Problem 63 - College Physics

Question

From the kinetic-molecular theory of an ideal gas (Chapter 15\()\) we know that the average kinetic energy of an atom is \(\frac{3}{2} k T\) . What is the wavelength of a photon that has this energy for a temperature of \(27^{\circ} \mathrm{C} ?\)

Step-by-Step Solution

Answer

The wavelength of the photon is \(3.20 \times 10^{-5}\, \mathrm{m}\).

Step 1: Convert Celsius to Kelvin

To use the kinetic energy formula, we need the temperature in Kelvin. Convert \(27^{\circ} \mathrm{C}\) to Kelvin by adding 273.15: \(T = 27 + 273.15 = 300.15\, \mathrm{K}\).

Step 2: Calculate average kinetic energy

Using the formula for average kinetic energy, \(KE = \frac{3}{2} k T\), where \(k\) is the Boltzmann constant \(= 1.38 \times 10^{-23} \mathrm{J/K}\). Substitute \(T = 300.15\, \mathrm{K}\): \(KE = \frac{3}{2} \times 1.38 \times 10^{-23} \times 300.15\).

Step 3: Simplify kinetic energy calculation

Perform the multiplication: \(KE = 3 \times 1.38 \times 10^{-23} \times 150.075 = 6.21 \times 10^{-21}\, \mathrm{J}\).

Step 4: Relate energy to wavelength of a photon

Using the relation \(E = \frac{hc}{\lambda}\), where \(h = 6.626 \times 10^{-34}\, \mathrm{Js}\) is Planck's constant and \(c = 3.00 \times 10^8\, \mathrm{m/s}\) is the speed of light. Solve for \(\lambda\): \(\lambda = \frac{hc}{E}\).

Step 5: Calculate the wavelength of the photon

Substitute \(E = 6.21 \times 10^{-21}\, \mathrm{J}\) into the wavelength formula: \(\lambda = \frac{6.626 \times 10^{-34} \times 3.00 \times 10^8}{6.21 \times 10^{-21}}\).

Step 6: Simplify the wavelength calculation

Perform the calculation: \(\lambda = \frac{1.9878 \times 10^{-25}}{6.21 \times 10^{-21}} = 3.20 \times 10^{-5}\, \mathrm{m}\).

Key Concepts

Ideal Gas Law

The Ideal Gas Law is a fundamental principle in chemistry and physics. It provides a relationship between pressure, volume, temperature, and the number of moles of a gas. The equation for the Ideal Gas Law is expressed as:

• \( PV = nRT \)

Where:

• \( P \) stands for pressure (in atmospheres, for example).
• \( V \) is the volume (in liters).
• \( n \) is the number of moles of the gas.
• \( R \) is the ideal gas constant \( (8.314 \, \mathrm{J/(mol \cdot K)}) \).
• \( T \) is the temperature in Kelvin.

This law applies to ideal gases, which are hypothetical gases that perfectly follow the gas laws under all conditions of temperature and pressure. Although no real gases are truly ideal, this law is a useful approximation for many gases under normal conditions. Understanding this law is crucial in finding out how variables react with changes in others, such as how increasing temperatures might affect pressure in a closed system.

Kinetic Energy

Kinetic energy is the energy that an object possesses due to its motion. In the context of gases, it refers to the movement of molecules. The kinetic-molecular theory helps to explain the behavior of gases, where it assumes that gas particles are in constant, random motion. The average kinetic energy of gas particles can be calculated with the formula:

• \( KE = \frac{3}{2} k T \)

Where:

• \( KE \) is the kinetic energy.
• \( k \) is the Boltzmann constant (\( 1.38 \times 10^{-23} \, \mathrm{J/K} \)).
• \( T \) is the temperature in Kelvin.

This equation shows that the kinetic energy of atoms increases as the temperature rises. It highlights the direct relationship between temperature and molecular movement.

Photon Wavelength

The wavelength of a photon is inversely related to its energy. Photons are the basic units of light and electromagnetic radiation, and they have both particle-like and wave-like properties. The energy of a photon can be expressed as:

• \( E = \frac{hc}{\lambda} \)

Where:

• \( E \) is the energy of the photon.
• \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \, \mathrm{Js} \)).
• \( c \) is the speed of light (\( 3.00 \times 10^8 \, \mathrm{m/s} \)).
• \( \lambda \) is the wavelength.

When the energy of a photon is known, its wavelength can be calculated using the formula \( \lambda = \frac{hc}{E} \). This relationship is crucial in fields such as spectroscopy, where the knowledge of wavelengths helps identify substances and analyze their properties.

Temperature Conversion

Temperature conversion is essential in science, particularly when using formulas that require specific units. Often, Celsius must be converted to Kelvin, as thermodynamic equations like those involving the kinetic-molecular theory use Kelvin. To convert Celsius to Kelvin, use the following formula:

• \( T_{\text{Kelvin}} = T_{\text{Celsius}} + 273.15 \)

For example, if you have a temperature of \(27^{\circ} \mathrm{C} \), to convert to Kelvin, add 273.15, resulting in \(300.15 \, \mathrm{K} \). This conversion ensures that temperature values are absolute, allowing for consistent and accurate calculations in physical equations.