Error Propogation for Half-Life

In summary: I found this solution here and have been trying to see how it was arrived at. See the image of the solution below. The equation you have arrived at certainly is consistent with the first part of this solution and makes sense to me, however I am still unclear on how that carries over to the second part.Many thanks for your help.
  • #1
Alejandro Golob
2
0
In this homework question we are told to calculate the half-life of an isotope based on count-rates before and after a given time interval, the relevant equation is given below.

Half-life = t1/2 = tln(2)/ln (A/A0)

The second part asks to determine the standard deviation of the half-life due to counting statistics. We have an uncertainty for the value A and the value A0.

I have found the following solution but cannot quite follow it, so was hoping someone might be able to explain/walk through it with me.

d/dt (t1/2) = − (ln (2) t/(A/A0))/ln((A/A0)2)

Plugging in t = 24 and R = 2.875 yields 5.248 hr−1

σ2t1/2 = (5.248)2 ((9.14/(41.4)2 + 16.83/(118.3)2) *(2.875)2

Thus, we find that expected standard deviation of half-life is σ(t1/2) = 1.22 h

Thanks in advance for any help.
 
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  • #2
d/dt does not make sense at that point. If you calculate the time-derivative of the expression for the half-life, you get a different result.
Alejandro Golob said:
Plugging in t = 24 and R = 2.875
24 what, and what is R?
The time-derivative of a time should be dimensionless.

You can calculate the uncertainty on the ratio first, and then propagate this to the half-life measurement. ##-\frac{\ln( 2) t }{A/A_0 \ln(A/A_0)^2}## can be useful there as some part of a different formula.

Can you give the numbers you use for your calculations?
 
  • #3
Thank you for the reply.

t=24hr and R=A/A0=(41.4minute-1)/118.3minute-1

I certainly agree with what you are saying about the time derivative of a time, I wasn't sure on this myself. The half-life itself is not a function of time anyways, so not sure how the time derivative makes sense or why it is invoked. It is possible that this is an error. I found this solution here and have been trying to see how it was arrived at. See the image of the solution below. The equation you have arrived at certainly is consistent with the first part of this solution and makes sense to me, however I am still unclear on how that carries over to the second part.
upload_2015-12-1_15-26-29.png

Many thanks for your help.

Best Regards,
 

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  • #4
Alejandro Golob said:
I certainly agree with what you are saying about the time derivative of a time, I wasn't sure on this myself. The half-life itself is not a function of time anyways, so not sure how the time derivative makes sense or why it is invoked. It is possible that this is an error.
If you look at the right hand side of the d/dt(t1/2) equation you can see that the derivative taken was with respect to R, not t.
 
  • #5
There are more issues. The ##\sigma S_i## calculations mix standard deviations (left and middle) and variance (right).
##\sigma R## should have been calculated in a clearer way, and so on. It is the left two brackets in the last formula.
5.248 seems to be a bad approximation of the expression above, I get 5.188.
 

1. What is error propagation for half-life?

Error propagation for half-life is a mathematical concept used in scientific research to estimate the uncertainty or error associated with the half-life of a substance. It takes into account the uncertainties in the initial and final measurements of the substance's decay rate, as well as any other factors that may affect the accuracy of the measurement.

2. Why is error propagation important for half-life calculations?

Half-life calculations are often used in various fields of science, such as chemistry and physics, to determine the stability and decay of substances. Error propagation is important because it allows us to quantify the uncertainty associated with the half-life measurement, making the results more reliable and accurate. It also helps to identify any sources of error in the measurement process.

3. How is error propagation calculated for half-life?

Error propagation is calculated using the formula: Δt₁/τ = (ln(2)/τ)Δt₀, where Δt₁ is the uncertainty in the final measurement, Δt₀ is the uncertainty in the initial measurement, and τ is the half-life of the substance. This formula takes into account the uncertainties in both measurements and the natural logarithm of 2, which is used to convert the decay rate into half-life.

4. Can error propagation affect the accuracy of half-life measurements?

Yes, error propagation can affect the accuracy of half-life measurements. The larger the uncertainties in the initial and final measurements, the larger the error propagation will be. This can lead to a larger uncertainty in the final half-life value and may affect the overall accuracy of the measurement. It is important to minimize uncertainties and use proper error propagation calculations to obtain more accurate results.

5. Are there any limitations to error propagation for half-life?

Yes, there are a few limitations to error propagation for half-life. This method assumes that the uncertainties in the initial and final measurements are independent of each other and that the decay rate follows a first-order reaction. It also does not take into account any systematic errors that may be present in the measurement process. Therefore, it is important to use error propagation in conjunction with other error analysis methods to obtain the most accurate results.

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