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Modern II: Simple Decay Problem

  • #1

Homework Statement


An unstable sysemt may have two or more decay modes, with a Rate R1 for decay through mode 1 and R2 for decay through mode 2. Use the meaning of the rate to show that the total lifetime [tex]\tau[/tex] of the system is given by 1/[tex]\tau[/tex]=1/[tex]\tau[/tex]1+1/[tex]\tau[/tex]2.


Homework Equations





The Attempt at a Solution


Homework Statement





Homework Equations





The Attempt at a Solution


I tried to manipulate the basic equation for decay, but I hit a dead end. I think the part I am stuck on is how to insert the fact that there are two different rates into the equation. Thanks
 

Answers and Replies

  • #2
fzero
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Can you write down the basic differential equation for decay? What do the 2 terms in the equation mean? The interpretation should suggest how to modify the equation when there is a 2nd mode of decay.
 
  • #3
Yeah you mean dN(t)/dt=-RN(t)? The R is rate, the N is the amount of substance, and t is time.

Also, N(t)=N*exp(-Rt)
 
  • #4
fzero
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Right but what does dN/dt mean? And what does it mean to say that it's equal to -RN? Once you consider the concepts, understanding the effect of a 2nd decay mode should follow quickly.
 
  • #5
It means the rate that the substance is disappearing. That rate is proportional the the amount present at any time.
 
  • #6
fzero
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So if there's one process that proceeds at rate [tex]R_1[/tex] and another with rate [tex]R_2[/tex], what is the total rate that the substance is disappearing?
 
  • #7
Eh I'm not sure. I know they don't just simply add.
 
  • #8
fzero
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Let's put it this way. We first compute the amount of substance that has decayed due to process one over an infinitesimal amount of time dt. Call this [tex]dN_1[/tex]. We next compute the amount that decays due to process two, [tex]dN_2[/tex]. When we compute this we can neglect the small change in [tex]N[/tex] due to process one. We then use these to determine the total decay [tex]dN[/tex].
 
  • #9
In this case, wouldn't they just add? Perhaps I was wrong in my previous statement.
 
  • #10
fzero
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The contributions to [tex]dN/dt[/tex] do add. You should check what that says about the total rate of decay (easy) and then connect that information to the lifetimes.
 
  • #11
So my guess is that the total rate of decay will also add, thus lifetimes add.
 
  • #12
fzero
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The lifetimes don't directly add, there's a step or two of math to obtain the desired formula.
 
  • #13
But if R adds, and R=1/Tao, why can't R+R1=1/Tao + 1/Tao1?
 
  • #14
fzero
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But if R adds, and R=1/Tao, why can't R+R1=1/Tao + 1/Tao1?
That's correct. You wrote "lifetimes add" without the correct equation and I wanted to alert you without feeding you the answer.
 
  • #15
Ahh ok thanks.

There's an extension of this problem that says to use a general result of the problem we just solved to show that the total lifetime of the system is approximately given by the shortest among the lifetimes of the different modes.

Would that just be because the smaller the Tao, the more significant it is when 1/Tao is considered?
 
  • #16
fzero
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Ahh ok thanks.

There's an extension of this problem that says to use a general result of the problem we just solved to show that the total lifetime of the system is approximately given by the shortest among the lifetimes of the different modes.

Would that just be because the smaller the Tao, the more significant it is when 1/Tao is considered?
Essentially. You can make it precise by expanding an expression for the total lifetime in terms of a small quantity.
 
  • #17
You mean actually try a number and see what happens?
 
  • #18
fzero
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You mean actually try a number and see what happens?
No I mean find an appropriate small quantity to use to expand in a Taylor series. You would then compare the leading (largest) term in the series with the prediction made in the problem. You'll want to write down the expression for the total lifetime and look at different ways of factoring it to start out.
 

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