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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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Key Concepts

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

Conservation of Mass
Understanding the concept of conservation of mass is crucial in nuclear equations. It tells us that the total mass before and after a nuclear reaction must remain the same. This means that the sum of the mass numbers on the left side of an equation equals the sum on the right side.
Let's take an example:
  • In an equation involving \({ }_{47}^{106}\mathrm{Ag}\), the mass number on the left is 106.
  • If an electron, \({ }_{-1}^{0}\mathrm{e}\), is combined with it, the mass number remains the same as electrons have negligible mass.
  • So, the mass number on the right also needs to be 106 to satisfy the conservation of mass, leading us to deduce the mass number of the missing particle.
Conservation of mass ensures that despite transformations in the nucleus, the mass remains constant.
Atomic Number
The atomic number is essential in identifying elements. It equals the number of protons in an atom's nucleus. This number is a unique identifier for an element. In nuclear equations, the atomic number helps ensure that the total number of protons is conserved during reactions.
For instance:
  • In the reaction \({ }_{106}^{263}\mathrm{Sg} \rightarrow { }_{2}^{4}\mathrm{He} + ?\), we start with an atomic number of 106.
  • When part of the nucleus transforms into a helium nucleus \({ }_{2}^{4}\mathrm{He}\) which has an atomic number of 2, the remaining component must have an atomic number of 104. This guides us to the missing element, \({ }_{104}\mathrm{Rf}\).
Understanding atomic numbers thus allows us to find the correct elements participating in a nuclear change.
Periodic Table
The periodic table is like a map for elements. Each element is assigned its unique atomic number and symbol, helping us to easily identify them during nuclear reactions. When solving nuclear equations, one often needs to refer to the periodic table to determine which element corresponds to a given atomic number.
Here's how it helps:
  • In a nuclear equation, the element with atomic number 84 is Polonium (symbol Po).
  • Knowing this helps in finding the products of nuclear reactions, as one can quickly identify the missing element.
  • This identification process helps ensure equations are balanced by correctly matching atomic numbers and symbols from the periodic table.
Thus, the periodic table is a fundamental tool in accurately solving nuclear equations.
Balancing Equations
Balancing equations is a critical part of chemistry, including nuclear equations. It ensures both the conservation of mass and atomic numbers. When you're faced with a nuclear reaction, balancing means ensuring both sides have equivalent totals for atomic and mass numbers.
Consider:
  • If you start with \({ }_{84}^{216}\mathrm{Po}\) and it changes into \({ }_{82}^{212}\mathrm{Pb}\), a helium nucleus must be produced to balance the mass and atomic numbers. This is the particle \({ }_{2}^{4}\mathrm{He}\).
  • The equation must account for everything, ensuring no protons or neutrons are missing after a transformation.
Balancing makes sure that all components before and after a reaction are accounted for correctly.

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Most popular questions from this chapter

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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Cobalt-60 is a strong gamma emitter that has a half-life of \(5.26 \mathrm{yr}\). The cobalt- 60 in a radiotherapy unit must be replaced when its radioactivity falls to \(75 \%\) of the original sample. If an original sample was purchased in June \(2016,\) when will it be necessary to replace the cobalt- \(60 ?\)

It takes 180 minutes for a 200 -mg sample of an unknown radioactive substance to decay to \(112 \mathrm{mg}\). What is the halflife of this substance?

A 10.00 -g plant fossil from an archaeological site is found to have a \({ }^{14} \mathrm{C}\) activity of 3094 disintegrations over a period of ten hours. A living plant is found to have a \({ }^{14} \mathrm{C}\) activity of 9207 disintegrations over the same period of time for an equivalent amount of sample with respect to the total contents of carbon. Given that the half-life of \({ }^{14} \mathrm{C}\) is 5715 years, how old is the plant fossil?

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