# Experimental average lifetime of particles

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• lukka98
In summary, my professor explained that by observing a large number of particles for a short period of time, we can estimate the average lifetime of the particles with a certain confidence level. This is because the probability of decay for each particle is small, but when observing a large number of particles, there is a high chance that at least one will decay. The smaller the average lifetime, the higher the decay probability per unit time, but this is still a small probability overall. Therefore, we can confidently say that the average lifetime of the particles is at least a certain value, such as
lukka98
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).

lukka98 said:
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.

lukka98
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.

Last edited:
ohwilleke and lukka98
PeroK said:
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.

PeroK said:
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 ?

lukka98 said:
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

lukka98
PS by small time, that means a time in which ##N## does not change significantly.

lukka98
PeroK said:
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.

lukka98 said:
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.

lukka98
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.

PeroK said:
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

lukka98 said:
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.

lukka98
PeroK said:
##\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##?

lukka98 said:
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.

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

PeroK and lukka98

## 1. What is the experimental average lifetime of particles?

The experimental average lifetime of particles is a measurement that determines the average amount of time a particle exists before decaying or undergoing a specific process. It is an important factor in understanding the behavior and properties of particles in various scientific fields such as particle physics and chemistry.

## 2. How is the experimental average lifetime of particles determined?

The experimental average lifetime of particles is determined through various experimental methods, such as particle accelerators and detectors. These experiments involve observing the decay of particles and recording their lifetimes, which are then used to calculate the average lifetime.

## 3. What factors can affect the experimental average lifetime of particles?

There are several factors that can affect the experimental average lifetime of particles, including the type of particle, its energy, and the environment it is in. Other factors such as temperature, pressure, and interactions with other particles can also impact the lifetime of particles.

## 4. Why is the experimental average lifetime of particles important in scientific research?

The experimental average lifetime of particles is important in scientific research as it provides valuable information about the fundamental properties and behavior of particles. It can also help scientists understand and predict the behavior of particles in different environments and conditions, leading to advancements in various fields of science and technology.

## 5. Can the experimental average lifetime of particles change over time?

Yes, the experimental average lifetime of particles can change over time due to various factors such as changes in the environment or the decay of particles. Scientists continue to conduct experiments to improve their understanding of particles and their lifetimes, leading to possible changes in the experimental average lifetime of particles over time.

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