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KT (energy) vs Blackbody Radiation for molecule in a box

  1. Jan 5, 2016 #1
    I'm trying to understand the meaning of kT (energy) in molecular systems, how to define the temperature of an individual molecule, and how a molecule recieves thermal energy or dissipates thermal energy.

    Here is my 'gendanken': One molecule is floating in a box, in vacuum. The temperature of all the walls of the box are room temperature, 298 K.
    I think the walls are emitting black-body radiation. If we use Wien's Law, the peak wavelength of this radiation is about 9.7 μm. So let us say the molecule is being bombarded with photons with wavelength of 9.7 μm, this corresponds to the photons having an energy of about 0.128 eV.

    What does it mean if the molecule is in thermal equilibrium with the box? Is the molecule also emitting photons at 0.128 eV? Where is this thermal energy stored - in vibrations of the bonds of the molecule, or in the electrons of the molecule?
    0.128 eV is about 4.9* kT, where kT is ~ 0.026 eV. I have read that the energy associated with a degree-of-freedom is about kT/2. Does this mean that the number of degrees of freedom available to the molecule depend on the temperature of the molecule? What if it is a monatomic gas molecule - He or Ar - it can't have 10 degrees of freedom!

    Is it reasonable that a blackbody radiator at room temperature emits photons with an energy of 5*kT?
  2. jcsd
  3. Jan 5, 2016 #2


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    Staff: Mentor

    No, because the molecule is not a blackbody. It definitely cannot absorb and emit at all frequencies.

    In all degrees of freedom, including translational. Note that because of quantum mechanics, not all degrees of freedom will be excited at a given temperature. At 298 K, a molecule has basically only translational and rotational degrees of freedom that are active, there is no vibrational or electronic excitation possible.

    That's the equipartition theorem. Every quadratic degree of freedom has an average energy of kT/2 at equilibrium.

    See above for the first question. For the second, an atom at 298 K has only translational degrees of freedom, so its average kinetic energy is 3kT/2.

    I'm not sure what you mean here.
  4. Jan 11, 2016 #3
    Thanks for your help DrC. I want to revisit my gendanken and replace the non-blackbody molecule in the vacuum box with a blackbody solid metal, and figure out the appropriate energy distribution for electrons.
    Let's say there is a sphere of solid metal floating in vacuum in a box with walls at room temperature. The metal sphere is large enough that it has a continuous density of states. The walls of the box are radiating as a black body. So, from Wien's law, I am assuming that the solid metal sphere is constantly being bombarded with photons with energy of 0.128 eV emitted by the box walls.

    I want to know the probability of occupation of an energy level of an electron 0.128 eV above the fermi level of the metal sphere.
    So I'll use the fermi-dirac distribution function, F(e)=1/(1+exp((E-mu)/kT) where kT = 0.026 eV for room temperature.
    E-mu = 0.128 eV, so F(0.128)=1/(1+exp(0.128/0.026))= 0.0072 or about 0.7%.

    So even though the metal sphere is constantly being bombarded with black-body photons at 0.128 eV, the probability of that electron energy level being occupied is less than 1%?
    Is this because the relaxation rate of electrons from the excited state to the fermi level is very quick? or is this because an incident photon is more likely to hit an electron with less energy than the fermi level, and so most of the incident photons excite electrons to energy levels lower than E-mu=0.128eV?

    If instead my sphere is filled with bosons, is the Bose-Einstein distribution the correct distribution to use?
    What particle should my sphere be filled with if the Boltzmann distribution correctly describes the energy distribution?
  5. Jan 16, 2016 #4


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    You cannot. The concept of temperature applies to a system of many particles, not to a single molecule. A molecule will have at a given time certain energy.

    Actually, inside a box (a cavity) you have a blackbody radiation, that is continuous from zero energy to infinity. The distribution is described by Planck's equation.

    A monoatomic gas has only 3 translational degrees of freedom. It has also electronic excitations but these are virtually impossible at room temperature.
    Another thing, there is no absorption of photons by molecular translation. The only possible energy exchange is during collision with the cavity walls.

    Yes, it is, Check the Planck's law, it gives you exactly the intensity of the radiation vs energy.

    No, not really. Fast kinetics means fast absorption and emission. In fact, it was the knowledge of absorption and emission kinetics that led Einstein to predict stimulated emission.
    Actually, a typical metal will have a very low absorption coefficient for 0.128 eV radiation (infrared) and most of the radiation would be reflected. It still does not affect the equilibrium distribution but how fast it would be reached.
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