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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

Explain in some detail what happens if control rods are too effective at absorbing neutrons so that each fission event produces too few unabsorbed neutrons.

Since each nuclear plant delivers \(\sim 1 \mathrm{GW}\) of electrical power, at \(\sim 40 \%\) thermodynamic efficiency this means a thermal generation rate of \(2.5\) GW. How many nuclear plants would we need to supply all 18 TW of our current energy demand? Since a typical lifetime is 50 years before decommissioning, how many days, on average would it be between new plants coming online (while old ones are retired) in a steady state?

A particular nuclide is found to have lost 3 neutrons and 1 proton after a decay chain. What combination of \(\alpha\) and \(\beta\) decays could account for this result?

If a friend creates a nucleus whose half-life is 4 hours and gives it to you at noon, what is the probability that it will not have decayed by noon the following day?

Based on the calculation that 18 TW would require an annual cube of seawater \(300 \mathrm{~m}\) on a side to provide enough deuterium, what is your personal share as one of 8 billion people on earth, in liters? Could you lift this yourself? One cubic meter is \(1,000 \mathrm{~L}\).
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