Problem 48 - Chemistry: An Atoms-Focused Approach

Question

Fluorine- 18 is often introduced into specific drug molecules for use as imaging agents. a. Write a balanced nuclear equation for the decay of \(^{18} \mathrm{F}\). b. Calculate the binding energy for \(^{18} \mathrm{F}\). The exact mass of \(^{18} \mathrm{F}\) is \(2.98915 \times 10^{-26} \mathrm{kg}\)

Step-by-Step Solution

Answer

a. Write the balanced nuclear equation for the decay of Fluorine-18. The nuclear equation for the decay of \(^{18}\mathrm{F}\) is: $$ ^{18} \mathrm{F} \rightarrow ^{18}\mathrm{O} + e^+ + \nu_e $$ b. Calculate the binding energy for \(^{18}\mathrm{F}\), given its exact mass. The binding energy for \(^{18}\mathrm{F}\) is approximately \(147.667\,\mathrm{MeV}\).

Step 1: Identify the decay process and write the balanced nuclear equation

Fluorine-18 (\(^{18}\mathrm{F}\)) undergoes beta-plus decay (also known as positron emission). In this process, a proton inside the nucleus transforms into a neutron, emitting a positron and a neutrino. The equation for the decay of \(^{18}\mathrm{F}\) can be written as: $$ ^{18} \mathrm{F} \rightarrow ^{18}\mathrm{O} + e^+ + \nu_e $$ Here, \(^{18}\mathrm{F}\) decays into \(^{18}\mathrm{O}\) (Oxygen-18), along with a positron (\(e^+\)) and a neutrino (\(\nu_e\)).

Step 2: Calculate the binding energy for \(^{18}\mathrm{F}\)

The binding energy of a nucleus can be calculated using the mass defect (\(\Delta M\)), which is the difference between the mass of the nucleus and the sum of masses of its individual nucleons (protons and neutrons). First, we calculate the mass of the individual nucleons in \(^{18}\mathrm{F}\). Since it has 9 protons and 9 neutrons (18 nucleons), we'll need the mass of a single proton (\(m_p\)) and the mass of a single neutron (\(m_n\)). The given mass of \(^{18}\mathrm{F}\) is \(2.98915 \times 10^{-26}\,\mathrm{kg}\). The mass of one proton is \(m_p = 1.6726219 \times 10^{-27}\,\mathrm{kg}\). The mass of one neutron is \(m_n = 1.6749275 \times 10^{-27}\,\mathrm{kg}\). Now, we calculate the sum of masses of these nucleons: $$ 9m_p + 9m_n = 9(1.6726219 \times 10^{-27}\,\mathrm{kg}) + 9(1.6749275 \times 10^{-27}\,\mathrm{kg}) = 3.0154518 \times 10^{-26}\, \mathrm{kg} $$ Next, we find the mass defect (\(\Delta M\)) as the difference between the sum of nucleon masses and the exact mass of \(^{18}\mathrm{F}\): $$ \Delta M = 3.0154518 \times 10^{-26}\,\mathrm{kg} - 2.98915 \times 10^{-26}\,\mathrm{kg} = 2.63018 \times 10^{-28}\,\mathrm{kg} $$ We convert this mass defect into energy using Einstein's mass-energy equivalence formula: \(E = \Delta M \times c^2\). The speed of light (\(c\)) is \(3.0 \times 10^8\, \mathrm{m\,s^{-1}}\). $$ E = 2.63018 \times 10^{-28}\,\mathrm{kg}\cdot (3.0 \times 10^8\,\mathrm{m\,s^{-1}})^2 = 2.36716 \times 10^{-11}\,\mathrm{J} $$ Now, to convert this energy into more practical units (MeV), we'll use the conversion factor \(1\,\mathrm{MeV} = 1.60218 \times 10^{-13}\, \mathrm{J}\): $$ E = \frac{2.36716 \times 10^{-11}\,\mathrm{J}}{1.60218 \times 10^{-13}\, \mathrm{J\,MeV^{-1}}} = 147.667\,\mathrm{MeV} $$ So, the binding energy for \(^{18}\mathrm{F}\) is approximately \(147.667\,\mathrm{MeV}\).

Key Concepts

Beta Decay

In nuclear physics, beta decay is a fascinating process. This involves the transformation of neutrons into protons or vice versa within an atomic nucleus. There are two primary kinds of beta decay: beta-minus (β-) decay and beta-plus (β+) decay. Let's focus on beta-plus decay, as it applies to the example of Fluorine-18.

Beta-plus decay, or positron emission, involves a proton changing into a neutron. During this transformation, a positron (which is a positively charged electron) and a neutrino are emitted. The equation for Fluorine-18 beta-plus decay is:

• Fluorine-18 decays into Oxygen-18 with the emission of a positron and a neutrino:
  
  \(^{18} \mathrm{F} \rightarrow ^{18}\mathrm{O} + e^+ + u_e\)

In this balanced equation, Fluorine-18 (\(^{18}\mathrm{F}\)) experiences a change in its nucleus while simultaneously ejecting a positron (\(e^+\)) and a neutrino (\(u_e\)). Understanding beta decay is crucial in many applications, such as medical imaging processes with Fluorine-18 as part of PET scans.

Binding Energy

The concept of binding energy illuminates why atomic nuclei are stable. Binding energy is the energy needed to separate a nucleus into its individual protons and neutrons. It's a crucial measure of nuclear stability and is derived from the mass defect in a nucleus.

To find the binding energy of Fluorine-18,

• Calculate the Sum of Proton and Neutron Masses: The mass of each proton is \(1.6726219 \times 10^{-27}\,\mathrm{kg}\) and each neutron is \(1.6749275 \times 10^{-27}\,\mathrm{kg}\). For Fluorine-18 with 9 protons and 9 neutrons, the sum is \(3.0154518 \times 10^{-26}\, \mathrm{kg}\).
• Determine Mass Defect: Subtract the exact mass of the nucleus (\(^{18}\mathrm{F}\)) from the sum of the nucleon masses.
• Use Einstein’s Equation: Convert the mass defect into energy using \(E = \Delta M \times c^2\).

The binding energy calculated results in approximately \(147.667\,\mathrm{MeV}\). This high energy indicates a stable nucleus, as a significant amount of energy is required to disrupt it.

Mass Defect

Mass defect is a fundamental concept in nuclear chemistry. It refers to the difference between the combined mass of a nucleus's individual protons and neutrons and the actual measured mass of the nucleus itself. This difference arises because some mass is converted into binding energy, which holds the nucleus together.

For Fluorine-18,

• Sum of Nucleon Masses: Calculate by multiplying the individual masses of protons and neutrons and then summing them.
• Calculate Mass Defect: Subtract the mass of the entire nucleus from this sum.

This resulting mass defect is found using:\[\Delta M = 3.0154518 \times 10^{-26}\,\mathrm{kg} - 2.98915 \times 10^{-26}\,\mathrm{kg} = 2.63018 \times 10^{-28}\,\mathrm{kg}\]This value of \(\Delta M\) is vital as it is directly associated with how much energy is released when the nucleus is formed from its individual nucleons. The larger the mass defect, the more energy is released, signifying a more stable nucleus. Understanding mass defect is crucial for the study of nuclear reactions and energy production.