Experimental average lifetime of particles

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  • #1
lukka98
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My professor said that if I take 10^24 particles, for example, and I observe them for 1 years, I can say that their average lifetime is at least 10^24 years.
I am not able to understand why, I though the law of decay is N(t) = N0*e^(-ta), so if after 1 year N(t) = N0, I can say ta ~ 0 , so tau >> t. ( tau = 1/a).
 

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  • #2
PeroK
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My professor said that if I take 10^24 particles, for example, and I observe them for 1 years, I can say that their average lifetime is at least 10^24 years.
You mean if none of them decays in a year?

For decay, for each particle the decay probability does not change over time, so we have:$$\Delta N = -\lambda N \Delta t$$where ##\lambda## is the decay rate and ##\Delta t## is small.

You can see from that if ##N \Delta t > \frac 1 \lambda## then you would expect some decay in ##\Delta t##.

Also, for this distribution ##\frac 1 \lambda = \tau##, where ##\tau## is the average lifetime.
 
  • #3
PeroK
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PS actually, we can say something stronger than that. The probability that each particle decays is ##\lambda##, so the probability that it doesn't is ## 1- \lambda##. And the probability that no particle decays is $$p(0) = (1 - \lambda)^N$$
If we test the hypothesis that ##\tau = 10^{24}## years. I.e. that ##\lambda = 10^{-24}## per year, then:
$$p(0) = (1 - \frac 1 {10^{24}})^{10^{24}} \approx e^{-1} \approx 0.37$$
The hypothesis that the half-life ##10^{24}## years or greater is valid - but only with 63% confidence. So, it could be a bit less.

You could always go for 95% confidence, which results in ##\lambda \approx 0.33 \times 10^{-23}##. In other words, we can say with 95% confidence that the average lifetime is greater than ##3 \times 10^{23}## years.
 
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  • #4
lukka98
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PS actually, we can say something stronger than that. The probability that each particle decays is ##\lambda##, so the probability that it doesn't is ## 1- \lambda##. And the probability that no particle decays is $$p(0) = (1 - \lambda)^N$$
If we test the hypothesis that ##\tau = 10^{24}## years. I.e. that ##\lambda = 10^{-24}## per year, then:
$$p(0) = (1 - \frac 1 {10^{24}})^{10^{24}} \approx e^{-1} \approx 0.37$$
The hypothesis that the half-life ##10^{24}## years or greater is valid - but only with 63% confidence. So, it could be a bit less.

You could always go for 95% confidence, which results in ##\lambda \approx 0.33 \times 10^{-23}##. In other words, we can say with 95% confidence that the average lifetime is greater than ##3 \times 10^{23}## years
Thank you.
 
  • #5
lukka98
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PS actually, we can say something stronger than that. The probability that each particle decays is ##\lambda##, so the probability that it doesn't is ## 1- \lambda##. And the probability that no particle decays is $$p(0) = (1 - \lambda)^N$$
If we test the hypothesis that ##\tau = 10^{24}## years. I.e. that ##\lambda = 10^{-24}## per year, then:
$$p(0) = (1 - \frac 1 {10^{24}})^{10^{24}} \approx e^{-1} \approx 0.37$$
The hypothesis that the half-life ##10^{24}## years or greater is valid - but only with 63% confidence. So, it could be a bit less.

You could always go for 95% confidence, which results in ##\lambda \approx 0.33 \times 10^{-23}##. In other words, we can say with 95% confidence that the average lifetime is greater than ##3 \times 10^{23}## years.
this answer give rise me a question ( maybe stupid):
##\lambda## is the probability per unit time that a particle decay, but if a particle have an average lifetime of ##\[ 10^{-20} s]\## so the decay probability for unit time ( i think per second..?) is greater than 1, but this is nosense, so what is wrong ?
 
  • #6
PeroK
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this answer give rise me a question ( maybe stupid):
##\lambda## is the probability per unit time that a particle decay, but if a particle have an average lifetime of ##\[ 10^{-20] s]\## so the decay probability for unit time ( i think per second..?) is greater than 1, but this is nosense, so what is wrong ?
Technically ##\lambda \Delta t## is the probability for a small time ##\Delta t##. In this case, with a low decay rate, one year is small. In fact, ##\lambda## arises through the differential equation: $$\frac{dN}{dt} = -\lambda N$$ See, for example:

http://hyperphysics.phy-astr.gsu.edu/hbase/Nuclear/halfli2.html
 
  • #7
PeroK
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PS by small time, that means a time in which ##N## does not change significantly.
 
  • #8
lukka98
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PS by small time, that means a time in which ##N## does not change significantly.
So for example if ##\tau## is 1 years, ##\frac{1}{\lambda}## is 1, and the probability to having a decay in 1 year is ... 1?
If i would calculate the probability of a decay of a particle of average lifetime of 1 years, after 1 years, how I do this? I can't be 1 I think.
 
  • #9
PeroK
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So for example if ##\tau## is 1 years, ##\frac{1}{\lambda}## is 1, and the probability to having a decay in 1 year is ... 1?
If i would calculate the probability of a decay of a particle of average lifetime of 1 years, after 1 years, how I do this? I can't be 1 I think.
No. If the lifetime is that short, then you can't add probabilities like that. It's like the chance of throwing a six on a die is always ##1/6##. The chances of a six by at least the second throw is ##1/6 + (5/6)(1/6)## etc.

Where the ##5/6## is the probability of not getting a six on the first throw.

This calculation shows that you only approach the limit of probability 1 as the number of throws tends to infinity. It's not certain after 6 throws.
 
  • #10
PeroK
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PS using dice is actually a good analogy, where throwing a six is equivalent to a particle decaying probabilistically. You start with a large number of dice, throw them all and take away any that show six. Then repeat. Each throw represents a relatively small unit of time, where the probability of decay anytime during that interval is ##1/6##. The proportion of dice left after ##n## throws is ##(5/6)^n##, the average lifetime of a die is six throws (that's an exercise to show that, if you want), and the half-life is approx four throws (another exercise).

The continuous decay model would assume the limit as the number of faces on the die tends to infinity, as does the frequency with which the dice are thrown.
 
  • #11
lukka98
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PS using dice is actually a good analogy, where throwing a six is equivalent to a particle decaying probabilistically. You start with a large number of dice, throw them all and take away any that show six. Then repeat. Each throw represents a relatively small unit of time, where the probability of decay anytime during that interval is ##1/6##. The proportion of dice left after ##n## throws is ##(5/6)^n##, the average lifetime of a die is six throws (that's an exercise to show that, if you want), and the half-life is approx four throws (another exercise).

The continuous decay model would assume the limit as the number of faces on the die tends to infinity, as does the frequency with which the dice are thrown.
Excuse me, If I can I have a thing, an example:

## ^{215}Po ## has a ## \lambda = 3.9 * 10^{2} s^{-1}##;
If I take the variation, according to equation that describe the decay, ## \Delta N(t) = - \lambda N(t) \Delta t ##, for example having ##N(t) = 10 ##. I have ##\Delta N(t) = -3900 ## for ##\Delta t = 1 s##.
What is wrong? I am a lot confused
 
  • #12
PeroK
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Excuse me, If I can I have a thing, an example:

## ^{215}Po ## has a ## \lambda = 3.9 * 10^{2} s^{-1}##;
If I take the variation, according to equation that describe the decay, ## \Delta N(t) = - \lambda N(t) \Delta t ##, for example having ##N(t) = 10 ##. I have ##\Delta N(t) = -3900 ## for ##\Delta t = 1 s##.
What is wrong? I am a lot confused
##\Delta t## is far too large for that decay rate. ##\Delta t## must be small for the approximation to hold.

The equation ## \Delta N(t) = - \lambda N(t) \Delta t ## is only an approximation for small ##\Delta t##.

As ##N(t)## reduces, so does the number of particles decaying per unit time.
 
  • #13
lukka98
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##\Delta t## is far too large for that decay rate. ##\Delta t## must be small for the approximation to hold.

The equation ## \Delta N(t) = - \lambda N(t) \Delta t ## is only an approximation for small ##\Delta t##.

As ##N(t)## reduces, so does the number of particles decaying per unit time.
In last, then I stop, so ##\lambda ## is equal to the "probability per unit time that the particle decays" only for "small" ##\lambda##?
 
  • #14
vanhees71
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Excuse me, If I can I have a thing, an example:

## ^{215}Po ## has a ## \lambda = 3.9 * 10^{2} s^{-1}##;
If I take the variation, according to equation that describe the decay, ## \Delta N(t) = - \lambda N(t) \Delta t ##, for example having ##N(t) = 10 ##. I have ##\Delta N(t) = -3900 ## for ##\Delta t = 1 s##.
What is wrong? I am a lot confused
This is the decay law for infinitesimally small ##\Delta t##. By making ##\Delta t \rightarrow 0## you get
$$\lim_{\Delta t \rightarrow 0} \frac{\Delta N}{\Delta t} =\frac{\mathrm{d} N}{\mathrm{d} t}=-\lambda N.$$
This is a differential equation for the number of particles at time ##t##. It's easy to solve. Just divide by ##N##:
$$\frac{1}{N} \frac{\mathrm{d} N}{\mathrm{d} t}=-\lambda.$$
The left-hand side can be written as
$$\frac{\mathrm{d}}{\mathrm{d} t} \ln(N)=-\lambda.$$
Now integrate from ##t=0## to ##t##, you get
$$\ln[N(t)]-\ln[N(0)]=\ln \left ( \frac{N(t)}{N(0)} \right)=-\lambda t$$
or, solved for ##N(t)##
$$N(t)=N(0) \exp(-\lambda t).$$
This is the exponential decay law.

For your example you have
$$N(1 \; \text{s})=10 \exp(-390) \approx 4.2 \cdot 10^{-69},$$
which is 0 for all practical purposes.
 
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