Question

In each of the following radioactive decay processes, supply the missing particle. a. \(^{60} \mathrm{Co} \rightarrow^{60} \mathrm{Ni}+?\) b. \(^{97} \mathrm{Tc}+? \rightarrow^{97} \mathrm{Mo}\) c. \(^{99} \mathrm{Tc} \rightarrow^{99} \mathrm{Ru}+?\) d. \(^{239} \mathrm{Pu} \rightarrow^{235} \mathrm{U}+?\)

Short answer

The missing particles for each radioactive decay process are: a. \(^{60}\mathrm{Co}\rightarrow ^{60}\mathrm{Ni} + ^0_{-1}\beta\) (beta particle) b. \(^{97}\mathrm{Tc} + ^0_{-1}\beta\rightarrow^{97}\mathrm{Mo}\) (electron) c. \(^{99}\mathrm{Tc}\rightarrow ^{99}\mathrm{Ru} + ^0_{-1}\beta\) (beta particle) d. \(^{239}\mathrm{Pu}\rightarrow ^{235}\mathrm{U} + ^4_2\alpha\) (alpha particle)

Step-by-step solution

Identify the initial and final elements

In this decay process, we have Cobalt-60 (\(^{60}\mathrm{Co}\)) decaying into Nickel-60 (\(^{60}\mathrm{Ni}\)) and producing another particle.

Balance the atomic mass number (A) and atomic number (Z)

To balance the decay equation, the number of protons and neutrons must be conserved between the initial and final states. This can be achieved by finding the difference in atomic numbers and mass numbers. For Cobalt-60 and Nickel-60: Change in Atomic Number: \(Z_2 - Z_1 = 28 - 27 = 1\) Change in Mass Number: \(A_2 - A_1 = 60 - 60 = 0\) Thus, the emitted particle must have an atomic number of 1 and a mass number of 0. This particle is known as a beta particle (electron, \(-1\beta\)). The final decay equation is: \(^{60}\mathrm{Co}\rightarrow ^{60}\mathrm{Ni} + ^0_{-1}\beta\) b. \(^{97}\mathrm{Tc}+? \rightarrow^{97}\mathrm{Mo}\)

Identify the initial and final elements

In this decay process, we have Technetium-97 (\(^{97}\mathrm{Tc}\)) absorbing a particle before decaying into Molybdenum-97 (\(^{97}\mathrm{Mo}\)).

Balance the atomic mass number (A) and atomic number (Z)

To balance the decay equation, we need to find the particle that is being absorbed. For Technetium-97 and Molybdenum-97: Change in Atomic Number: \(Z_1 - Z_2 = 43 - 42 = 1\) Change in Mass Number: \(A_1 - A_2 = 97 - 97 = 0\) Thus, the absorbed particle must have an atomic number of 1 and a mass number of 0. The absorbed particle is an electron (\(-1\beta\)). The final decay equation is: \(^{97}\mathrm{Tc} + ^0_{-1}\beta\rightarrow^{97}\mathrm{Mo}\) c. \(^{99}\mathrm{Tc}\rightarrow^{99}\mathrm{Ru}+?\)

Identify the initial and final elements

In this decay process, we have Technetium-99 (\(^{99}\mathrm{Tc}\)) decaying into Ruthenium-99 (\(^{99}\mathrm{Ru}\)) and producing another particle.

Balance the atomic mass number (A) and atomic number (Z)

For Technetium-99 and Ruthenium-99: Change in Atomic Number: \(Z_2 - Z_1 = 44 - 43 = 1\) Change in Mass Number: \(A_2 - A_1 = 99 - 99 = 0\) Thus, the emitted particle must have an atomic number of 1 and a mass number of 0. It is a beta particle (electron, \(-1\beta\)). The final decay equation is: \(^{99}\mathrm{Tc}\rightarrow ^{99}\mathrm{Ru} + ^0_{-1}\beta\) d. \(^{239}\mathrm{Pu}\rightarrow^{235}\mathrm{U}+?\)

Identify the initial and final elements

In this decay process, we have Plutonium-239 (\(^{239}\mathrm{Pu}\)) decaying into Uranium-235 (\(^{235}\mathrm{U}\)) and producing another particle.

Balance the atomic mass number (A) and atomic number (Z)

For Plutonium-239 and Uranium-235: Change in Atomic Number: \(Z_1 - Z_2 = 94 - 92 = 2\) Change in Mass Number: \(A_1 - A_2 = 239 - 235 = 4\) Thus, the emitted particle must have an atomic number of 2 and a mass number of 4. This particle is an alpha particle (\(^4_2\alpha\), containing 2 protons and 2 neutrons). The final decay equation is: \(^{239}\mathrm{Pu}\rightarrow ^{235}\mathrm{U} + ^4_2\alpha\)

Key concepts

Beta Decay

In radioactive decay processes, beta decay is one of the most common forms where a beta particle is emitted or absorbed. A beta particle is essentially an electron, denoted as \(^0_{-1}\beta\). This particle carries a single negative charge, meaning it has an atomic number of \(-1\) and a mass number of \(0\).

During beta decay, an element's atomic number changes while its mass number remains the same. More specifically, the process involves a neutron in the nucleus transforming into a proton, emitting a beta particle in the process. This results in the atomic number increasing by 1, while the mass number does not change.

• Example A: Cobalt-60 \(^{60}\mathrm{Co}\) transforms into Nickel-60 \(^{60}\mathrm{Ni}\), emitting a beta particle.
• Example C: Technetium-99 \(^{99}\mathrm{Tc}\) transforms into Ruthenium-99 \(^{99}\mathrm{Ru}\), also emitting a beta particle.

These examples highlight the characteristic property of beta decay: the emitted particle's atomic numbers allow the conservation of charge in nuclear equations, thereby balancing the overall equation.

Alpha Decay

Alpha decay is another type of radioactive decay where an alpha particle is emitted from a nucleus. This particle consists of two protons and two neutrons, carrying a 2+ charge. In terms of its representation, an alpha particle is denoted as \(^4_2\alpha\).

When alpha decay occurs, the atomic number of the original element decreases by 2, and the mass number decreases by 4. This results in the formation of a new element that is two places lower in the periodic table and has a slightly lighter nucleus.

• Example D: Plutonium-239 \(^{239}\mathrm{Pu}\) transforms into Uranium-235 \(^{235}\mathrm{U}\) while emitting an alpha particle.

Alpha decay commonly occurs in heavier elements with atomic numbers greater than 82. These elements are inherently unstable, and alpha decay helps them achieve a more stable state.

Alpha particles, because they have more mass than beta particles, do not penetrate materials as deeply, making them easier to shield against with simple materials like paper. Yet, they produce a large impact on the material they do collide with.

Nuclear Equations

In radioactive decay, nuclear equations are used to represent the transformation of one element into another. They balance the count of protons and neutrons before and after decay, ensuring the conservation of mass and atomic numbers in the nuclei involved.

The components of a nuclear equation include:

• The starting (parent) nucleus
• The particle emitted or absorbed
• The resulting (daughter) nucleus

By writing these components with their appropriate atomic and mass numbers, we can ensure that both sides of the equation reflect equal values.

For instance, in a beta decay equation such as \(^{60}\mathrm{Co}\rightarrow ^{60}\mathrm{Ni} + ^0_{-1}\beta\), the cobalt atom decays to nickel by emitting an electron. The atomic numbers are balanced: 27 (for cobalt's protons) becomes 28 (for nickel's protons) plus \(-1\) (accounting for the beta particle). Similarly, \(^{239}\mathrm{Pu}\rightarrow ^{235}\mathrm{U} + ^4_2\alpha\) demonstrates conservation in an alpha decay where the sum of atomic and mass numbers remains constant across the equation.

Remember, understanding the changes and stability achieved through these decay processes is crucial for grasping nuclear chemistry's immense practical and theoretical implications.