Problem 125 - Chemistry: An Atoms-Focused Approach

Question

The absorption of a neutron by \(^{11} \mathrm{B}\) produces a radioactive nuclide that decays by either \(\alpha\) decay or \(\beta\) decay. Write balanced nuclear equations describing the decay reactions.

Step-by-Step Solution

Answer

Answer: The balanced nuclear equations are: Alpha decay: $$_{5}^{12} \mathrm{B} \rightarrow _{3}^{8} \mathrm{Li} + _{2}^{4} \mathrm{He}$$ Beta-minus decay: $$_{5}^{12} \mathrm{B} \rightarrow _{6}^{12} \mathrm{C} + _{-1}^{0} \mathrm{e}$$ Beta-plus decay: $$_{5}^{12} \mathrm{B} \rightarrow _{4}^{12} \mathrm{Be} + _{1}^{0} \mathrm{e}$$

Step 1: Determine the nuclide after neutron absorption

When \(^{11} \mathrm{B}\) absorbs a neutron, it becomes \(^{12} \mathrm{B}\). The nuclear reaction is: $$_{5}^{11} \mathrm{B} + _{0}^{1} \mathrm{n} \rightarrow _{5}^{12} \mathrm{B}$$

Step 2: Write the alpha-decay equation

In alpha decay, the nucleus emits an alpha particle (\( _{2}^{4} \mathrm{He}\)). When \(^{12} \mathrm{B}\) undergoes alpha decay, the equation can be written as: $$_{5}^{12} \mathrm{B} \rightarrow _{3}^{8} \mathrm{X} + _{2}^{4} \mathrm{He}$$ We can now determine the element X by its atomic number, in this case, 3 (Lithium, Li): $$_{5}^{12} \mathrm{B} \rightarrow _{3}^{8} \mathrm{Li} + _{2}^{4} \mathrm{He}$$

Step 3: Write the beta-decay equation

In beta decay, the nucleus emits a beta particle (\( _{-1}^{0} \mathrm{e}\) or \( _{1}^{0} \mathrm{e}\), depending on whether it's a beta-minus decay or a beta-plus decay). Here, we are not given which type of beta decay occurs, so let's write the equations for both beta-minus and beta-plus decay. Beta-minus decay: $$_{5}^{12} \mathrm{B} \rightarrow _{6}^{12} \mathrm{X} + _{-1}^{0} \mathrm{e}$$ In this case, the element X has an atomic number of 6 (Carbon, C): $$_{5}^{12} \mathrm{B} \rightarrow _{6}^{12} \mathrm{C} + _{-1}^{0} \mathrm{e}$$ Beta-plus decay: $$_{5}^{12} \mathrm{B} \rightarrow _{4}^{12} \mathrm{X} + _{1}^{0} \mathrm{e}$$ In this case, the element X has an atomic number of 4 (Beryllium, Be): $$_{5}^{12} \mathrm{B} \rightarrow _{4}^{12} \mathrm{Be} + _{1}^{0} \mathrm{e}$$ As a result, we have obtained the balanced nuclear equations for each type of decay.

Key Concepts

Radioactive Decay

Radioactive decay is a process where an unstable atomic nucleus loses energy by emitting radiation. It's a bit like a rebalancing act for atoms. During this process, particles or energy are released, and the original atom is transformed into a different element or isotope. This transformation is spontaneous, meaning it happens naturally without any external influence. The result is typically the creation of a more stable nucleus. Atoms undergoing radioactive decay are called radioactive isotopes or radioisotopes. The specific type of decay depends on the nuclear structure of the isotope involved. Key types of radioactive decay include alpha decay, beta decay, and gamma decay, each characterized by different emitted particles or radiation.

• Alpha decay: This involves the emission of an alpha particle.
• Beta decay: This involves the emission of a beta particle, which can be either an electron or a positron.
• Gamma decay: This involves the release of energy in the form of gamma rays.

Each type of decay follows rules concerning the conservation of energy, charge, and atomic mass number.

Alpha Decay

Alpha decay is one of the most common types of radioactive decay. During this process, an unstable nucleus emits an alpha particle, which consists of two protons and two neutrons, the same as a helium nucleus. Because of this, alpha particles can be written as \[ _{2}^{4} \text{He} \]The emission of an alpha particle leads to a decrease in the mass number of the original atom by 4 units and a decrease in the atomic number by 2 units. This results in the formation of a new element.For example, when \(_5^{12} \text{B}\) undergoes alpha decay, the nuclear equation is:\[_{5}^{12} \text{B} \rightarrow _{3}^{8} \text{Li} + _{2}^{4} \text{He}\]Here, boron (\(_5^{12} \text{B}\)) transforms into a lithium isotope (\(_3^{8} \text{Li}\)) by emitting an alpha particle. Alpha decay typically happens in elements heavier than lead. It's a very effective way to reduce the size of a nucleus and stabilizes heavier elements.

Beta Decay

Beta decay is another common type of radioactive decay where an unstable nucleus emits a beta particle. This can either be a beta-minus particle (\(_{-1}^{0} \text{e}\)), which is an electron, or a beta-plus particle (\(_{1}^{0} \text{e}\)), which is a positron. This process allows a neutron in the nucleus to transform into a proton or the reverse, changing the element to another.**Beta-minus decay** involves the transformation where a neutron is converted into a proton, and a beta-minus particle (electron) is emitted:\[_{5}^{12} \text{B} \rightarrow _{6}^{12} \text{C} + _{-1}^{0} \text{e}\]In this reaction, boron (\(_5^{12} \text{B}\)) becomes carbon (\(_6^{12} \text{C}\)) with an emitted electron.**Beta-plus decay** involves the conversion of a proton into a neutron, emitting a beta-plus particle (positron):\[_{5}^{12} \text{B} \rightarrow _{4}^{12} \text{Be} + _{1}^{0} \text{e}\]Here, boron (\(_5^{12} \text{B}\)) is transformed into beryllium (\(_4^{12} \text{Be}\)) by releasing a positron.In beta decay, the mass number remains constant since the nucleus simply rearranges its internal particles. This type of decay plays a key role in the radioactive decay chains that lead to the formation of stable atoms.