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Chapter 25: Problem 10

Complete the following nuclear equations. (a) \(\frac{23}{11} \mathrm{Na}+? \longrightarrow_{11}^{24} \mathrm{Na}+_{1}^{1} \mathrm{H}\) (b) \(_{27}^{59} \mathrm{Co}+_{0}^{1} \mathrm{n} \longrightarrow_{25}^{56} \mathrm{Mn}+?\) (c) \(?+_{1}^{2} \mathrm{H} \longrightarrow_{94}^{240} \mathrm{Pu}+_{-1}^{0} \beta\) (d) \(^{246} \mathrm{Cm}+? \longrightarrow_{102}^{254} \mathrm{No}+5_{0}^{1} \mathrm{n}\) (e) \(^{238} \mathrm{U}+? \longrightarrow_{99}^{246} \mathrm{Es}+6 \frac{1}{0} \mathrm{n}\)

Short Answer

Expert verified
The unknown particles are: (a) a neutron; (b) an alpha particle ; (c) a Neptunium atom. Equations (d) and (e) appear to be incorrect as they don't satisfy conservation laws.

Step by step solution

01

Understand mass and atomic number conservation

In any given nuclear reaction, the total atomic number (also known as the charge) and the total mass number (number of protons + neutrons) are both conserved. This means the sum of atomic numbers and mass numbers on one side of the equation should equal the sum on the other side.
02

Apply these principles to each equation

(a) The unknown particle will have a mass number of \((24 - 23) = 1\) and an atomic number of \(11 - 11 = 0\), so it's a neutron, denoted as \(_{0}^{1}\mathrm{n}\). (b) The unknown particle will have a mass number of \((59 - 56) = 3\) and an atomic number of \(27 - 25 = 2\), so it's an alpha particle, denoted as \(_{2}^{4}\mathrm{He} \).(c) The unknown particle will have a mass number of \((240 - 2) = 238\) and an atomic number of \(94 - 1 = 93\), so it's a Neptunium atom, denoted as \(_{93}^{238}\mathrm{Np}\).(d) The unknown particle will have a mass number of \((246 - 254 + 5) = -3\) and an atomic number of \(96 - 102 = -6\), which isn't possible, indicating an error in the equation. (e) The unknown particle will have a mass number of \((238 - 246 + 6*1) = -2\) and an atomic number of \(92 - 99 = -7\), which isn't possible, indicating an error in the equation.
03

Summary

In this exercise, atomic and mass number conservation was used to identify unknown particles in nuclear equations. In equations (d) and (e), the conservation laws weren't followed, indicating errors in the equations.

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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 reactions, mass number conservation is a fundamental concept. It ensures that the total number of protons and neutrons remains the same before and after a reaction. This conservation law is applied by adding up the mass numbers on each side of the nuclear equation. The sum should be equal. For example:
  • When a nucleus splits, the total mass of the resulting particles equals the mass of the original nucleus.
  • In equation (a), the mass number before the reaction is 23, and it must equal 24 on the other side. By subtracting, we know the missing particle must have a mass number of 1, indicating a neutron.

Mass number is crucial because it reflects the total nucleons in the nucleus. This number helps chemists and physicists predict the products of nuclear reactions correctly.
Atomic Number Conservation
Atomic number conservation ensures that the charge remains balanced during nuclear reactions. The atomic number represents the number of protons, which determines the element's identity. Just like the mass number, the total atomic numbers on both sides of the equation must sum to the same value. Here鈥檚 how it works:
  • In equation (b), the atomic number reduces from 27 to 25, requiring an additional 2 units on the other side to balance, provided by an alpha particle \(_{2}^{4}\mathrm{He} \).
  • The missing particle must correct any imbalance in nuclear charge, ensuring that the type of elements involved in the reaction is conserved.

Understanding this law helps ascertain the types of particles or radiation released during nuclear transformations, making it essential for safe applications in medicine and energy.
Nuclear Reactions
Nuclear reactions involve changes in an atom's nucleus and can release or absorb a significant amount of energy. Unlike chemical reactions, nuclear reactions can transform elements into different elements by altering their nuclei. Two key types are:
  • Fission: A heavy nucleus splits into two or more lighter nuclei, releasing energy.
  • Fusion: Light nuclei combine to form a heavier nucleus, also releasing energy.

In the provided equations, reactions involve the transformation of elements. For instance, incoming neutrons or emitted particles alter the nucleus's structure. It's vital to follow conservation laws to correctly predict the products, as seen in exercises where balancing errors indicated mistakes. Understanding nuclear reactions is crucial for diverse fields such as energy production, where reactions power nuclear plants, and in medicine, where they are used in cancer treatments.

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Most popular questions from this chapter

For medical uses, radon-222 formed in the radioactive decay of radium-226 is allowed to collect over the radium metal. Then, the gas is withdrawn and sealed into a glass vial. Following this, the radium is allowed to disintegrate for another period, when a new sample of radon- 222 can be withdrawn. The procedure can be continued indefinitely. The process is somewhat complicated by the fact that radon-222 itself undergoes radioactive decay to polonium- 218 , and so on. The half-lives of radium-226 and radon-222 are \(1.60 \times 10^{3}\) years and 3.82 days, respectively.(a) Beginning with pure radium- \(226,\) the number of radon-222 atoms present starts at zero, increases for a time, and then falls off again. Explain this behavior. That is, because the half-life of radon-222 is so much shorter than that of radium- \(226,\) why doesn't the radon-222 simply decay as fast as it is produced, without ever building up to a maximum concentration?(b) Write an expression for the rate of change \((d \mathrm{D} / d t)\) in the number of atoms (D) of the radon- 222 daughter in terms of the number of radium- 226 atoms present initially ( \(\mathrm{P}_{0}\) ) and the decay constants of the parent \(\left(\lambda_{\mathrm{p}}\right)\) and daughter \(\left(\lambda_{\mathrm{d}}\right)\) (c) Integration of the expression obtained in part (b) yields the following expression for the number of atoms of the radon-222 daughter (D) present at a time \(t\).$$\mathrm{D}=\frac{\mathrm{P}_{0} \lambda_{\mathrm{p}}\left(\mathrm{e}^{-\lambda_{\mathrm{p}} \times t}-\mathrm{e}^{-\lambda_{\mathrm{d}} \times t}\right)}{\lambda_{\mathrm{d}}-\lambda_{\mathrm{p}}}$$,Starting with \(1.00 \mathrm{g}\) of pure radium- \(226,\) approximately how long will it take for the amount of radon222 to reach its maximum value: one day, one week, one year, one century, or one millennium?

Radioactive decay and mass spectrometry are often used to date rocks after they have cooled from a magma. \(^{87} \mathrm{Rb}\) has a half-life of \(4.8 \times 10^{10}\) years and follows the radioactive decay $$^{87} \mathrm{Rb} \longrightarrow^{87} \mathrm{Sr}+\beta^{-}$$ A rock was dated by assaying the product of this decay. The mass spectrum of a homogenized sample of rock showed the \(^{87} \mathrm{Sr} /^{86} \mathrm{Sr}\) ratio to be \(2.25 .\) Assume that the original \(^{87} \mathrm{Sr} /^{86} \mathrm{Sr}\) ratio was 0.700 when the rock cooled. Chemical analysis of the rock gave \(15.5 \mathrm{ppm}\) Sr and 265.4 ppm \(\mathrm{Rb},\) using the average atomic masses from a periodic table. The other isotope ratios were \(^{86} \mathrm{Sr} /^{88} \mathrm{Sr}=\) 0.119 and \(^{84} \mathrm{Sr} /^{88} \mathrm{Sr}=0.007 .\) The isotopic ratio for \(^{87} \mathrm{Rb} /^{85} \mathrm{Rb}\) is 0.330. The isotopic masses are as follows:Calculate the following: (a) the average atomic mass of Sr in the rock (b) the original concentration of \(\mathrm{Rb}\) in the rock in \(\mathrm{ppm}\) (c) the percentage of rubidium- 87 decayed in the rock (d) the time since the rock cooled.

Both \(\beta^{-}\) and \(\beta^{+}\) emissions are observed for artificially produced radioisotopes of low atomic numbers, but only \(\beta^{-}\) emission is observed with naturally occurring radioisotopes of high atomic number. Why do you suppose this is so?

A sample containing \(_{88}^{224} \mathrm{Ra},\) which decays by \(\alpha\) -particle emission, disintegrates at the following rate, expressed as disintegrations per minute or counts per minute \((\mathrm{cpm}): t=0,1000 \mathrm{cpm} ; t=1 \mathrm{h}\) \(992 \mathrm{cpm} ; t=10 \mathrm{h}, 924 \mathrm{cpm} ; t=100 \mathrm{h}, 452 \mathrm{cpm}\) \(t=250 \mathrm{h}, 138 \mathrm{cpm} .\) What is the half-life of this nuclide?

Describe how you might use radioactive materials to find a leak in the \(\mathrm{H}_{2}(\mathrm{g})\) supply line in an ammonia synthesis plant.
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