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Chapter 15: Problem 24

A particular fission of \({ }^{235} \mathrm{U}+\mathrm{n}(\) total \(A=236)\) breaks up. One fragment has \(Z=54\) and \(N=86\), making it \({ }^{140} \mathrm{Xe} .\) If no extra neutrons are produced in this event, what must the other fragment be, so all numbers add up? Refer to a periodic table (e.g., Fig. B.1; p. 375 ) to learn which element has the corresponding \(Z\) value, and express the result in the notation \({ }^{\mathrm{A}} \mathrm{X}\).

Short Answer

Expert verified
Answer: The unknown fragment has an atomic number (Z) of 38 and a mass number (A) of 96, and its corresponding element is Strontium (Sr). Therefore, the unknown fragment is \({ }^{96}Sr\).

Step by step solution

01

Calculate the Sum of Mass Numbers of Reactants and Products

First, we need to calculate the sum of mass numbers of the reactants and compare it with the sum of mass numbers in the products. The reactants are Uranium-235 (\({ }^{235}U\)) and a neutron (n). Sum of Mass Numbers in Reactants = Mass number of \({ }^{235}U\) + Mass number of neutron = 235 + 1 = 236
02

Calculate the Sum of Mass Numbers of Known Fragments

Next, calculate the mass numbers of the known fragments. In the products, we are given that one fragment is Xenon-140 (\({ }^{140}Xe\)). Sum of Mass Numbers in Known Fragments = Mass number of \({ }^{140}Xe\) = 140
03

Find the Mass Number of the Unknown Fragment

Since the total mass number should be conserved, we can find the mass number of the unknown fragment by subtracting the sum of mass numbers of known fragments from the total sum of mass numbers of reactants. Mass Number of Unknown Fragment = Sum of Mass Numbers in Reactants - Sum of Mass Numbers in Known Fragments = 236 - 140 = 96
04

Calculate the Sum of Atomic Numbers of Reactants and Products

Now we need to calculate the sum of atomic numbers of the reactants and compare it with the sum of atomic numbers in the products. Sum of Atomic Numbers in Reactants = Atomic number of \({ }^{235}U\) + Atomic number of neutron = 92 + 0 = 92
05

Calculate the Atomic Number of the Unknown Fragment

Since the total atomic number should be conserved, we can find the atomic number of the unknown fragment by subtracting the atomic number of the known fragment from the total sum of atomic numbers of reactants. Atomic Number of Unknown Fragment = Sum of Atomic Numbers in Reactants - Atomic Number of \({ }^{140}Xe\) = 92 - 54 = 38
06

Identify the Element of the Unknown Fragment

Now that we have the atomic number of the unknown fragment (38), we can refer to a periodic table to identify the corresponding element. In this case, the element is Strontium (Sr).
07

Write the Unknown Fragment in the Notation \({ }^{\mathrm{A}} \mathrm{X}\)

We now know the atomic number (Z = 38) and mass number (A = 96) of the unknown fragment. We can write it in the notation \({ }^{\mathrm{A}} \mathrm{X}\) as: \({ }^{96}Sr\) So the other fragment produced in the fission event is Strontium-96 (\({ }^{96}Sr\)).

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Most popular questions from this chapter

The world uses energy at a rate of \(18 \mathrm{TW}\), amounting to almost \(6 \times 10^{20}\) J per year. What is the mass-equivalent \(^{83}\) of this amount of annual energy? What context can you provide for this amount of mass?

Based on the fractional mass loss associated with turning four hydrogen atoms into a helium atom, what fraction of the sun's mass would it lose over its lifetime by converting all its hydrogen into helium, under the simplifying assumption that it starts its life as \(100 \%\) hydrogen?

A large boulder whose mass is \(1,000 \mathrm{~kg}\) having a specific heat capacity of \(1,000 \mathrm{~J} / \mathrm{kg} /{ }^{\circ} \mathrm{C}\) is heated from \(0^{\circ} \mathrm{C}\) to a glowing \(1,800^{\circ} \mathrm{C}\). How much more massive is it, assuming no atoms have been added or subtracted?

Operating approximately 450 nuclear plants over about 60 years at a total thermal level of \(1 \mathrm{TW}\), we have had two major radioactive releases into the environment. If we went completely down the nuclear road and get all \(18 \mathrm{TW}^{89}\) this way, what rate of accidents might we expect, if the rate just scales with usage levels?

Problem 15 indicated that we need the mass-equivalent of fewer than 10 tons 87 of material to support the world's annual energy needs. But given realities that only \(0.08 \%\) of mass is converted to energy in nuclear reactions, that only \(0.72 \%\) of natural uranium is fissile \({ }^{235} \mathrm{U}\), and that only half of the \({ }^{235} \mathrm{U}\) is retrievable \(^{88}\) and "burned" in reactors, how many tons of uranium must be mined per year to support 18 TW via conventional fission, assuming for the sake of this problem that 5 tons of mass need to convert to energy via \(E=m c^{2} ?\)
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