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Chapter 28: Problem 63

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 \(27^{\circ} \mathrm{C} ?\)

Short Answer

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

Step by step solution

01

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: \(T = 27 + 273.15 = 300.15\, \mathrm{K}\).
02

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 \(T = 300.15\, \mathrm{K}\): \(KE = \frac{3}{2} \times 1.38 \times 10^{-23} \times 300.15\).
03

Simplify kinetic energy calculation

Perform the multiplication: \(KE = 3 \times 1.38 \times 10^{-23} \times 150.075 = 6.21 \times 10^{-21}\, \mathrm{J}\).
04

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 \(\lambda\): \(\lambda = \frac{hc}{E}\).
05

Calculate the wavelength of the photon

Substitute \(E = 6.21 \times 10^{-21}\, \mathrm{J}\) into the wavelength formula: \(\lambda = \frac{6.626 \times 10^{-34} \times 3.00 \times 10^8}{6.21 \times 10^{-21}}\).
06

Simplify the wavelength calculation

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

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

X rays with initial wavelength 0.0665 \(\mathrm{nm}\) undergo Compton scattering. What is the longest wavelength found in the scattered \(x\) rays? At which scattering angle is this wavelength observed?

\(\bullet\) Removing vascular lesions. A pulsed dye laser emits light of wavelength 585 nm in 450\(\mu\) s pulses. Because this wave- length is strongly absorbed by the hemoglobin in the blood, the method is especially effective for removing various types of blemishes due to blood, such as port-wine- colored birth-marks. To get a reasonable estimate of the power required for such laser surgery, we can model the blood as having the same specific heat and heat of vaporization as water \((4190 \mathrm{J} / \mathrm{kg} \cdot \mathrm{K}\) , \(2.256 \times 10^{6} \mathrm{J} / \mathrm{kg} ) .\) Suppose that each pulse must remove 2.0\(\mu g\) of blood by evaporating it, starting at \(33^{\circ} \mathrm{C}\) . (a) How much energy must each pulse deliver to the blemish? (b) What must be the power output of this laser? (c) How many photons does each pulse deliver to the blemish?

(a) What is the least amount of energy, in electron volts, that must be given to a hydrogen atom which is initially in its ground level so that it can emit the \(\mathrm{H}_{\alpha}\) line in the Balmer series? (b) How many different possibilities of spectral-line emissions are there for this atom when the electron starts in the \(n=3\) level and eventually ends up in the ground level? Calculate the wavelength of the emitted photon in each case.

Doorway diffraction. If your wavelength were \(1.0 \mathrm{m},\) you would undergo considerable diffraction in moving through a doorway. (a) What must your speed be for you to have this wavelength? (Assume that your mass is 60.0 \(\mathrm{kg} .\) ) (b) At the speed calculated in part (a), how many years would it take you to move 0.80 \(\mathrm{m}\) (one step)? Will you notice diffraction effects as you walk through doorways?

\(\bullet\) Protons are accelerated from rest by a potential difference of 4.00 \(\mathrm{kV}\) and strike a metal target. If a proton produces one photon on impact, what is the minimum wavelength of the resulting \(\mathrm{x}\) rays? How does your answer compare to the minimum wavelength if 4.00 \(\mathrm{keV}\) electrons are used instead? Why do x-ray tubes use electrons rather than protons to produce \(x\) rays?
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