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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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Most popular questions from this chapter

Explain why more energy is released in a fusion process than in a fission process.

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.

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.

\(^{222} \mathrm{Rn}\) is an \(\alpha\) -particle emitter with a half-life of 3.82 days. Is it hazardous to be near a flask containing this isotope? Under what conditions might \(^{222} \mathrm{Rn}\) be hazardous?

The immediate decay product of element 118 is thought to be element \(116 .\) Write a complete nuclear equation for this reaction.
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