Calculating the Excited State Lifetime of 223Ra Nucleus

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SUMMARY

The discussion focuses on calculating the excited state lifetime of the 223Ra nucleus following its decay from 227Th. The relevant equation used is the energy-time uncertainty relation, expressed as ΔEΔt = ħ/2, where ΔE is the natural line width of the emitted gamma ray, measured at 0.6 keV. Participants clarify that the 80 keV gamma photon does not determine the lifetime; instead, the line width is critical for this calculation. The final expression for the lifetime (Δt) is derived as Δt = ħ/(2 * 0.6 keV).

PREREQUISITES
  • Quantum mechanics principles, specifically the energy-time uncertainty relation.
  • Understanding of nuclear decay processes and excited states.
  • Familiarity with gamma radiation and its properties.
  • Basic knowledge of Planck's constant (ħ) and its application in physics.
NEXT STEPS
  • Study the derivation and applications of the energy-time uncertainty principle in quantum mechanics.
  • Research the properties and significance of gamma radiation in nuclear physics.
  • Explore the decay processes of other isotopes and their excited states.
  • Learn about the calculation of lifetimes in quantum systems using different methods.
USEFUL FOR

Physics students, nuclear physicists, and researchers interested in quantum mechanics and nuclear decay processes will benefit from this discussion.

phyguy321
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Homework Statement


the nucleus of 227Th decays to 223Ra and \alpha. the daughter nucleus is left in a short lived excited state and decays down to the ground state with the emission of an 80 keV gamma ray. the natural line width of this gamma ray is .6 keV. what is the lifetime of the excited state of the 223Ra nucleus?


Homework Equations


\DeltaE \Deltat = \hbar/2 where \Deltat is the lifetime \tau


The Attempt at a Solution


can i just solve for \Deltat as the lifetime of the excited state? letting \DeltaE = 80 keV? I am not sure what to do with the line width, it's not in my book anywhere.
 
Physics news on Phys.org
The 80 keV photon isn't the thing that determines the lifetime of the state. Check this out and see if it doesn't clear things up.

http://www.mwit.ac.th/~Physicslab/hbase/quantum/parlif.html
 
Last edited by a moderator:
so the .6 keV is our uncertainty energy?
so \Deltat =hbar/2E
 

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