Beta Decay: Videos & Practice Problems

Beta decay is a type of radioactive decay that occurs when an unstable atomic nucleus emits a beta particle, which is essentially an electron. In this process, the beta particle can be represented by the symbol \( e^- \), indicating that it has a negligible atomic mass (effectively 0) and an atomic number of -1. This negative atomic number reflects the fact that an electron is the antiparticle of a proton, which has an atomic number of +1.

To illustrate beta decay, consider the example of mercury-201 (\( \text{Hg}_{80}^{201} \)). In this case, mercury has an atomic number of 80, meaning it contains 80 protons. When mercury-201 undergoes beta decay, it emits a beta particle. The conservation of atomic mass and atomic number must be maintained during this process. Since the emitted electron has no mass, the atomic mass of the resulting element remains 201. However, the atomic number changes due to the emission of the electron.

To determine the new atomic number after the emission, we can set up the equation: \( 80 + (-1) = 81 \). This means that the new element formed after the beta decay will have an atomic number of 81, which corresponds to thallium (Tl). Therefore, the complete beta decay reaction can be represented as:

\[ \text{Hg}_{80}^{201} \rightarrow \text{Tl}_{81}^{201} + e^- \]

This example highlights the key aspects of beta decay, including the emission of a beta particle and the resulting change in atomic number while maintaining the atomic mass.

In the study of radioactive particles, alpha particles are recognized as the largest type, while beta particles are smaller in comparison. This size difference plays a crucial role in their behavior and effects on biological systems. Due to their larger size, alpha particles possess a higher ionizing power, meaning they can cause more damage when they interact with matter, particularly if ingested. Conversely, beta particles, being smaller, exhibit a lower ionizing power, which translates to less damage upon ingestion.

However, the smaller size of beta particles also grants them greater penetrating power, allowing them to penetrate the skin more effectively than alpha particles. This characteristic necessitates specific protective measures to prevent beta particles from entering the body. To effectively shield against beta radiation, a sheet of metal or a large block of wood is required, as these materials can absorb the energy of the beta particles and prevent them from causing harm.

In nuclear chemistry, beta decay is a type of radioactive decay in which a beta particle (an electron or positron) is emitted from an atomic nucleus. This process results in the transformation of one element into another while maintaining the same atomic mass. Understanding how to write balanced nuclear equations for beta emissions is crucial for studying nuclear reactions.

For example, consider magnesium-25, which has an atomic mass of 25 and an atomic number of 12. During beta decay, magnesium-25 emits a beta particle. The atomic mass remains unchanged at 25, while the atomic number increases by one, resulting in the formation of aluminum-25. The balanced nuclear equation can be represented as:

$$^{25}_{12}\text{Mg} \rightarrow ^{25}_{13}\text{Al} + \beta^-$$

In this equation, the beta particle is denoted as $$\beta^-$$, indicating that it is an electron emitted from the nucleus.

Another example involves ruthenium-102, which has an atomic mass of 102 and an atomic number of 44. Upon emitting a beta particle, the atomic mass remains 102, while the atomic number increases to 45, resulting in rhodium-102. The balanced nuclear equation for this decay is:

$$^{102}_{44}\text{Ru} \rightarrow ^{102}_{45}\text{Rh} + \beta^-$$

These examples illustrate the process of beta decay, where the emission of a beta particle leads to the transformation of one element into another, highlighting the importance of understanding atomic numbers and mass in nuclear reactions.

In the process of nuclear decay, understanding the transformations that occur during alpha and beta decay is crucial for solving problems related to the formation of new elements. In this scenario, we start with Thorium-232, which has an atomic number of 90, and we need to determine how many alpha and beta decays have occurred to produce Lead-208, with an atomic number of 82.

Alpha decay involves the emission of an alpha particle, which consists of 2 protons and 2 neutrons (essentially a helium nucleus). This process results in a decrease of the atomic mass by 4 and the atomic number by 2. Therefore, for each alpha decay, the element loses 4 units of mass and 2 units of atomic number. Conversely, beta decay involves the emission of a beta particle, represented as \( \frac{0}{-1} e \), which increases the atomic number by 1 without affecting the atomic mass.

To find the total number of decays, we first calculate the change in atomic mass from Thorium-232 to Lead-208. The difference in atomic mass is:

\( 232 - 208 = 24 \)

This indicates that a total decrease of 24 in atomic mass must be accounted for by alpha decays. Since each alpha decay reduces the atomic mass by 4, we can determine the number of alpha decays:

\( \frac{24}{4} = 6 \)

Thus, 6 alpha decays have occurred, resulting in a new atomic number:

\( 90 - (6 \times 2) = 78 \)

This means that after 6 alpha decays, we have transformed Thorium-232 into Platinum-208 (atomic number 78). Next, we need to convert Platinum-78 into Lead-82. Since each beta decay increases the atomic number by 1, we need to calculate how many beta decays are necessary to increase the atomic number from 78 to 82:

\( 82 - 78 = 4 \)

This indicates that 4 beta decays are required. Therefore, the complete transformation from Thorium-232 to Lead-208 involves 6 alpha decays and 4 beta decays.

In summary, the process can be represented as:

\( ^{232}_{90} \text{Th} \rightarrow 6 \text{α} + 4 \text{β} \rightarrow ^{208}_{82} \text{Pb} \)

Understanding the effects of alpha and beta decay on atomic mass and atomic number is essential for solving nuclear decay problems effectively.