How Does Temperature Affect Radioactive Decay Rates?

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SUMMARY

The discussion centers on the relationship between temperature and radioactive decay rates, specifically how heating a radioactive substance can reduce its radioactivity due to relativistic thermal motion of atoms. The formula presented, λ' = λ / (3/2 kT/mc^2 + 1), illustrates the decay rate adjustment, where λ is the original decay rate, k is the Boltzmann constant, T is temperature, and m is the mass of the atom. However, significant temperature increases, on the order of billions of degrees, are required to observe measurable effects on decay rates, making practical applications in typical environments negligible.

PREREQUISITES
  • Understanding of radioactive decay and its mathematical representation
  • Familiarity with the Boltzmann constant and its role in thermodynamics
  • Knowledge of relativistic physics and thermal motion
  • Basic concepts of nuclear interactions and atomic mass
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Physicists, nuclear engineers, and researchers interested in the effects of temperature on radioactive materials and decay processes.

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

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 Boltzmann 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!
 
To the first order of magnitude you're looking at something like this

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

where \lambda 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.
 
DiamondGeezer said:
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?
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.
http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/spht.html

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

Thermal energies of atoms are on the order of 0.02 eV at about room temperature.
 
hamster143 said:
To the first order of magnitude you're looking at something like this

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

where \lambda 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.

Let's see that first approximation again:

\lambda' = \frac{\lambda}{ \frac{\frac{3}{2}kT}{mc^2} +1}

Is that correct?
 

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