Question

Write balanced nuclear equations for the following: (a) Formation of Mn-52 by positron emission. (b) Formation of Ac-228 by \(\beta\) -emission. (c) Formation of \(\mathrm{Np}-232\) by \(\alpha\) -decay.

Short answer

Question: Write balanced nuclear equations for the following processes: (a) formation of Mn-52 by positron emission, (b) formation of Ac-228 by β-emission, and (c) formation of Np-232 by α-decay. Answer: (a) \[_{24}^{52}\mathrm{Cr}\rightarrow\,_{}^{}\mathrm{e^{+}}+\,_{25}^{52}\mathrm{Mn}\] (b) \[_{90}^{228}\mathrm{Th}\rightarrow\,_{}^{}\mathrm{e^{-}}+\,_{89}^{228}\mathrm{Ac}\] (c) \[_{94}^{236}\mathrm{Pu}\rightarrow\,_{}^{}\mathrm{\alpha}+\,_{93}^{232}\mathrm{Np}\]

Step-by-step solution

Identify initial and final nuclei and emitted particles for each process

For each process, we need to identify the initial nucleus, the final nucleus, and the particle that is emitted. (a) Formation of Mn-52 by positron emission. Initial nucleus: Unknown Final nucleus: Mn-52 (Manganese-52) Emitted particle: Positron (e+) (b) Formation of Ac-228 by β-emission. Initial nucleus: Unknown Final nucleus: Ac-228 (Actinium-228) Emitted particle: Beta particle (β− or e−) (c) Formation of Np-232 by α-decay. Initial nucleus: Unknown Final nucleus: Np-232 (Neptunium-232) Emitted particle: Alpha particle (α or 4He)

Write down the balanced nuclear equations

For each process, we will write down the balanced nuclear equation, making sure to conserve the mass and atomic numbers on both sides of the equation. (a) Formation of Mn-52 by positron emission. Initial nucleus: We need to find an element that, when undergoing positron emission, forms Mn-52. Since a positron has a positive charge and a mass number of 0, the mass number will remain the same, and the atomic number will decrease by 1 in the process. Therefore, the initial nucleus will be 52Cr (Chromium-52). Balanced equation: \[_{24}^{52}\mathrm{Cr}\rightarrow\,_{}^{}\mathrm{e^{+}}+\,_{}^{}\mathrm{Mn}\] (b) Formation of Ac-228 by β-emission. Initial nucleus: We need to find an element that, when undergoing beta emission, forms Ac-228. Since a beta particle has a negative charge and a mass number of 0, the mass number will remain the same, and the atomic number will increase by 1 in the process. Therefore, the initial nucleus will be 228Th (Thorium-228). Balanced equation: \[_{90}^{228}\mathrm{Th}\rightarrow\,_{}^{}\mathrm{e^{-}}+\,_{}^{}\mathrm{Ac}\] (c) Formation of Np-232 by α-decay. Initial nucleus: We need to find an element that, when undergoing alpha decay, forms Np-232. Since an alpha particle has a positive charge of 2 and a mass number of 4, the mass number will decrease by 4 and the atomic number will decrease by 2 in the process. Therefore, the initial nucleus will be 236Pu (Plutonium-236). Balanced equation: \[_{94}^{236}\mathrm{Pu}\rightarrow\,_{}^{}\mathrm{\alpha}+\,_{}^{}\mathrm{Np}\]

Key concepts

Positron Emission

Positron emission is a type of radioactive decay where a nucleus emits a positron, which is a positively charged electron, also represented as \text{e}^+. This process occurs in proton-rich nuclei, when a proton within the nucleus is transformed into a neutron while releasing a positron and an electron neutrino.

In the exercise, the nucleus of Chromium-52 (\text{Cr}) undergoes positron emission to form Manganese-52 (\text{Mn}). A defining characteristic of positron emission is that the mass number (the total number of protons and neutrons) stays the same because both a proton and neutron contribute to the mass, but the atomic number (the number of protons) decreases by one, as a proton is lost.

The balanced nuclear equation for this process can be written as: \[ _{24}^{52}\text{Cr} \rightarrow \text{e}^+ + _{25}^{52}\text{Mn} \] This demonstrates the conservation of both mass and charge during positron emission, accounting for the atomic numbers (24 to 25) and mass numbers (both 52) on each side of the equation.

Beta Emission

Beta emission, also known as beta decay, is the process by which an unstable nucleus releases a beta particle, which is an electron (or positron in the case of beta-plus decay). In beta-minus decay (\text{β}^-), a neutron is converted into a proton while ejecting an electron and an antineutrino.

The example given in the exercise involves Thorium-228 (\text{Th}) transforming into Actinium-228 (\text{Ac}) through beta emission. Just like in positron emission, the mass number remains unchanged since the neutron inside the nucleus gets converted into a proton. However, the atomic number increases by one because the new proton adds to the number of protons in the nucleus.

The balanced equation for this process is: \[ _{90}^{228}\text{Th} \rightarrow \text{β}^- + _{89}^{228}\text{Ac} \] Beta emission highlights the transformation within the nucleus and adherence to the laws of conservation where neither mass nor electric charge is lost.

Alpha Decay

Alpha decay is a type of radioactive decay where an unstable atomic nucleus emits an alpha particle, which consists of two protons and two neutrons, and is identical to the nucleus of a helium atom (\text{He}). The process leads to the reduction of both mass and atomic numbers of the original nucleus.

As shown in the exercise, Plutonium-236 (\text{Pu}) undergoes alpha decay to produce Neptunium-232 (\text{Np}). The emission of an alpha particle, which has a mass number of 4 and an atomic number of 2, results in a decrease in mass by four units and a decrease in atomic number by two units.

The balanced nuclear equation is written as: \[ _{94}^{236}\text{Pu} \rightarrow \text{α} + _{93}^{232}\text{Np} \] This equation reflects the conversion that occurs during alpha decay, highlighting the changes in both mass and atomic numbers, and demonstrating the conservation of mass and charge in nuclear reactions.