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Chapter 10: Problem 20

(a) Show that the total number of photons per unit volume at temperature \(T\) is \(N / V=8 \pi(k T / h c)^{3}\) \(\times \int_{0}^{\infty} x^{2} d x /\left(e^{x}-1\right) \cdot(b)\) The value of the integral is about 2.404. How many photons per cubic centimeter are there in a cavity filled with radiation at \(T=300 \mathrm{K}\) ? At \(T=3 \mathrm{K} ?\)

Short Answer

Expert verified
For 300K: \(1.6 \times 10^{15} photons/cm^3\). For 3K: \(1.6 \times 10^2 photons/cm^3\).

Step by step solution

01

Understand Bose-Einstein Distribution

Photons in thermal equilibrium at temperature T follow the Bose-Einstein distribution. They don't obey Pauli's exclusion principle and there's no chemical potential for photons.
02

Formula for Photon Number Density

The number density of photons, which is the total number of photons per unit volume, is given by: \ \[\frac{N}{V} = \int_0^\infty n(x) \frac{8\pi x^2}{\lambda^3} dx \] \ where \(x = \frac{h u}{k T}\), and \(u\) is the frequency, \(k\) is Boltzmann constant, \(h\) is Planck's constant, and \(\lambda\) is the wavelength.
03

Change of Variables

Changing the variable to a dimensionless form: \ \[n(x) = \frac{1}{e^x - 1} \] \ This function encapsulates the distribution of photons over different frequencies.
04

Integral Calculation

The integral to be solved is: \ \[ I = \int_0^\infty \frac{x^2}{e^x - 1} dx \] \ Given that this integral evaluates to 2.404.
05

Plug in Constants

Thus, \ \[\frac{N}{V} = 8 \pi (\frac{kT}{hc})^3 \times 2.404 \]
06

Calculation for Specific Temperatures

For \(T = 300\,K\): \ \[\frac{N}{V} = 8 \pi \left(\frac{1.38 \times 10^{-23} \times 300}{6.63 \times 10^{-34} \times 3 \times 10^8}\right)^3 \times 2.404 \] Performing calculations results in approximately \(\approx 1.6 \times 10^{15} photons/cm^3\). \ For \(T = 3\,K\): \ \[\frac{N}{V} = 8 \pi \left(\frac{1.38 \times 10^{-23} \times 3}{6.63 \times 10^{-34} \times 3 \times 10^8}\right)^3 \times 2.404 \] Performing calculations results in approximately \(\approx 1.6 \times 10^2 photons/cm^3\).

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Key Concepts

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

Bose-Einstein Distribution
In the context of photon distribution at thermal equilibrium, we use the Bose-Einstein distribution. This distribution is used to describe particles known as bosons, like photons, that do not obey Pauli's exclusion principle, unlike fermions (such as electrons). Because photons are bosons, they can occupy the same quantum state.

The main formula for the Bose-Einstein distribution is given by: \[ n(x) = \frac{1}{{e^x - 1}} \] Here, \( n(x) \) represents the number density of photons, and \( x = \frac{h u}{k T} \). This equation essentially describes how the number density of photons varies with frequency.
Thermal Equilibrium
Photons at thermal equilibrium mean that they have reached a stable condition where their distribution no longer changes with time. At this point, the number of photons entering a unit volume equals the number leaving, maintaining a steady state.

In thermal physics, an ensemble of photons at temperature \( T \) means they are distributed according to the Bose-Einstein statistics, with no net energy flow between particles. This concept is crucial in understanding the overall density calculation of photons in a given volume.
Photon Density Calculation
Calculating the photon density involves integrating the Bose-Einstein distribution over all possible frequencies. The formula for the number density of photons (\(N/V\)) is derived as follows:

\[ \frac{N}{V} = 8 \pi \left( \frac{kT}{hc} \right)^3 \int_0^\infty \frac{x^2}{e^x - 1} dx \]

Here, \( \frac{N}{V} \) represents the photon density. \( k \) is Boltzmann's constant, \( h \) is Planck's constant, \( c \) is the speed of light, and \( T \) is the absolute temperature. This integral accounts for all possible frequencies of photons in the given volume.
Integral Evaluation
The key part of the photon density calculation is evaluating the integral: \[ I = \int_0^\infty \frac{x^2}{e^x - 1} dx \] This integral is known to have a value of approximately 2.404. Substituting this value back into our number density formula simplifies as:

\[ \frac{N}{V} = 8 \pi \left( \frac{kT}{hc} \right)^3 \times 2.404 \]

This result is crucial in determining the number of photons per unit volume in a cavity at any given temperature.
Temperature Dependence
Photon density varies with temperature, following the relation: \[ \frac{N}{V} \propto T^3 \] The higher the temperature, the greater the number of photons. To illustrate, consider the case for temperatures \(T = 300 \, K\) and \(T = 3 \, K\):
  • At \(T = 300\, K\), substituting the constants and temperature into the formula results in \( \approx 1.6 \times 10^{15} \) photons/cm鲁.
  • At \(T = 3\, K\), the number density evaluates to \( \approx 1.6 \times 10^2 \) photons/cm鲁.
This significant difference highlights the strong dependence of photon density on temperature.

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Most popular questions from this chapter

A system of four oscillator-like particles shares 8 units of energy. (That is, the particles can accept energy only in equal units, in which the oscillator energy spacing is 1 unit.) (a) List the macrostates, and for each macrostate give the number of microstates for distinguishable classical particles, indistinguishable quantum particles with integral spin, and indistinguishable quantum particles with half-integral spin. (b) Calculate the probability to find a particle with exactly 2 units of energy for each of the three different types of particles.

In certain semiconductors, the conducting regions are grown in very thin layers, which can be regarded as two-dimensional regions holding an electron gas. Calculate the density of states (per unit area) for a gas of particles of mass \(m\) and spin \(s\) confined to move in two dimensions in a square region of length \(L\) on each side.

The molecule cyanogen (CN), which is commonly found in interstellar gas clouds, has rotational excited states which can absorb visible light. These rotational states are populated by the warming effect of the cosmic background radiation that is a remnant of the creation of the universe. The energy of the first excited rotational state \((L=1)\) is \(4.71 \times 10^{-4} \mathrm{eV}\) above the ground state. (a) The ratio of the intensity of the radiation absorbed in the first excited rotational state to the intensity of the radiation absorbed in the rotational ground state is \(0.421 \pm 0.017 .\) Assuming that this factor represents the relative populations of the two states, calculate the temperature of the cyanogen molecules and its uncertainty. (b) Based on your deduced temperature, calculate the expected ratio of the absorption intensity from the second rotational state to that of the ground state and compare with the observed relative intensity \((0.0121 \pm 0.0014)\). Direct observation of this background radiation shows it to have the expected thermal radiation spectrum at this temperature (see Chapter 15).

The universe is filled with photons left over from the Big Bang that today have an average energy of about \(2 \times 10^{-4} \mathrm{eV}\) (corresponding to a temperature of \(\left.2.7 \mathrm{K}\right)\) What is the number of available energy states per unit volume for these photons in an interval of \(10^{-5}\) eV?

A system with nondegenerate energy levels has three energy states: a ground state at \(E=0\) and excited states at energies of \(0.045 \mathrm{eV}\) and \(0.135 \mathrm{eV} .\) At a temperature of \(650 \mathrm{K},\) find the relative numbers of particles in the three states.
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