Problem 20 - Chemistry: An Atoms-Focused Approach

Question

Write a balanced nuclear equation for a. Beta emission by \(^{161} \mathrm{Tb}\) b. Alpha emission by \(^{255} \mathrm{Lr}\) c. Electron capture by \(^{67} \mathrm{Ga}\) d. Positron emission by \(^{72} \mathrm{As}\)

Step-by-Step Solution

Answer

Question: Write the balanced nuclear equations for the following radioactive decay processes: a. Beta emission by \(^{161} \mathrm{Tb}\) b. Alpha emission by \(^{255} \mathrm{Lr}\) c. Electron capture by \(^{67} \mathrm{Ga}\) d. Positron emission by \(^{72} \mathrm{As}\) Answer: a. \(^{161} \mathrm{Tb}_{65} \rightarrow ^{161} \mathrm{Dy}_{66} + ^{0} \mathrm{e}_{-1}\) b. \(^{255} \mathrm{Lr}_{103} \rightarrow ^{251} \mathrm{No}_{101} + ^{4} \mathrm{He}_{2}\) c. \(^{67} \mathrm{Ga}_{31} + ^{0} \mathrm{e}_{-1} \rightarrow ^{67} \mathrm{Zn}_{30}\) d. \(^{72} \mathrm{As}_{33} \rightarrow ^{72} \mathrm{Ge}_{32} + ^{0} \mathrm{e}_{1}\)

Step 1: 1. Identify radioactive decay process

In this situation, we are dealing with beta emission (also known as beta decay). In beta decay, a neutron is transformed into a proton, and an electron (known as a beta particle) is emitted.

Step 2: 2. Write initial equation

The initial equation will look like this: $$^{161} \mathrm{Tb} \rightarrow ^{A} \mathrm{X} + ^{0} \mathrm{e}_{-1} $$

Step 3: 3. Balance the equation

We balance the equation by ensuring that the atomic number and mass number are conserved. For this decay process, the atomic number will increase by one while the mass number remains the same: $$^{161} \mathrm{Tb}_{65} \rightarrow ^{161} \mathrm{Dy}_{66} + ^{0} \mathrm{e}_{-1} $$ b. Alpha emission by \(^{255} \mathrm{Lr}\)

Step 4: 1. Identify radioactive decay process

In this case, we are dealing with alpha emission (also known as alpha decay). In alpha decay, an alpha particle, consisting of 2 protons and 2 neutrons (atomic mass 4 and atomic number 2), is emitted.

Step 5: 2. Write initial equation

The initial equation will look like this: $$^{255} \mathrm{Lr} \rightarrow ^{A} \mathrm{X} + ^{4} \mathrm{He}_{2}$$

Step 6: 3. Balance the equation

We balance the equation by ensuring that the atomic number and mass number are conserved. For this decay process, the atomic number will decrease by two and the mass number will decrease by four: $$^{255} \mathrm{Lr}_{103} \rightarrow ^{251} \mathrm{No}_{101} + ^{4} \mathrm{He}_{2}$$ c. Electron capture by \(^{67} \mathrm{Ga}\)

Step 7: 1. Identify radioactive decay process

In this situation, we are dealing with electron capture. In electron capture, a proton in the nucleus captures an electron from an inner electron shell and is converted into a neutron.

Step 8: 2. Write initial equation

The initial equation will look like this: $$^{67} \mathrm{Ga} + ^{0} \mathrm{e}_{-1} \rightarrow ^{A} \mathrm{X} $$

Step 9: 3. Balance the equation

We balance the equation by ensuring that the atomic number and mass number are conserved. For this decay process, the atomic number will decrease by one and the mass number remains the same: $$^{67} \mathrm{Ga}_{31} + ^{0} \mathrm{e}_{-1} \rightarrow ^{67} \mathrm{Zn}_{30}$$ d. Positron emission by \(^{72} \mathrm{As}\)

Step 10: 1. Identify radioactive decay process

In this case, we are dealing with positron emission (also known as beta decay). In positron emission, a proton is transformed into a neutron, and a positron (the antiparticle of the electron with a positive charge) is emitted.

Step 11: 2. Write initial equation

The initial equation will look like this: $$^{72} \mathrm{As} \rightarrow ^{A} \mathrm{X} + ^{0} \mathrm{e}_{1}$$

Step 12: 3. Balance the equation

We balance the equation by ensuring that the atomic number and mass number are conserved. For this decay process, the atomic number will decrease by one while the mass number remains the same: $$^{72} \mathrm{As}_{33} \rightarrow ^{72} \mathrm{Ge}_{32} + ^{0} \mathrm{e}_{1}$$

Key Concepts

Beta Emission

Beta emission, also commonly referred to as beta decay, is a fascinating process in nuclear chemistry. During this type of radioactive decay, a neutron inside the nucleus is transformed into a proton.
This transformation results in the emission of an electron, known as a beta particle.
One characteristic of beta decay is that while the atomic number increases by one, the mass number remains unchanged. This is because the total number of nucleons (protons and neutrons) in the nucleus does not change.
Consider the example of beta emission by \(^{161} \mathrm{Tb}\_{65}\). The balanced nuclear equation is:- Before decay: \(^{161} \mathrm{Tb}_{65}\)- Emission: \(^{0} \mathrm{e}_{-1}\) (beta particle)- After decay: \(^{161} \mathrm{Dy}_{66}\)In this process, terbium (Tb) is transformed into dysprosium (Dy) by increasing the atomic number by one, from 65 to 66.

Alpha Emission

Alpha emission, also known as alpha decay, involves the release of an alpha particle from the nucleus. This particle includes 2 protons and 2 neutrons, giving it a mass of 4 and an atomic number of 2.
The emission of an alpha particle reduces the parent nucleus's mass number by 4 and the atomic number by 2.Let's look at alpha emission by \(^{255} \mathrm{Lr}_{103}\) to illustrate the concept:- Before decay: \(^{255} \mathrm{Lr}_{103}\)- Emission: \(^{4} \mathrm{He}_{2}\) (alpha particle)- After decay: \(^{251} \mathrm{No}_{101}\)Through this decay process, lawrencium (Lr) becomes nobelium (No), marked by a drop in both atomic and mass numbers.

Electron Capture

Electron capture is a unique form of radioactive decay where an inner electron is captured by the nucleus. During this process, a proton is converted into a neutron.
This transforms the element into another by decreasing its atomic number by one, while the mass number stays constant since the nucleon count doesn't change.Take, for example, electron capture by \(^{67} \mathrm{Ga}_{31}\):- Before capture: \(^{67} \mathrm{Ga}_{31}\)- Captured electron: \(^{0} \mathrm{e}_{-1}\)- After decay: \(^{67} \mathrm{Zn}_{30}\)In this scenario, gallium (Ga) becomes zinc (Zn), demonstrating a reduction in atomic number by one.

Positron Emission

In positron emission, a proton is transformed into a neutron and gives off a positron, which is the antimatter equivalent of an electron but with a positive charge.
This type of decay reduces the atomic number by one while the mass number stays unchanged, as the total nucleon count remains the same.Positron emission can be exemplified by the decay of \(^{72} \mathrm{As}_{33}\):- Before emission: \(^{72} \mathrm{As}_{33}\)- Emission: \(^{0} \mathrm{e}_{1}\) (positron)- After decay: \(^{72} \mathrm{Ge}_{32}\)In this instance, arsenic (As) changes into germanium (Ge), showcasing a decrease in the atomic number. Positron emission is a key process in numerous scientific applications, including medical imaging techniques like PET scans.