. This means that the sum of atomic numbers on the reactant side must equal the sum on the product side, and the same goes for mass numbers. Each element is indicated by its chemical symbol, and specific isotopes are identified by numerical superscripts and subscripts, representing the mass number and atomic number respectively. Understanding the structure and balancing of nuclear equations is crucial in the field of nuclear physics and chemistry.
Atomic Number
The atomic number, represented as a subscript in nuclear equations, is a fundamental property of an element. It defines the number of protons in the nucleus of every atom belonging to that element. For instance, plutonium (_{94}^{239} ext{Pu}) has an atomic number of 94, meaning each plutonium atom contains 94 protons. This number is vital when balancing nuclear equations because it ensures the conservation of electrical charge. During nuclear reactions, although elements may transform into different isotopes or even entirely different elements, the overall atomic balance is preserved, maintaining the integrity of nuclear equations.
Mass Number
Mass number is prominently displayed as a superscript in nuclear reactions. This number tells us the total number of protons and neutrons located within an atom's nucleus. It is crucial to note that the mass number differs from the atomic mass, which includes slight decimal variations due to binding energy and isotopic distribution. Within the context of nuclear fission, balancing the mass number on both sides of a nuclear equation is key. For example, in the equation involving fermium (_{100}^{256} ext{Fm}), the mass number on the reactant side needs to equal the total mass numbers on the product side, taking into account additional particles such as neutrons or missing atomic fragments. This ensures the principle of mass conservation is adhered to, even when individual atoms undergo significant transformations.
Neutrons
Neutrons are neutral particles found in the nucleus of an atom alongside protons. Unique to neutrons is their lack of electric charge, which is why their atomic number is always 0 in nuclear equations. As they contribute to the mass number, however, they have a mass number of 1. In nuclear fission reactions, neutrons play a pivotal role. They can initiate the process by colliding with a heavy nucleus, causing it to split and release further neutrons, thus facilitating a chain reaction. In the exercises outlined, neutrons are explicitly shown, either as initial participants in the reaction or as products. For example, when plutonium and a neutron interact, they produce multiple neutrons, confirming the critical position neutrons hold in sustaining nuclear reactions.

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Most popular questions from this chapter

The energy from solar radiation falling on Earth is \(1.07 \times 10^{16} \mathrm{~kJ} / \mathrm{min} .(\mathbf{a})\) How much loss of mass from the Sun occurs in one day from just the energy falling on Earth? (b) If the energy released in the reaction $$ { }^{235} \mathrm{U}+{ }_{0}^{1} \mathrm{n} \longrightarrow{ }_{56}^{141} \mathrm{Ba}+{ }_{36}^{92} \mathrm{Kr}+3{ }_{0}^{1} \mathrm{n} $$ \(\left(\begin{array}{c}235 \\ \mathrm{U} \text { nuclear mass, } 234.9935 \mathrm{u} ;{ }^{141} \mathrm{Ba} \text { nuclear mass, }\end{array}\right.\) \(140.8833 \mathrm{u} ;{ }^{92} \mathrm{Kr}\) nuclear mass, \(91.9021 \mathrm{u}\) ) is taken as typical of that occurring in a nuclear reactor, what mass of uranium- 235 is required to equal \(0.10 \%\) of the solar energy that falls on Earth in 1.0 day?

Which of the following statements best explains why alpha emission is relatively common, but proton emission is extremely rare? (a) Alpha particles are very stable because of magic numbers of protons and neutrons. (b) Alpha particles occur in the nucleus. (c) Alpha particles are the nuclei of an inert gas. (d) An alpha particle has a higher charge than a proton.

Iodine-131 is a convenient radioisotope to monitor thyroid activity in humans. It is a beta emitter with a half-life of 8.02 days. The thyroid is the only gland in the body that uses iodine. A person undergoing a test of thyroid activity drinks a solution of NaI, in which only a small fraction of the iodide is radioactive. (a) Why is NaI a good choice for the source of iodine? (b) If a Geiger counter is placed near the person's thyroid (which is near the neck) right after the sodium iodide solution is taken, what will the data look like as a function of time? (c) A normal thyroid will take up about \(12 \%\) of the ingested iodide in a few hours. How long will it take for the radioactive iodide taken up and held by the thyroid to decay to \(0.01 \%\) of the original amount?

Iodine-131 is used as a nuclear medicine to treat hyperthyroidism. The half- life of \({ }^{131}\) I is 8.04 days. How long will it take for a \(500 \mathrm{mg}\) sample of \({ }^{131}\) I to decay into \(1 \%\) of its original mass?

How much time is required for a 5.00-g sample of \({ }^{233} \mathrm{~Pa}\) to decay to \(0.625 \mathrm{~g}\) if the half-life for the beta decay of \({ }^{233} \mathrm{~Pa}\) is 27.4 days?

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