Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

Radionuclides occur with half-lives in a vast range of over 37 magnitudes as listed in this site. In question 7584, Lubos Motl explained how Gyr half-lives were determined. This method doesn't appear applicable to evaluate sub-microsecond half-lives. How are extremely short half-lives determined?

share|improve this question

1 Answer 1

up vote 3 down vote accepted

Line width.

Any kind of fast decay is measured by its line width.

You make a high-precision energy measurements of a statistical number of decays, measure the width of the resulting distribution, subtract off contributions due to Doppler broadening and your instrument's intrinsic resolution and what you are left with goes into

$$ \Delta E \Delta t = \frac{\hbar}{2} $$

This is a staple method of experimental particle physics.

As you can see, very short half-lives translate to very broad lines. Other examples include the decay of the $J/\Psi$ (and indeed most strong decays).

share|improve this answer
1  
I'm always amazed at the utility of Heisenberg's principle. –  Michael Luciuk Feb 22 '12 at 4:05
    
But how can you be sure that it gives you an equality in Heisenberg's principle? My question is the following: Wouldn't this kind of reasoning give you a upper bound to the half-life instead of giving you a precise measurement of the same? –  user23873 May 8 '13 at 19:17
1  
Upper bound turns to an exact equality (with maybe some known factor of order unity) if we know the exact distribution function along mentioned variables. For example, Gaussian wavepacket along $x$ gives you an equality $\Delta p\Delta x=\hbar/2$. And for the decay processes we know that the exact function is the law of decay $\exp(-t/\tau)=\exp(-\Gamma t/\hbar)$. Thus we can use the Heisenberg's relation in its equality sense. –  firtree May 9 '13 at 3:26

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.