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Chapter 21: Problem 30

Complete and balance the following nuclear equations by supplying the missing particle: (a) \({ }_{47}^{106} \mathrm{Ag}+{ }_{-1}^{0} \mathrm{e} \longrightarrow ?\) (b) \({ }_{106}^{263} \mathrm{Sg} \longrightarrow{ }_{2}^{4} \mathrm{He}+?\) (c) \({ }_{84}^{216} \mathrm{Po} \longrightarrow{ }_{82}^{212} \mathrm{~Pb}+?\) (d) ${ }_{5}^{10} \mathrm{~B}+? \longrightarrow{ }_{3} \mathrm{Li}+{ }_{2}^{4} \mathrm{He}$ (e) \({ }^{220} \mathrm{Rn} \longrightarrow{ }_{2}^{4} \mathrm{He}+?\)

Short Answer

Expert verified
The short answer to the given nuclear equations is: (a) \({ }_{47}^{106}\mathrm{Ag} + { }_{-1}^{0}\mathrm{e} \longrightarrow { }_{46}^{106}\mathrm{Pd}\) (b) \({ }_{106}^{263}\mathrm{Sg} \longrightarrow{ }_{2}^{4}\mathrm{He} + { }_{104}^{259}\mathrm{Rf}\) (c) \({ }_{84}^{216}\mathrm{Po} \longrightarrow{ }_{82}^{212}\mathrm{~Pb} + { }_{2}^{4}\mathrm{He}\) (d) \({ }_{5}^{10}\mathrm{~B} + { }_{0}^{1}\mathrm{n} \longrightarrow{ }_{3}\mathrm{Li} + { }_{2}^{4}\mathrm{He}\) (e) \({ }_{86}^{220}\mathrm{Rn} \longrightarrow{ }_{2}^{4}\mathrm{He} + { }_{84}^{216}\mathrm{Po}\)

Step by step solution

01

Analyze the given equation

We will use the conservation laws of atomic number and mass number. In this equation, we have an unknown particle we need to find.
02

Apply conservation laws to find the missing particle

According to the conservation law of atomic numbers, the total number of protons must be conserved on both sides of the equation. So, the atomic number of the missing particle is equal to 47 + (-1) = 46. According to the conservation law of mass numbers, the total number of nucleons must be conserved on both sides of the equation. So, the mass number of the missing particle is equal to 106 + 0 = 106.
03

Find the name of the missing particle

We check the periodic table and find the element corresponding to the atomic number 46 is Palladium(Pd). So, the missing particle is \({ }_{46}^{106}\mathrm{Pd}\). The balanced equation is: \({ }_{47}^{106}\mathrm{Ag} + { }_{-1}^{0}\mathrm{e} \longrightarrow { }_{46}^{106}\mathrm{Pd}\). (b) \({ }_{106}^{263}\mathrm{Sg} \longrightarrow{ }_{2}^{4}\mathrm{He} +?\)
04

Analyze the given equation

We will use the conservation laws of atomic number and mass number. In this equation, we have an unknown particle we need to find.
05

Apply conservation laws to find the missing particle

According to the conservation law of atomic numbers, the total number of protons must be conserved on both sides of the equation. So, the atomic number of the missing particle is equal to 106 - 2 = 104. According to the conservation law of mass numbers, the total number of nucleons must be conserved on both sides of the equation. So, the mass number of the missing particle is equal to 263 - 4 = 259.
06

Find the name of the missing particle

We check the periodic table and find the element corresponding to the atomic number 104 is Rutherfordium(Rf). So, the missing particle is \({ }_{104}^{259}\mathrm{Rf}\). The balanced equation is: \({ }_{106}^{263}\mathrm{Sg} \longrightarrow{ }_{2}^{4}\mathrm{He} + { }_{104}^{259}\mathrm{Rf}\). (c) \({ }_{84}^{216}\mathrm{Po} \longrightarrow{ }_{82}^{212}\mathrm{~Pb} +?\)
07

Analyze the given equation

We will use the conservation laws of atomic number and mass number. In this equation, we have an unknown particle we need to find.
08

Apply conservation laws to find the missing particle

According to the conservation law of atomic numbers, the total number of protons must be conserved on both sides of the equation. So, the atomic number of the missing particle is equal to 84 - 82 = 2. According to the conservation law of mass numbers, the total number of nucleons must be conserved on both sides of the equation. So, the mass number of the missing particle is equal to 216 - 212 = 4.
09

Find the name of the missing particle

We check the periodic table and find the element corresponding to the atomic number 2 is Helium(He). So, the missing particle is \({ }_{2}^{4}\mathrm{He}\). The balanced equation is: \({ }_{84}^{216}\mathrm{Po} \longrightarrow{ }_{82}^{212}\mathrm{~Pb} + { }_{2}^{4}\mathrm{He}\). (d) \({ }_{5}^{10}\mathrm{~B} + ? \longrightarrow{ }_{3}\mathrm{Li} + { }_{2}^{4}\mathrm{He}\)
10

Analyze the given equation

We will use the conservation laws of atomic number and mass number. In this equation, we have an unknown particle we need to find.
11

Apply conservation laws to find the missing particle

According to the conservation law of atomic numbers, the total number of protons must be conserved on both sides of the equation. So, the atomic number of the missing particle is equal to 3 + 2 - 5 = 0. According to the conservation law of mass numbers, the total number of nucleons must be conserved on both sides of the equation. So, the mass number of the missing particle is equal to 4 + 7 - 10 = 1.
12

Find the name of the missing particle

The missing particle will be a neutron, as it has no protons (atomic number = 0) and a mass number of 1. So, the missing particle is \({ }_{0}^{1}\mathrm{n}\). The balanced equation is: \({ }_{5}^{10}\mathrm{~B} + { }_{0}^{1}\mathrm{n} \longrightarrow{ }_{3}\mathrm{Li} + { }_{2}^{4}\mathrm{He}\). (e) \({ }^{220}\mathrm{Rn} \longrightarrow{ }_{2}^{4}\mathrm{He} +?\)
13

Analyze the given equation

We will use the conservation laws of atomic number and mass number. In this equation, we have an unknown particle we need to find and the atomic number of Radon is missing, which is 86.
14

Apply conservation laws to find the missing particle

According to the conservation law of atomic numbers, the total number of protons must be conserved on both sides of the equation. So, the atomic number of the missing particle is equal to 86 - 2 = 84. According to the conservation law of mass numbers, the total number of nucleons must be conserved on both sides of the equation. So, the mass number of the missing particle is equal to 220 - 4 = 216.
15

Find the name of the missing particle

We check the periodic table and find the element corresponding to the atomic number 84 is Polonium(Po). So, the missing particle is \({ }_{84}^{216}\mathrm{Po}\). The balanced equation is: \({ }_{86}^{220}\mathrm{Rn} \longrightarrow{ }_{2}^{4}\mathrm{He} + { }_{84}^{216}\mathrm{Po}\).

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Most popular questions from this chapter

Tests on human subjects in Boston in 1965 and \(1966,\) following the era of atomic bomb testing, revealed average quantities of about \(2 \mathrm{pCi}\) of plutonium radioactivity in the average person. How many disintegrations per second does this level of activity imply? If each alpha particle deposits $8 \times 10^{-13} \mathrm{~J}\( of energy and if the average person weighs \)75 \mathrm{~kg},$ calculate the number of grays and sieverts of radiation in 1 yr from such a level of plutonium.

The atomic masses of nitrogen-14, titanium- 48 , and xenon129 are $13.999234 \mathrm{u}, 47.935878 \mathrm{u},\( and \)128.904779 \mathrm{u},$ respectively. For each isotope, calculate (a) the nuclear mass, (b) the nuclear binding energy, (c) the nuclear binding energy per nucleon.

What is the function of the control rods in a nuclear reactor? What substances are used to construct control rods? Why are these substances chosen?

It has been suggested that strontium-90 (generated by nuclear testing) deposited in the hot desert will undergo radioactive decay more rapidly because it will be exposed to much higher average temperatures. (a) Is this a reasonable suggestion? (b) Does the process of radioactive decay have an activation energy, like the Arrhenius behavior of many chemical reactions (Section 14.5\() ?\)

In 2002 , a team of scientists from Russia and the United States reported the creation of the first atom of element 118 , which is named oganesson, and whose symbol is Og. The synthesis involved the collision of californium- 249 atoms with accelerated ions of an atom which we will denote X. In the synthesis, an oganesson-294 is formed together with three neutrons. $$ { }_{98}^{249} \mathrm{Cf}+\mathrm{X} \longrightarrow{ }_{118}^{294} \mathrm{Og}+3{ }_{0}^{1} \mathrm{n} $$ (a) What are the identities of isotopes X? (b) Isotope \(X\) is unusual in that it is very long-lived (its half-life is on the order of \(10^{19} \mathrm{yr}\) ) in spite of having an unfavorable neutron-to-proton ratio (Figure 21.1). Can you propose a reason for its unusual stability? (c) Oganesson-294 decays into livermorium-290 by alpha decay. Write a balanced equation for this.
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