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Life time in spontaneous emission for laser

  1. Mar 18, 2007 #1
    As we know, if we have a group of atoms at a moment N(t) in an exited state, it will exponentially decrease according to the relation:

    N(t)=N(0) exp(-At)=N(0) exp(-t/(tau))

    Where A is the Einsteins constant for the spontaneos emission,

    In my book they defined tau in the relation I explained in PDF


    Can Any one tell me WHERE DID HE BROUGHT THE SECOND RELATION FROM????

    Thanks for reading
     

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  2. jcsd
  3. Mar 20, 2007 #2
    Why no body is saying anything????????????????? This is always happening !!! Are my questions that difficult???? or not clear??? or Neglegible????

    Why no discussions are going in my questions !!!
     
  4. Mar 20, 2007 #3
    Heh. Your question is highly specific and technical (so few people will be able to answer it at all), your post is not well formatted (un-tex'd equations, excess punctuation and SHOUTING make reading unpleasant) and you require us to download some big attached file before we even know what your question is (which you can plainly see, few people will bother with). If you expect something from others, you might consider making it easy, fun, interesting or profitable for them to help you.
     
    Last edited: Mar 20, 2007
  5. Mar 23, 2007 #4
    Oh god !!!! Why shouting? does capitalizing means shouting? Capitalizing here means like bold or underline of somthing like this, to provide the reader with the simple way of questioning, and the briefed question, I can understand the highly specific question and technical !! But we can't say 177 kb is too much to download, If you have 28.8 kbps modem you'll need 1 min to download it, the page we are reading now got a size greater than 200 Kbs

    Can you guide me how can we make some mathematical problem like this interesting? i don't think it's possible, maybe i'll need to provide a prize for the correct answer
     
  6. Mar 23, 2007 #5
    Google netiquette. (And just personally, I also find the swearing uncomfortable.)
    This is really the wrong attitude when you're asking others to help you (they should be the ones to decide what is too much). In the case of this page, we already have the images cached, and your pdf adds nothing that you couldn't have included in your post.
    If this isn't interesting to you, why do you care? (This isn't the homework forum.)

    OK, so, "remaining number" is "initial number" multiplied by some exponential function of time and intrinsic activity.. basic decay stuff so far (you could probably verify that formula for youself, checking the units of activity, easy to forget if you need the odd factor of ln2.. and I've no idea how you thought this involves Einstein.. oh, you're applying it to spontaneous emission, nevermind).

    Now, [itex]\tau[/itex] often represents either half-life, or mean lifespan for any given particle. It looks like the latter: you've written it as, per initial particle, the mean value of time itself, weighted according to "the number of remaining particles and their activity" (which is basically just the number that decay at that moment). Its calculating the average amount of time that a randomly chosen particle will last before it decays. (And then it's shown conveniently that this is inversely proportional to activity.)

    Perhaps we could write the integral more suggestively as: [tex]\tau = \frac{\displaystyle\sum^{all\ decays\ \delta N}\ t\ \delta N} {N_{total}}[/tex]
    and remember that [itex]\frac{dN}{dt}[/itex] works out as -AN, in order to express everything (specifically including dN) in terms of one parameter (time).
     
    Last edited: Mar 23, 2007
  7. Mar 23, 2007 #6
    As one might guess from my user name, I am probably qualified to answer this question, but I find the original poster's impatience quite off-putting, and the original post doesn't really inspire me to respond to it. Letting us know the author(s) and title of the book you are using would be helpful, for example, so that we can see how it explains (or doesn't explain) this topic. That said, I will probably study cesiumfrog's answer later and contribute any additional insights I might be able to offer.
     
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