Problem 4 - Chemistry Matter and Change

Question

Calculate Table 24.2 gives approximate energy values in units of MeV. Convert each value into joules using the following conversion factor \(1 \mathrm{MeV}=1.6 \times 10^{-13} \mathrm{J}\)

Step-by-Step Solution

Answer

Assuming three energy values: \(3.5 \mathrm{MeV}\), \(5.0 \mathrm{MeV}\), and \(2.7 \mathrm{MeV}\), we convert them to Joules using the conversion factor \(1 \mathrm{MeV} = 1.6 \times 10^{-13} \mathrm{J}\). The energy values in Joules are: For \(3.5 \mathrm{MeV}\): \(5.6 \times 10^{-13} \mathrm{J}\) For \(5.0 \mathrm{MeV}\): \(8.0 \times 10^{-13} \mathrm{J}\) For \(2.7 \mathrm{MeV}\): \(4.32 \times 10^{-13} \mathrm{J}\)

Step 1: Identify the given energy values and conversion factor

First, we need to identify the given energy values and the conversion factor. The conversion factor is provided as \(1 \mathrm{MeV} = 1.6 \times 10^{-13} \mathrm{J}\). Since we don't have the actual Table 24.2, we will assume some arbitrary energy values in MeV to use in the following steps. For the purpose of this explanation, let's consider three energy values: \(3.5 \mathrm{MeV}\), \(5.0 \mathrm{MeV}\), and \(2.7 \mathrm{MeV}\).

Step 2: Convert the MeV to Joules using the conversion factor

Now that we have the energy values in MeV and the conversion factor, we will multiply each energy value by the conversion factor to obtain the value in Joules. \(E_1 = 3.5 \mathrm{MeV} \times 1.6 \times 10^{-13} \mathrm{J/MeV}\) \(E_2 = 5.0 \mathrm{MeV} \times 1.6 \times 10^{-13} \mathrm{J/MeV}\) \(E_3 = 2.7 \mathrm{MeV} \times 1.6 \times 10^{-13} \mathrm{J/MeV}\)

Step 3: Calculate the energy values in Joules

Perform the multiplication for each energy value: \(E_1 = 3.5 \times 1.6 \times 10^{-13} \mathrm{J} = 5.6 \times 10^{-13} \mathrm{J}\) \(E_2 = 5.0 \times 1.6 \times 10^{-13} \mathrm{J} = 8.0 \times 10^{-13} \mathrm{J}\) \(E_3 = 2.7 \times 1.6 \times 10^{-13} \mathrm{J} = 4.32 \times 10^{-13} \mathrm{J}\)

Step 4: Present the results

The energy values in Joules are: For \(3.5 \mathrm{MeV}\), the equivalent energy in Joules is \(5.6 \times 10^{-13} \mathrm{J}\) For \(5.0 \mathrm{MeV}\), the equivalent energy in Joules is \(8.0 \times 10^{-13} \mathrm{J}\) For \(2.7 \mathrm{MeV}\), the equivalent energy in Joules is \(4.32 \times 10^{-13} \mathrm{J}\)

Key Concepts

MeV to Joules conversion

Understanding how to convert energy from Mega-electronvolts (MeV) to Joules is crucial when dealing with nuclear chemistry and physics problems. MeV is a common unit used to measure energy at an atomic and nuclear level. However, in most physics and chemistry problems, energies are often expressed in Joules (J), derived from the International System of Units.
To convert from MeV to Joules, we utilize the conversion factor:
**1 MeV = 1.6 × 10^-13 J**.

• This means that for every MeV of energy, it can be transformed into an equivalent energy measure in Joules by simply multiplying the MeV value by 1.6 × 10^-13.
• For instance, 3.5 MeV would convert to Joules as: \( 3.5 \text{ MeV} \times 1.6 \times 10^{-13} \text{ J/MeV} = 5.6 \times 10^{-13} \text{ J} \)

The conversion allows us to compare or further analyze energy values in a context that is unified across various scientific disciplines.

Nuclear Chemistry

Nuclear chemistry focuses on the reactions and changes in atomic nuclei. These reactions may involve the splitting of a heavy nucleus (fission) or the combination of light nuclei (fusion). Both processes release substantial amounts of energy.
Here's why MeV is commonly used in nuclear chemistry:

• It is a convenient unit due to the immense energy changes that occur at the nuclear level. For comparison, chemical reactions typically release energy in kilojoules (kJ), whereas nuclear reactions are thousands to millions of times more energetic.
• Why does this matter? Because the energy released in these reactions can be quantitatively compared using conversions like MeV to Joules to understand their impact fully.

Nuclear chemistry importantly helps us understand not only how these processes occur but also how the energy released can be harnessed for practical applications, such as power generation in nuclear reactors.

Energy calculations

Performing energy calculations is a foundational skill in both chemistry and physics. Calculations enable scientists to predict outcomes, speed up reactions, or even enhance efficiency in various systems.
In energy conversions, following the formula for conversion and understanding how to apply little numbers, such as exponents, is vital:

• When converting MeV values to Joules, using exponents is essential as it simplifies the representation of very large or very small numbers.
• For example, multiplying a value like 5 MeV by the conversion factor using exponents accurately gives: \( 5 \times 1.6 \times 10^{-13} \text{ J} = 8.0 \times 10^{-13} \text{ J} \)

Mastering these calculations prepares students and professionals to engage more effectively in fields where energy transformations are critical, ensuring precision and understanding at a deeper scientific level.

Physics in Chemistry

Physics and chemistry are closely intertwined. Concepts from physics are often used to explain chemical phenomena, and vice versa. Understanding the physical principles underlying chemical reactions sheds light on everything from reaction kinetics to energy exchanges.
Some areas where physics intersects with chemistry include:

• Thermodynamics: How energy flows in chemical reactions.
• Quantum Mechanics: Explains atomic behavior and energy levels, critical for understanding why MeV is useful in nuclear reactions.
• Conservation Laws: Physics principles, such as conservation of energy, help predict and comprehend outcomes in chemical processes.

In both the microscopic and macroscopic worlds, these principles guide the understanding of natural phenomena, making them an essential part of education in scientific fields.