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Chapter 21: Problem 55

Complete the following nuclear equations. a. \(\quad \frac{13 i}{52} \mathrm{Te} \rightarrow \frac{131}{53} \mathrm{I}+?\) b. \(?+\frac{122}{54} X e+-_{-1}^{0} \beta\) c. \(?+_{2}^{4} \mathrm{He} \rightarrow_{7}^{13} \mathrm{N}+_{0}^{1} \mathrm{n}\) d. \(?+_{1}^{1} \mathrm{H} \rightarrow_{31}^{67} \mathrm{Ga}+2_{0}^{1} \mathrm{n}\)

Short Answer

Expert verified
\) b) \(?+\frac{122}{54} X e+-_{-1}^{0} \beta\) c) \(?+_{2}^{4} \mathrm{He} \rightarrow_{7}^{13} \mathrm{N}+_{0}^{1} \mathrm{n}\) d) \(?+_{1}^{1} \mathrm{H} \rightarrow_{31}^{67} \mathrm{Ga}+2_{0}^{1}\mathrm{n}\) Answer: a) \(\frac{13 i}{52} \mathrm{Te} \rightarrow \frac{131}{53} \mathrm{I}+\frac{-79}{-40}\beta\) b) \(\frac{55}{122} \mathrm{X} +\frac{122}{54} X e+-_{-1}^{0} \beta\) c) \(\frac{5}{9} \mathrm{X} +_{2}^{4} \mathrm{He} \rightarrow_{7}^{13} \mathrm{N}+_{0}^{1} \mathrm{n}\) d) \(\frac{30}{64} \mathrm{X} +_{1}^{1} \mathrm{H} \rightarrow_{31}^{67} \mathrm{Ga}+2_{0}^{1}\mathrm{n}\)

Step by step solution

01

Identify given particles in each equation

To complete the nuclear equations, observe the atomic and mass numbers of the given particle(s) on both sides of the arrow. Identify the particle that should be there to balance the equation.
02

Complete Equation a

Given: \(\frac{52}{13} \mathrm{Te} \rightarrow \frac{131}{53} \mathrm{I}+?\) Since atomic numbers need to be conserved, the total atomic numbers must be the same on both sides. Unknown atomic number: \(13 = 53 + x \Rightarrow x = -40\) Since mass numbers need to be conserved, the total mass numbers must be the same on both sides. Unknown mass number: \(52 = 131 + y \Rightarrow y = -79\) So, the equation becomes: \(\frac{13 i}{52} \mathrm{Te} \rightarrow \frac{131}{53} \mathrm{I}+\frac{-79}{-40}\beta\)
03

Complete Equation b

Given: \(?+\frac{122}{54} X e+-_{-1}^{0} \beta\) Unknown atomic number: \(x = 54 - (-1) \Rightarrow x = 55\) Unknown mass number: \(y = 122 - 0 \Rightarrow y = 122\) So, the equation becomes: \(\frac{55}{122} \mathrm{X} +\frac{122}{54} X e+-_{-1}^{0} \beta\)
04

Complete Equation c

Given: \(?+_{2}^{4} \mathrm{He} \rightarrow_{7}^{13} \mathrm{N}+_{0}^{1} \mathrm{n}\) Unknown atomic number: \(x = 7 - 2 \Rightarrow x = 5\) Unknown mass number: \(y = 13 - 4 \Rightarrow y = 9\) So, the equation becomes: \(\frac{5}{9} \mathrm{X} +_{2}^{4} \mathrm{He} \rightarrow_{7}^{13} \mathrm{N}+_{0}^{1} \mathrm{n}\)
05

Complete Equation d

Given: \(?+_{1}^{1} \mathrm{H} \rightarrow_{31}^{67} \mathrm{Ga}+2_{0}^{1}\mathrm{n}\) Unknown atomic number: \(x = 31 - 1 \Rightarrow x = 30\) Unknown mass number: \(y = 67 - 1 - 2(1) \Rightarrow y = 64\) So, the equation becomes: \(\frac{30}{64} \mathrm{X} +_{1}^{1} \mathrm{H} \rightarrow_{31}^{67} \mathrm{Ga}+2_{0}^{1}\mathrm{n}\)

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Key Concepts

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

Mass Number Conservation
In nuclear equations, one of the fundamental principles is the conservation of mass number. The mass number is the total number of protons and neutrons in the nucleus of an atom. When writing nuclear equations, it's crucial that the sum of the mass numbers on the reactant side equals the sum on the product side.

This principle ensures that the same number of nucleons (protons and neutrons) are present before and after the reaction.
  • For example, in the original exercise for Equation a, we started with a mass number of 52 on the left. To balance the mass number after decay, the sum of 131 plus the unknown must equal 52, leading to a calculation of the missing mass number as -79.
  • The negative value suggests that the initial attempt did not correctly align with physical reality, highlighting the importance of ensuring nuclear equations remain plausible and accurate.
Atomic Number Conservation
The conservation of atomic number is another key tenet in nuclear reactions. The atomic number represents the number of protons in an atom and essentially defines the chemical element.

Similar to the conservation of mass number, nuclear reactions must also have equal totals of atomic numbers on both sides of the equation. This rule ensures that the type of atoms remain consistent throughout the equation.
  • For Equation b from the original exercise, where an atom is undergoing beta decay, the atomic number conservation principle requires that the sum of atomic numbers before and after the reaction match. In this case, the emission of a beta particle caused an increase in atomic number from 54 to 55, since the beta particle essentially converts a neutron to a proton.
  • Understanding the changes in atomic numbers can help identify the elements involved and ensure the equation remains valid.
Beta Decay
Beta decay is a type of radioactive decay where an unstable nucleus emits a beta particle. A beta particle can be either an electron (\(\beta^-\)) or a positron (\(\beta^+\)). This process leads to a change in the atomic number of the nucleus involved, but the mass number remains unaffected because it is primarily concerned with changing neutron-to-proton ratios.

  • In beta minus (\(\beta^-\)) decay, a neutron is converted into a proton and an electron, the latter being ejected as a beta particle. This increases the atomic number by one.
  • In beta plus (\(\beta^+\)) decay, a proton is turned into a neutron and a positron, reducing the atomic number by one. As seen in Equation b in the step-by-step solution, a beta minus decay resulted in an increase in the atomic number of the element, changing the element in the process while conserving mass numbers.
Understanding beta decay is essential because it plays a significant role in the transformation of elements, especially in nuclear chemistry and physics.

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Most popular questions from this chapter

Describe an anti-proton.

All of the following fusion reactions produce \(^{32}\) S. Calculate the energy released in each reaction from the masses of the isotopes: \(^{4} \mathrm{He}(4.00260 \mathrm{amu}),^{6} \mathrm{Li}(6.01512 \mathrm{amu}),^{12} \mathrm{C}\),\((12.000 \mathrm{amu}),^{14} \mathrm{N}(14.00307 \mathrm{amu}),^{16} \mathrm{O}(15.99491 \mathrm{amu})\),\(^{24} \mathrm{Mg}(23.98504 \mathrm{amu}),^{28} \mathrm{Si}(27.97693 \mathrm{amu}),^{32} \mathrm{S}\),\((31.97207 \mathrm{amu})\). a. \(16+^{16} \mathrm{O} \rightarrow^{32} \mathrm{S}\) b. \(^{23} \mathrm{Si}+^{4} \mathrm{He} \rightarrow^{32} \mathrm{S}\) \(c_{1}^{14} N+^{12} C+^{6} L i \rightarrow^{32} S\) d. \(^{24} \mathrm{Mg}+2^{4} \mathrm{He} \rightarrow^{32} \mathrm{S}\)

What is the effect of \(\beta\) decay on the ratio of neutrons to protons in a nucleus?

Calculate the neutron-to-proton ratios for each of the following and predict the decay pathways of the following radioactive isotopes: \((\mathrm{a})^{238} \mathrm{U} ;(\mathrm{b})^{186} \mathrm{Re} ;(\mathrm{c})^{86} \mathrm{Y}\).

The absorption of a neutron by \(^{11} \mathrm{B}\) produces a radioactive nuclide that decays by either \(\alpha\) decay or \(\beta\) decay. Write balanced nuclear equations describing the decay reactions.
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