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Is there a formula linking radioactive decay, temperature and perhaps, heat capacity?

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Is there a formula linking radioactive decay, temperature and perhaps, heat capacity?

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vanesch

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

But by that time, they are undergoing already a lot of nuclear interactions!

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[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.

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Astronuc

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

Is there a formula linking radioactive decay, temperature and perhaps, heat capacity?

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

http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/spht.html

http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/temper.html

Thermal energies of atoms are on the order of 0.02 eV at about room temperature.A convenient operational definition of temperature is that it is a measure of the average translational kinetic energy associated with the disordered microscopic motion of atoms and molecules.

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Let's see that first approximation again:

[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.

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

Is that correct?

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