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Temperature and Radioactivity

  1. Sep 9, 2008 #1
    I know that if a radioactive substance is heated, then the radioactivity is reduced because of the relativistic thermal motion of the atoms.

    Is there a formula linking radioactive decay, temperature and perhaps, heat capacity?
  2. jcsd
  3. Sep 10, 2008 #2


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    The problem is that in order for this to be significant, the thermal energy of the particles needs to be of the order of their rest masses. Now, the rest mass of a single proton or neutron (expressed in energy units) is of the order of 1 GeV. At a temperature of 11 000 K, the average thermal energy of a particle is about 1 eV (that's given by the ratio of the Boltzman constant and the elementary charge). So in order for hydrogen atoms to have thermal energies which make them move relativistically in a significant way, we'd have to heat them to about 11 000 billion degrees. In order to do so for a radioactive nucleus with about 100 protons and neutrons, that's 100 times more even.

    But by that time, they are undergoing already a lot of nuclear interactions!
  4. Sep 10, 2008 #3
    To the first order of magnitude you're looking at something like this

    [tex] \lambda' = \lambda / (3/2 kT/mc^2 + 1)[/tex]

    where [tex]\lambda[/tex] is the rate of decay, k is Boltzmann constant, T is temperature, and m is mass of the atom.

    The effect is there but it's tiny. You'd have to heat the substance to billions of degrees to get anything remotely measurable.
  5. Sep 10, 2008 #4


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    Relativistic thermal motion would imply an extraordinarily high temperature - some thing beyond normal experience in the terrestrial environment.

    Radiation is a nuclear property as opposed to temperature and heat capacity (or specific heat) which are atomic or interatomic properties.

    Thermal energies of atoms are on the order of 0.02 eV at about room temperature.
  6. Sep 10, 2008 #5
    Let's see that first approximation again:

    [tex] \lambda' = \frac{\lambda}{ \frac{\frac{3}{2}kT}{mc^2} +1}[/tex]

    Is that correct?
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