Problem 66 - General Chemistry: Principles and Modern Applications

Question

Calculate the minimum kinetic energy (in megaelectronvolts) that \(\alpha\) particles must possess to produce the nuclear reaction $$_{2}^{4} \mathrm{He}+^{14}_{7} \mathrm{N} \longrightarrow^{17}_{8} \mathrm{O}+_{1}^{1} \mathrm{H}.$$ The nuclidic masses are \(_{2}^{4} \mathrm{He}=4.00260 \mathrm{u}\); \(_{7}^{14} \mathrm{He}=14.00307\mathrm{u}\);\(_{1}^{1} \mathrm{H}=1.00783 \mathrm{u}\);\(_{8}^{17} \mathrm{H}=16.99913 \mathrm{u}\);

Step-by-Step Solution

Answer

The minimum kinetic energy that α particles must possess is 289.61 MeV.

Step 1: Calculate individual mass energy of particles

First, we need to calculate the mass energy of each particle, because energy is conserved in nuclear reactions. The mass energy can be calculated by using the relation E=m*c², where m is the mass of particle and c is the speed of light. The mass of particles must be converted into kg from atomic mass unit (u). The energy obtained will be in joules. As speed of light c = \(3 \times 10^8\) m/s, and 1 u = \(1.66 \times 10^{-27}\) kg, we find the energy of each particle: \(E_{^{4}He} = 4.00260 u \times 1.66 \times 10^{-27} kg/u \times (3 \times 10^{8} m/s)^2 = 5.582 \times 10^{-10}\) J, \(E_{^{14}N} = 14.00307 u \times 1.66 \times 10^{-27} kg/u \times (3 \times 10^{8} m/s)^2 = 1.5602 \times 10^{-9}\) J, \(E_{^{1}H} = 1.00783 u \times 1.66 \times 10^{-27} kg/u \times (3 \times 10^{8} m/s)^2 = 1.411 \times 10^{-10}\) J, \(E_{^{17}O} = 16.99913 u \times 1.66 \times 10^{-27} kg/u \times (3 \times 10^{8} m/s)^2 = 1.8911 \times 10^{-9}\) J

Step 2: Calculate mass defect, ∆m

The mass defect is the difference between the initial mass and final mass. It can be calculated as ∆m = (mass of reactants – mass of products) = (total energy of reactants - total energy of products)= ((5.582 \times 10^{-10}) J + (1.5602 \times 10^{-9}) J) - ((1.411 \times 10^{-10}) J + (1.8911 \times 10^{-10}) J) = 4.64 \times 10^{-10} J

Step 3: Converting joules to megaelectronvolts (MeV)

The energy calculated is in joules, but in nuclear reactions, the energy is usually represented in electronvolts (eV), specifically in megaelectronvolts (MeV). To convert J to MeV, 1 J = \(6.242 \times 10^{12}\) MeV. The energy in MeV is thus given by 4.64 \times 10^{-10} J * \(6.242 \times 10^{12}\) MeV/J = 289.61 MeV

Key Concepts

Kinetic Energy Calculation

Kinetic energy is the energy an object possesses due to its motion. In the context of nuclear reactions, we often calculate the kinetic energy of particles to understand their behavior and interactions. For nuclear reactions, the total kinetic energy of the particles involved is crucial as it influences how these particles interact and transform in reactions.To calculate the kinetic energy, the formula used is:\[E_k = \frac{1}{2} m v^2\]where:

• \(E_k\) is the kinetic energy,
• \(m\) is the mass of the particle,
• \(v\) is the velocity of the particle.

In the given problem, rather than using velocity, we are interested in the energy equivalence through a mass defect, which inherently considers the kinetic interactions of the involved particles. This concept helps us find the minimum kinetic energy that \(\alpha\) particles must possess to trigger the given nuclear reaction. Recognizing the role of mass and energy conservation is fundamental in these calculations. The mass energy equivalence principle, \(E = mc^2\), often comes into play when determining energy scales for particles in nuclear physics.

Mass Defect

Mass defect is a key concept in nuclear physics, crucial for understanding nuclear reactions. It refers to the difference between the sum of the individual masses of nucleons (protons and neutrons) that make up an atom and the actual mass of the nucleus. This "missing" mass has been converted to energy, which helps bind the nucleus together through the strong nuclear force.To calculate the mass defect \(\Delta m\) in the exercise, we determine the difference between the total mass of the reactants and the products of the nuclear reaction:\[\Delta m = (\text{Total mass of reactants}) - (\text{Total mass of products})\]In this calculation:

• Reactants: \(_{2}^{4} \mathrm{He}\) and \(^{14}_{7} \mathrm{N}\)
• Products: \(_{8}^{17} \mathrm{O}\) and \(^{1}_{1} \mathrm{H}\)

Using their respective nuclidic masses, we find the mass defect, indicating the amount of mass converted into binding energy in the nuclear reaction. Mass defect is fundamental for calculating the associated energy changes in nuclear processes, often sparking the reaction such as in the provided exercise where minimum kinetic energy required is sought.

MeV Conversion

In nuclear physics, energies are typically represented in electronvolts (eV), with megaelectronvolts (MeV) being a common unit due to the vast energy scales involved in nuclear reactions. Converting energy from joules to MeV allows for a more intuitive understanding of the scales involved and is widely used in literature and calculations.To perform this conversion, use the relation:\[1 \text{ J} = 6.242 \times 10^{12} \text{ MeV}\]This conversion factor comes from the definition of the electronvolt, which is the amount of kinetic energy gained by a single electron accelerating through an electric potential difference of one volt.In the context of the exercise, the calculated energy due to the mass defect is initially in joules. Multiplying this energy by the conversion factor provides the energy in MeV, allowing us to discuss and analyze the nuclear reaction efficiently.This process is vital for communicating findings and values in a compact and understandable manner, especially when working with high-energy systems like nuclear reactions.