- #1

lukka98

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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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- #1

lukka98

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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).

- #2

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You mean if none of them decays in a year?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.

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

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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.

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:

- #4

lukka98

- 30

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Thank you.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

- #5

lukka98

- 30

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this answer give rise me a question ( maybe stupid):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.

##\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

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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: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 ?

http://hyperphysics.phy-astr.gsu.edu/hbase/Nuclear/halfli2.html

- #7

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

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

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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.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.

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

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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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Excuse me, If I can I have a thing, an example:

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.

## ^{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

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##\Delta t## is far too large for that decay rate. ##\Delta t## must be small for the approximation to hold.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

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

- 30

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In last, then I stop, so ##\lambda ## is equal to the "probability per unit time that the particle decays" only for "small" ##\lambda##?##\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.

- #14

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This is the decay law for infinitesimally small ##\Delta t##. By making ##\Delta t \rightarrow 0## you getExcuse 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

$$\lim_{\Delta t \rightarrow 0} \frac{\Delta N}{\Delta t} =\frac{\mathrm{d} N}{\mathrm{d} t}=-\lambda N.$$

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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