Question

Write a balanced nuclear equation for the decay of each of the following nuclides to produce a beta particle. a. \(\quad \frac{136}{53}\)I b. \(\quad \frac{133}{51} S b\) c. \(\quad \frac{117}{49} \mathrm{In}\)

Short answer

a. \(\frac{136}{53}\text{I} \rightarrow \frac{136}{54}\text{Xe} + \frac{0}{-1}\beta\) b. \(\frac{133}{51}\text{Sb} \rightarrow \frac{133}{52}\text{Te} + \frac{0}{-1}\beta\) c. \(\frac{117}{49}\text{In} \rightarrow \frac{117}{50}\text{Sn} + \frac{0}{-1}\beta\)

Step-by-step solution

Identify the decay equation

For the decay of Iodine-136 into a beta particle, the balanced nuclear equation will be: \[\frac{136}{53}\text{I} \rightarrow \text{Product Nuclide} + \frac{0}{-1}\beta \]

Identify the product nuclide

Since the mass number remains constant, the product nuclide must have a mass number of 136. The atomic number must increase by one since a neutron is converted to a proton. Thus, the product nuclide has an atomic number of 54. We can represent the unknown product as follows: \[\frac{136}{54}\text{X}\]

Complete the decay equation

Replacing the product nuclide in the equation, we get: \[\frac{136}{53}\text{I} \rightarrow \frac{136}{54}\text{X} + \frac{0}{-1}\beta\] The balanced nuclear equation for the decay of Iodine-136 is: \[\frac{136}{53}\text{I} \rightarrow \frac{136}{54}\text{Xe} + \frac{0}{-1}\beta\] b. Decay of \(\frac{133}{51}\text{Sb}\):

Identify the decay equation

For the decay of Antimony-133 into a beta particle, the balanced nuclear equation will be: \[\frac{133}{51}\text{Sb} \rightarrow \text{Product Nuclide} + \frac{0}{-1}\beta \]

Identify the product nuclide

Since the mass number remains constant, the product nuclide must have a mass number of 133. The atomic number must increase by one since a neutron is converted to a proton. Thus, the product nuclide has an atomic number of 52. We can represent the unknown product as follows: \[\frac{133}{52}\text{X}\]

Complete the decay equation

Replacing the product nuclide in the equation, we get: \[\frac{133}{51}\text{Sb} \rightarrow \frac{133}{52}\text{X} + \frac{0}{-1}\beta\] The balanced nuclear equation for the decay of Antimony-133 is: \[\frac{133}{51}\text{Sb} \rightarrow \frac{133}{52}\text{Te} + \frac{0}{-1}\beta\] c. Decay of \(\frac{117}{49}\text{In}\):

Identify the decay equation

For the decay of Indium-117 into a beta particle, the balanced nuclear equation will be: \[\frac{117}{49}\text{In} \rightarrow \text{Product Nuclide} + \frac{0}{-1}\beta \]

Identify the product nuclide

Since the mass number remains constant, the product nuclide must have a mass number of 117. The atomic number must increase by one since a neutron is converted to a proton. Thus, the product nuclide has an atomic number of 50. We can represent the unknown product as follows: \[\frac{117}{50}\text{X}\]

Complete the decay equation

Replacing the product nuclide in the equation, we get: \[\frac{117}{49}\text{In} \rightarrow \frac{117}{50}\text{X} + \frac{0}{-1}\beta\] The balanced nuclear equation for the decay of Indium-117 is: \[\frac{117}{49}\text{In} \rightarrow \frac{117}{50}\text{Sn} + \frac{0}{-1}\beta\]

Key concepts

Beta Particle Emission

In nuclear chemistry, the process of beta particle emission plays a crucial role in the stability of an atom. A beta particle is essentially an electron or positron that is emitted from the nucleus of an unstable atom. When an atom undergoes beta decay, a neutron in the nucleus is transformed into a proton, and this change results in the emission of a beta particle from the nucleus.

During beta-minus decay, an electron (denoted as \(\beta^-\)) is released, whereas beta-plus decay involves the emission of a positron (denoted as \(\beta^+\)). The result of beta-minus decay is an increase in the atomic number by one, which indicates the formation of a new element. This new element is positioned one place further to the right on the periodic table.

In the context of the provided nuclear decay example, the beta particle emission can be visualized as:

• For Iodine-136 (\(\frac{136}{53}\)I), a neutron becomes a proton releasing an electron as a beta particle, leading to Xenon-136 (\(\frac{136}{54}\)Xe).
• For Antimony-133 (\(\frac{133}{51}\)Sb), a similar process results in the formation of Tellurium-133 (\(\frac{133}{52}\)Te).
• For Indium-117 (\(\frac{117}{49}\)In), the emission of a beta particle yields Tin-117 (\(\frac{117}{50}\)Sn).

The equation reflects the conservation of charge and mass, as the sum of mass numbers and the sum of atomic numbers on both sides of the equation must be equal.

Balancing Nuclear Reactions

Balancing nuclear reactions is an essential step in understanding the processes that occur within the nucleus of an atom. To achieve a balanced equation, one must ensure that the total number of protons (atomic number) and the total mass number (sum of protons and neutrons) remains constant before and after the reaction.

When balancing beta decay reactions, pay attention to these two rules:

• The mass number does not change since a beta particle has negligible mass.
• The atomic number increases by one because a neutron is converted into a proton, emitting an electron (beta particle).

Following the principles of conservation of mass and charge, the resulting element, also called the 'daughter nuclide,' will have the same mass number but an atomic number that is one unit higher than the original 'parent nuclide.' For instance, Iodine-136, with a mass number of 136 and atomic number 53, decays to form Xenon-136 with a mass number still of 136 but an atomic number of 54, thus respecting the balance of the nuclear reaction.

Nuclear Chemistry

Nuclear chemistry is a field that explores the changes in the nucleus of atoms that lead to the transmutation of elements and the release of nuclear energy. This area of chemistry is pivotal for various applications including, but not limited to, energy production in nuclear power plants, radiometric dating methods for determining the age of materials, and medical treatments such as radiotherapy.

In nuclear reactions like beta decay, the balance between protons and neutrons in the nucleus shifts, which can result in the atom changing into another element entirely, as seen in the provided exercise. This concept is the core of radioactive decay processes where unstable isotopes seek stability by emitting particles or radiation.

Nuclear reactions, such as the decay equations demonstrated in the solution for Iodine-136, Antimony-133, and Indium-117, not only show the transmutation of elements but also the intricacies of predicting and understanding the stability of isotopes and their resulting decay products. It's the careful study of these processes that allows scientists to harness nuclear reactions for beneficial uses, all while recognizing and mitigating the potential risks associated with radioactivity.