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Complete and balance the following nuclear equations by supplying the missing particle: (a) \({ }_{94}^{239} \mathrm{Pu}+{ }_{0}^{1} \mathrm{n} \longrightarrow{ }_{-1}^{0} \mathrm{e}+?\) (b) \({ }_{92}^{238} \mathrm{U}+{ }_{2}^{4} \mathrm{He} \longrightarrow 3{ }_{0}^{1} \mathrm{n}+?\) (c) \({ }^{218} \mathrm{At} \longrightarrow{ }_{-1}^{0} \mathrm{e}+?\) (d) \({ }_{62}^{146} \mathrm{Sm} \longrightarrow{ }_{60}^{142} \mathrm{Nd}+?\) (e) \({ }_{53}^{118} \mathrm{I}+{ }_{-1}^{0} \mathrm{e} \longrightarrow\) ?

Short Answer

Expert verified
a) Am-240; b) Pu-239; c) Rn-218; d) He-4; e) Te-118.

Step by step solution

01

Understand the Nuclear Equation

In a nuclear equation, the sums of the atomic numbers (protons) and mass numbers (nucleons) must be equal on both sides of the equation. This helps us find missing particles.
02

Balance Equation (a)

Given: \[ _{94}^{239} \text{Pu} + _{0}^{1} \text{n} \longrightarrow _{-1}^{0} \text{e} + ? \]Atomic number left = 94; right = -1 + x (x = atomic number of missing particle).Mass number left = 239 + 1; right = 0 + y (y = mass number of missing particle).\[\begin{align*}x &= 94 + 0 + 1 = 95\y &= 239 + 1 - 0 = 240\\end{align*}\]So, the missing particle is \( _{95}^{240} \text{Am} \) (Americium).
03

Balance Equation (b)

Given:\[ _{92}^{238} \text{U} + _{2}^{4} \text{He} \longrightarrow 3 \times (_{0}^{1} \text{n}) + ? \]Atomic number left = 92 + 2 = 94; right = 0 + x (x = atomic number of missing particle) + 3(0).Mass number left = 238 + 4 = 242; right = 1(3) + y (y = mass number of missing particle).\[\begin{align*}x &= 94 - 0 = 94\y &= 242 - 3 = 239\\end{align*}\]So, the missing particle is \( _{94}^{239} \text{Pu} \) (Plutonium).
04

Balance Equation (c)

Given:\[ _{85}^{218} \text{At} \longrightarrow _{-1}^{0} \text{e} + ? \]Atomic number left = 85; right = -1 + x (x = atomic number of missing particle).Mass number left = 218; right = 0 + y (y = mass number of missing particle).\[\begin{align*}x &= 85 + 1 = 86\y &= 218\\end{align*}\]So, the missing particle is \( _{86}^{218} \text{Rn} \) (Radon).
05

Balance Equation (d)

Given:\[ _{62}^{146} \text{Sm} \longrightarrow _{60}^{142} \text{Nd} + ? \]Atomic number left = 62; right = 60 + x (x = atomic number of missing particle).Mass number left = 146; right = 142 + y (y = mass number of missing particle).\[\begin{align*}x &= 62 - 60 = 2\y &= 146 - 142 = 4\\end{align*}\]So, the missing particle is \( _{2}^{4} \text{He} \) (Helium nucleus or Alpha particle).
06

Balance Equation (e)

Given:\[ _{53}^{118} \text{I} + _{-1}^{0} \text{e} \longrightarrow ? \]Atomic number left = 53 - 1(x = 52); right = x (x = atomic number of missing particle).Mass number left = 118; right = y (y = mass number of missing particle).\[\begin{align*}x &= 53 - 1 = 52\y &= 118\\end{align*}\]So, the missing particle is \( _{52}^{118} \text{Te} \) (Tellurium).

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Key Concepts

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

Atomic Number
The atomic number is a fundamental concept in chemistry and nuclear physics. It refers to the number of protons present in the nucleus of an atom. This number is crucial because it determines the identity of the element. For example, an atom with an atomic number of 1 is hydrogen, while an atomic number of 92 corresponds to uranium. In nuclear equations, ensuring that the sum of the atomic numbers on both sides of the equation is equal is key to identifying any missing particles. This balance maintains the conservation of charge during nuclear reactions, allowing scientists and students to accurately determine what particles result from these reactions.
Mass Number
The mass number of an atomic nucleus is the sum of its protons and neutrons, which are collectively known as nucleons. This number is crucial for understanding reactions in nuclear physics. Unlike atomic number, mass number can vary among atoms of the same element, resulting in different isotopes. For instance, carbon-12 and carbon-14 both have 6 protons (atomic number 6) but contain different numbers of neutrons, leading to mass numbers of 12 and 14 respectively.
When balancing nuclear equations, the mass number must also be conserved, meaning the total mass number on the left side of the equation must equal the total on the right side. This principle allows for the correct identification of missing particles in nuclear reactions, which is important when accounting for changes in an atom's mass following decay or other transformation processes.
Balancing Equations
Balancing nuclear equations is essential for understanding nuclear reactions as they illustrate how atoms transform into other atoms by gaining or losing nuclear particles. A balanced nuclear equation ensures that both atomic numbers and mass numbers match on each side, upholding the conservation laws of matter and charge.
  • First, you need to identify the given particles and their atomic and mass numbers on both the reactant and product sides of the equation.
  • Then, the missing particle is determined by calculating the difference needed to equalize both atomic and mass numbers on either side of the equation.
  • Finally, verify the calculation to ensure both numbers are correctly balanced, indicating a valid nuclear transformation.
By maintaining balance in nuclear equations, we can better understand the changes occurring within an atom during a nuclear reaction, which helps predict the behavior of isotopes and radioactive materials.
Nuclear Reactions
Nuclear reactions are processes in which nuclei of atoms interact, resulting in a change of element or isotope. These changes are often accompanied by the release or absorption of energy. Such reactions are distinct from chemical reactions, which involve only the electrons of an atom. There are different types of nuclear reactions, such as fission, fusion, decay, and transmutation. In fission, a nucleus splits into two or more smaller nuclei, while fusion refers to two or more smaller nuclei combining to form a larger one. Decay involves the spontaneous transformation of an unstable nucleus into a more stable one, releasing particles or radiation in the process.
  • Fission reactions are harnessed in nuclear reactors to produce energy.
  • Fusion is the process powering stars, including our sun.
  • Decay reactions allow for applications in dating ancient objects through carbon dating.
Understanding these reactions is crucial not only in physics and chemistry but also in various technological applications, from medicine to energy production.

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Most popular questions from this chapter

Some watch dials are coated with a phosphor, like \(\mathrm{ZnS}\), and a polymer in which some of the \({ }^{1} \mathrm{H}\) atoms have been replaced by \({ }^{3} \mathrm{H}\) atoms, tritium. The phosphor emits light when struck by the beta particle from the tritium decay, causing the dials to glow in the dark. The half-life of tritium is 12.3 yr. If the light given off is assumed to be directly proportional to the amount of tritium, by how much will a dial be dimmed in a watch that is 50 yr old?

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) \({ }_{5}^{10} \mathrm{~B}+? \longrightarrow{ }_{3}^{7} \mathrm{Li}+{ }_{2}^{4} \mathrm{He}\) (e) \({ }_{86}^{220} \mathrm{Rn} \longrightarrow{ }_{2}^{4} \mathrm{He}+?\)

Decay of which nucleus will lead to the following products: \((\mathbf{a})\) uranium- 235 by alpha decay; (b) aluminium-26 by positron emission; (c) deuterium by alpha decay; (d) yttrium-90 by beta decay?

The energy from solar radiation falling on Earth is \(1.07 \times 10^{16} \mathrm{~kJ} / \mathrm{min} .(\mathbf{a})\) How much loss of mass from the Sun occurs in one day from just the energy falling on Earth? (b) If the energy released in the reaction $$ { }^{235} \mathrm{U}+{ }_{0}^{1} \mathrm{n} \longrightarrow{ }_{56}^{141} \mathrm{Ba}+{ }_{36}^{92} \mathrm{Kr}+3{ }_{0}^{1} \mathrm{n} $$ \(\left(\begin{array}{c}235 \\ \mathrm{U} \text { nuclear mass, } 234.9935 \mathrm{u} ;{ }^{141} \mathrm{Ba} \text { nuclear mass, }\end{array}\right.\) \(140.8833 \mathrm{u} ;{ }^{92} \mathrm{Kr}\) nuclear mass, \(91.9021 \mathrm{u}\) ) is taken as typical of that occurring in a nuclear reactor, what mass of uranium- 235 is required to equal \(0.10 \%\) of the solar energy that falls on Earth in 1.0 day?

Write balanced nuclear equations for the following processes: (a) radon-198 undergoes alpha emission; (b) thorium-234 undergoes beta emission; (c) copper-61 undergoes positron emission; (d) silver-106 undergoes electron capture.

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