Calculating the Excited State Lifetime of 223Ra Nucleus

AI Thread Summary
The discussion focuses on calculating the excited state lifetime of the 223Ra nucleus following its decay from 227Th. The key equation used is ΔEΔt = ħ/2, where Δt represents the lifetime of the excited state. Participants clarify that the 80 keV gamma photon does not determine the lifetime; instead, the natural line width of 0.6 keV is relevant for the calculation. The uncertainty in energy (ΔE) is identified as 0.6 keV, leading to the conclusion that Δt can be calculated using the formula Δt = ħ/(2ΔE). The discussion emphasizes the importance of understanding the relationship between energy uncertainty and lifetime in quantum mechanics.
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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.
 
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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
 
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so the .6 keV is our uncertainty energy?
so \Deltat =hbar/2E
 
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