MHB Calculating New Emission Rate of Radioactive Source

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The discussion centers on calculating the new emission rate of a radioactive source that initially emits particles at a rate of 1 per second, with emissions following a Poisson distribution. It is established that the probability of 0 or 1 emission in 4 seconds is 0.8, leading to the calculation of the new rate, which is approximately 0.206 emissions per second. The calculation involves using the Poisson probability formula, resulting in a new average of 0.824 emissions in 4 seconds. Participants also discuss methods for arriving at this value, including the use of online calculators and Poisson distribution tables. The conversation highlights the mathematical approach to determining changes in emission rates.
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A radioactive source emits particles at an average rate of 1 pe second. Assume that the number of emissions follows a Poisson distribution. The emission rate changes such that the probability of 0 or 1 emission in 4 seconds becomes 0.8. What is the new rate? Thanks.
 
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araz said:
A radioactive source emits particles at an average rate of 1 pe second. Assume that the number of emissions follows a Poisson distribution. The emission rate changes such that the probability of 0 or 1 emission in 4 seconds becomes 0.8. What is the new rate? Thanks.

Hi araz, welcome to MHB! ;)

We have:
$$P(\text{0 or 1 emission in 4 seconds})=0.8 \\ \implies
P(\text{0 emission in 4 seconds}) + P(\text{1 emission in 4 seconds}) = \frac{\lambda^0 e^{-\lambda}}{0!} + \frac{\lambda^1 e^{-\lambda}}{1!} = (1+\lambda)e^{-\lambda} = 0.8 \\ \implies
\lambda \approx 0.824
$$
So the average number of emissions in 4 seconds is $0.824$, which is $0.206$ per second.
 
Klaas van Aarsen said:
Hi araz, welcome to MHB! ;)

We have:
$$P(\text{0 or 1 emission in 4 seconds})=0.8 \\ \implies
P(\text{0 emission in 4 seconds}) + P(\text{1 emission in 4 seconds}) = \frac{\lambda^0 e^{-\lambda}}{0!} + \frac{\lambda^1 e^{-\lambda}}{1!} = (1+\lambda)e^{-\lambda} = 0.8 \\ \implies
\lambda \approx 0.824
$$
So the average number of emissions in 4 seconds is $0.824$, which is $0.206$ per second.

Thank you Klaas. Did you use guess and check or numerical methods to get 0.824?
Regards
 
araz said:
Thank you Klaas. Did you use guess and check or numerical methods to get 0.824?
Regards

I used an online calculator to find it.
Alternatively guessing can work, or we can look it up in a Poisson distribution table.
 
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