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Chapter 25: Problem 9

Supply the missing information in each of the following nuclear equations representing a radioactive decay process.(a) \(160_?\mathrm{W} \longrightarrow\\{\mathrm{Hf}+?\) (b) \(38_? \mathrm{Cl} \longrightarrow_{?}^{?} \mathrm{Ar}+?\) (c) \(^{214} ? \longrightarrow_{?}^{?} \mathrm{Po}+_{-1}^{0} \boldsymbol{\beta}\) (d) \(_{17}^{32} \mathrm{Cl} \longrightarrow_{1}^{?} ?+?\)

Short Answer

Expert verified
The completed nuclear equations are: \n(a) $^{160}_{74} \mathrm{W} \longrightarrow ^{156}_{72} \mathrm{Hf} + ^{4}_{2} \mathrm{He}$ \n(b) $^{38}_{17} \mathrm{Cl} \longrightarrow ^{38}_{18} \mathrm{Ar} + _{-1}^{0}\mathrm{\beta}$\n(c) $^{214}_{85} \longrightarrow_{84}^{214} \mathrm{Po} + _{-1}^{0} \boldsymbol{\beta}$\n(d) $^{32}_{17} \mathrm{Cl} \longrightarrow_{1}^{0} \mathrm{e}+ ^{32}_{16} \mathrm{S}$

Step by step solution

01

Solve for (a)

Given the decay: $^{160}_{?} \mathrm{W} \longrightarrow \dots \mathrm{Hf}+?$, W represents Tungsten. An atomic number (the number of protons) must be supplied. For Tungsten, this number is 74. The decay product is Hf (Hafnium), and W undergoes alpha decay shedding an alpha particle (He) with atomic number 2 and mass number 4, so the equation becomes: $^{160}_{74} \mathrm{W} \longrightarrow ^{156}_{72} \mathrm{Hf} + ^{4}_{2} \mathrm{He}$.
02

Solve for (b)

Given the decay: $^{38}_{?} \mathrm{Cl} \longrightarrow_{?}^{?} \mathrm{Ar}+?$, Cl represents Chlorine. An atomic number must be supplied. For Chlorine, this number is 17. The decay product is Ar (Argon), and Cl undergoes beta decay, emitting a beta particle (\( \beta \)) with atomic number -1 and mass number 0, so the equation becomes: $^{38}_{17} \mathrm{Cl} \longrightarrow ^{38}_{18} \mathrm{Ar} + _{-1}^{0}\mathrm{\beta}$.
03

Solve for (c)

Given the decay: $^{214}_{?} \longrightarrow_{?}^{?} \mathrm{Po}+_{-1}^{0} \boldsymbol{\beta}$, Po is a product which is Polonium, the atomic number must be supplied. For Polonium, this number is 84. The original atom is not given, but we can infer it from the total atomic number and atomic mass. The original atom must have the same atomic mass 214 and an atomic number of 85 (since a beta particle with atomic number -1 is emitted), so the equation becomes: $^{214}_{85} \longrightarrow_{84}^{214} \mathrm{Po} + _{-1}^{0} \boldsymbol{\beta}$.
04

Solve for (d)

Given the decay: $^{32}_{17} \mathrm{Cl} \longrightarrow_{1}^{?} ?+?$, Cl represents Chlorine. The decay product is not given. Given that Chlorine emits a positron (with atomic number 1 and mass number 0), the product atom must be Sulfur (S) with atomic mass 32 and atomic number 16. So, the equation becomes: $^{32}_{17} \mathrm{Cl} \longrightarrow_{1}^{0} \mathrm{e}+ ^{32}_{16} \mathrm{S}$.

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Key Concepts

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

Radioactive Decay
Radioactive decay is a process where an unstable atomic nucleus loses energy by emitting radiation. This can include particles or electromagnetic waves. Students often encounter this concept when discussing isotopes, which are forms of the same element with different numbers of neutrons.
In nature, radioactive decay occurs at a consistent rate, known as a half-life, which is the time it takes for half of a sample of the radioactive substance to decay. This concept is crucial for understanding nuclear equations, which help depict the transformation of elements.
Key points to remember:
  • Radioactive decay involves emission of particles or radiation.
  • It leads to the transformation of an element into another.
  • Half-life is a measure of the decay rate.
Alpha Decay
Alpha decay is a common type of radioactive decay where an unstable nucleus releases an alpha particle. An alpha particle consists of 2 protons and 2 neutrons, essentially making it a helium nucleus. This results in the reduction of both the atomic mass and atomic number of the original element.
For example, when Tungsten (W) undergoes alpha decay, it releases an alpha particle, changing into Hafnium (Hf).
Key details about alpha decay:
  • Decreases the atomic number by 2.
  • Decreases the atomic mass number by 4.
  • Produces a new element lower in the periodic table.
Beta Decay
In beta decay, an unstable nucleus transforms by emitting a beta particle, which is a high-energy, high-speed electron or positron. When an electron is emitted, it's called beta-minus decay, whereas if a positron is emitted, it's called beta-plus decay.
During beta-minus decay, a neutron is converted into a proton, increasing the atomic number by 1, while the atomic mass remains unchanged. For instance, in Chlorine's decay to Argon, a beta particle is emitted, illustrating beta decay.
Characteristics of beta decay:
  • Increases or decreases the atomic number by 1 depending on the particle.
  • Does not affect the atomic mass number.
  • Transforms the element into the next or previous one in the periodic table.
Positron Emission
Positron emission, a form of beta-plus decay, occurs when a proton inside the nucleus is converted into a neutron while releasing a positron. This conversion decreases the atomic number by 1, turning the element into the one just before it in the periodic table.
For example, in the decay of Chlorine to Sulfur, a positron is emitted. This process helps balance out a nucleus with too many protons, adding to stabilization.
Important points about positron emission:
  • Decreases the atomic number by 1.
  • Atomic mass remains the same.
  • Results in a more stable nucleus.

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Most popular questions from this chapter

Iodine-129 is a product of nuclear fission, whether from an atomic bomb or a nuclear power plant. It is a \(\beta^{-}\) emitter with a half-life of \(1.7 \times 10^{7}\) years. How many disintegrations per second would occur in a sample containing \(1.00 \mathrm{mg}^{129} \mathrm{I} ?\)

Write nuclear equations to represent (a) the decay of \(^{214} \mathrm{Ra}\) by \(\alpha\) -particle emission (b) the decay of \(^{205}\) At by positron emission (c) the decay of \(^{212} \mathrm{Fr}\) by electron capture (d) the reaction of two deuterium nuclei (deuterons) to produce a nucleus of \(\frac{3}{2} \mathrm{He}\). (e) the production of \({243}_{97} \mathrm{Bk}\) get by the \(\alpha\) -particle bombardment of\({241}_{95} \mathrm{Am}\) (f) a nuclear reaction in which thorium-232 is bombarded with \(\alpha\) particles, producing a new nuclide and four neutrons.

Briefly describe each of the following ideas, phenomena, or methods: (a) radioactive decay series;(b) charged-particle accelerator; (c) neutron-to- proton ratio; (d) mass-energy relationship; (e) background radiation.

Explain the important distinctions between each pair of terms: (a) electron and positron; (b) half-life and decay constant; (c) mass defect and nuclear binding energy; (d) nuclear fission and nuclear fusion; (e) primary and secondary ionization.

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?
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