Calculating Decay Lifetime of Unstable Isotope

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SUMMARY

The discussion focuses on calculating the decay lifetime of an unstable isotope using the average energy of emitted gamma rays, specifically 100 keV, and a line-width of 5 x 10-6 eV. The relationship between energy and frequency, as described by the equation E = hf, is explored in the context of the Doppler shift effect. The time-energy Heisenberg uncertainty principle is identified as a relevant concept, with the equation Energy uncertainty x Time uncertainty = h / (4π) being highlighted as crucial for understanding the relationship between energy measurements and decay lifetime.

PREREQUISITES
  • Understanding of gamma ray emissions and their energy measurements
  • Familiarity with the Doppler shift effect in physics
  • Knowledge of the Heisenberg uncertainty principle, particularly the time-energy relationship
  • Basic grasp of quantum mechanics and particle physics concepts
NEXT STEPS
  • Study the time-energy Heisenberg uncertainty principle in detail
  • Explore the implications of the Doppler shift effect on gamma ray emissions
  • Research methods for calculating decay lifetimes of unstable isotopes
  • Examine the relationship between energy uncertainty and decay processes in quantum mechanics
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Physicists, nuclear scientists, and students studying quantum mechanics or particle physics, particularly those interested in isotope decay and gamma ray emissions.

square_imp
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My question relates to calculating the decay lifetime of an unstable isotope. The information given is the average energy of the emitted gamma ray from the decay has an average energy of say 100kev and a line-width of 5 x 10^-6ev. From this information I need to work out the average lifetime for the isotope.

From what I can gather the problem seems to be similar to the Doppler shift effect. From the equation E = hf the energy of the gamma ray is related to its frequency and therefore the change in frequency can be worked out. Does the line width mean the upper and lower boundary of the gamma ray energy with the 100keV in the middle of that range? The other thing is that the shift of the energy either means the source is moving or the energy of the gamma rays emitted is changing for some other reason. The relation between this and the lifetime is not obvious to me. Any help would be much appreciated. I am probably missing something obvious.
 
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square_imp said:
My question relates to calculating the decay lifetime of an unstable isotope. The information given is the average energy of the emitted gamma ray from the decay has an average energy of say 100kev and a line-width of 5 x 10^-6ev. From this information I need to work out the average lifetime for the isotope.

From what I can gather the problem seems to be similar to the Doppler shift effect. From the equation E = hf the energy of the gamma ray is related to its frequency and therefore the change in frequency can be worked out. Does the line width mean the upper and lower boundary of the gamma ray energy with the 100keV in the middle of that range? The other thing is that the shift of the energy either means the source is moving or the energy of the gamma rays emitted is changing for some other reason. The relation between this and the lifetime is not obvious to me. Any help would be much appreciated. I am probably missing something obvious.

To *me*, this seems to be an application of the time-energy Heisenberg uncertainty principle. Have you covered that?
 
From what I recall we have covered parts of the Heisenberg uncertainty principle, to note the position-momentum relationship. I will have a look and see what I can find about the time-energy relationship. I presume it will be similar to the position-momentum relationship.
 
From looking at the Heisenberg Principle again I find the relation as follows:

Energy uncertainty x Time uncertainty = Planks constant / 4 x pi

This is an equation describing that both the time and energy of a particle cannot be simultaneously accurately measured. The connection with my original problem I still cannot see really. :confused:
 

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