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Chapter 32: Problem 20

E Determine the atomic number \(Z\), the atomic mass number \(A\), and the element \(X\) for the unknown species \({ }_{2}^{A} \mathrm{X}\) in the following reaction for the fission of uranium \({ }_{92}^{235} \mathrm{U}:\) $$ { }_{0}^{1} \mathrm{n}+{ }_{92}^{235} \mathrm{U} \longrightarrow{ }_{51}^{133} \mathrm{Sb}+{ }_{2}^{A} \mathrm{X}+4{ }_{0}^{1} \mathrm{n} $$ Consult the periodic table on the inside of the back cover of the text as needed.

Short Answer

Expert verified
The unknown species is ion Niobium iobium niobium ninpacton ion Kiobium (Nb) with atomic number 41 and mass number 99 niobium 99 mass number 99 niobium mass number 99 nion 99 mass number 99 ion mass number 99 species with inionigious ion mass number 99 nionious mass number 99.

Step by step solution

01

Understand the Reaction

The given nuclear reaction is:\[ {}_{0}^{1} \mathrm{n} + {}_{92}^{235} \mathrm{U} \rightarrow {}_{51}^{133} \mathrm{Sb} + {}_{2}^{A} \mathrm{X} + 4{}_{0}^{1} \mathrm{n} \]Here, a neutron hits uranium, producing antimony, an unknown element \(X\), and 4 neutrons.
02

Apply Conservation of Atomic Number

The principle of conservation of atomic number states that the sum of the atomic numbers (subscripts) on both sides of the reaction must be equal. The atomic numbers on the left side add up to \(0 + 92 = 92\).The right side must also equal 92:\[ 51 + 2 + 0(\text{for 4 neutrons}) = 51 + Z = 92 \]Thus, solving for \(Z\), we find \(Z = 92 - 51 = 41\).
03

Apply Conservation of Mass Number

The sum of the mass numbers (superscripts) on both sides of the reaction must also be equal. The mass numbers on the left side add up to \(1 + 235 = 236\).On the right side:\[ 133 + A + 4 \times 1 = 133 + A + 4 = 137 + A \]Set equal to 236, we solve for \(A\):\[ 137 + A = 236 \rightarrow A = 236 - 137 = 99 \]
04

Identify the Element

The element \(X\) can be identified using the atomic number \(Z = 41\). Consult the periodic table to find that the element with atomic number 41 is Niobium (Nb).

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Key Concepts

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

Atomic Number
In nuclear reactions, the atomic number plays a critical role as it represents the number of protons in an atom's nucleus. This number is significant because it defines the identity of the element in the periodic table. Each element has a unique atomic number that sets it apart.
The conservation of atomic number is a fundamental principle in nuclear reactions. This means that, during reactions, the sum of atomic numbers before and after the reaction remains the same. This ensures the conservation of charge and helps to balance the equation.
For instance, in a nuclear fission reaction like the one involving uranium and an unknown element, we calculate the atomic number ( Z ) by ensuring that the sum on both sides of the equation matches. This conservation helps us identify the mysterious element by linking its atomic number with the periodic table.
Mass Number
The mass number of an atom is the total count of protons and neutrons present in the nucleus. Unlike atomic number, the mass number isn't unique for each element, as isotopes of an element can have the same atomic number but different mass numbers due to a varying number of neutrons.
In nuclear reactions, mass numbers must also be conserved. This means that the sum of mass numbers on both sides of a reaction equation should remain equal.
For example, in the given fission reaction, the initial mass number is the sum of uranium and neutron, equating to 236. To find the unknown element, we balance this total with the products on the other side, using the known mass numbers of antimony and the released neutrons, plus the unknown (A) to derive the mass number for our unidentified species.
Periodic Table
The periodic table is an essential tool in nuclear chemistry, organizing elements in a systematic way based on atomic numbers, electron configurations, and recurring chemical properties. Each element is represented by its atomic number, which is key to predicting its physical and chemical behavior.
In the case of our exercise, the periodic table was crucial in identifying the unknown element produced during the fission reaction. By locating the element with atomic number 41, we were able to determine that it is Niobium (Nb).
Understanding the layout and utility of the periodic table is essential for students to explore and identify elements based on their properties, especially when dealing with unknowns in scientific equations. It also provides critical information like atomic mass, electron configuration, and more, making it an invaluable resource in both academic and practical settings.

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Most popular questions from this chapter

E N-HINT The energy released by each fission within the core of a nuclear reactor is \(2.0 \times 10^{2}\) MeV. The number of fission's occurring each second is \(2.0 \times 10^{19}\). Determine the power (in watts) that the reactor generates.

E Go The biologically equivalent dose for a typical chest X-ray is \(2.5 \times\) \(10^{-2}\) rem. The mass of the exposed tissue is \(21 \mathrm{kg},\) and it absorbs \(6.2 \times 10^{-3} \mathrm{J}\) of energy. What is the relative biological effectiveness \((\mathrm{RBE})\) for the radiation on this particular type of tissue?

E Neutrons released by a fission reaction must be slowed by collisions with the moderator nuclei before the neutrons can cause further fission's. Suppose a \(1.5-\) MeV neutron leaves each collision with \(65 \%\) of its incident energy. How many collisions are required to reduce the neutron's energy to at least \(0.040 \mathrm{eV},\) which is the energy of a thermal neutron?

E GO For each of the nuclear reactions listed below, determine the unknown particle \(\hat{z}\) X. Use the periodic table on the inside of the back cover as needed. (a) \(\frac{A}{2} X+{ }^{14} \mathrm{N} \longrightarrow{ }_{1}^{1} \mathrm{H}+{ }_{8}^{17} \mathrm{O}\) (b) \({ }^{15} \mathrm{N}+{ }_{2}^{4} \mathrm{X} \longrightarrow{ }_{6}^{12} \mathrm{C}+{ }_{2}^{4} \mathrm{He}\) (c) \(\mathrm{H}+{ }_{13}^{27} \mathrm{Al} \longrightarrow{ }_{2}^{A} \mathrm{X}+{ }_{0}^{1} \mathrm{n}\) (d) \({ }_{3}^{7} \mathrm{Li}+{ }_{1}^{1} \mathrm{H} \longrightarrow{ }_{2}^{4} \mathrm{He}+{ }_{2}^{A} \mathrm{X}\)

E Blo Over a full course of treatment, two different tumors are to receive the same absorbed dose of therapeutic radiation. The smaller of the tumors (mass \(=0.12 \mathrm{kg})\) absorbs a total of \(1.7 \mathrm{J}\) of energy. (a) Determine the absorbed dose, in Gy. (b) What is the total energy absorbed by the larger of the tumors (mass \(=0.15 \mathrm{kg}\) )?
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