Distance Dependence of Radiation

1. Significant Figure revisited.

Example:

a. For N = 381 cpm, (Poisson distribution, if the observed value is N, then the uncertainty is sqrt(N).)

δN = sqrt(381) = 19.519.... ≈ 2 ⨉ 10¹ cpm. (Only ONE significant figure in scientific form.)

Then, N should be modified as N = 38 ⨉ 10¹ cpm. The best way to report the value is N = (38 ± 2) ⨉ 10¹ cpm, however, N = 380 ± 10 cpm is also acceptable (although this form indicates that there are 2 sig figs for the uncertainty).

b. For N = 91 cpm, δN = sqrt(91) = 9.539.... ≈ 1⨉10¹ cpm. Then N = (9 ± 1) ⨉ 10¹ cpm.

c. If you are getting a result of 3.07123 ± 0.04566,

First, the uncertainty should have one sig fig, so it rounds to 0.05;

Then, the value becomes 3.07, not 3 -- based on the fact that there is 2 decimal place in the uncertainty.

The final result is 3.05 ± 0.05.

Note: the decimal place of the value, is determined by the decimal place of the uncertainty, which should first be rounded to 1 significant figure.

2. Uncertainty of the linearized data.

a. For distance x with uncertainty δx, the natural log is x' = ln(x). Therefore δx' = |dx'/dx|δx = (δx)/x.

For example, if δx = 0.005 cm for a ruler, then

δx' = (δx)/x = 0.005/x_observed.

b. For Poisson data N with uncertainty δN, the natural log is N' = ln(N). Therefore δN' = (δN)/N.

Since δN_uncorrected = sqrt(N),

δN' = (δN_uncorrected)/N_uncorrected = sqrt(N_uncorrected)/N_uncorrected = 1/sqrt(N_uncorrected).

3. Final result

If 2 is not in the interval of [mean-error, mean+error], we cannot say that Gauss's Law can explain your data. To be more statistically correct (you don't need to say this), "Gauss's Law cannot be explained at the 68% confidence level". Go to the links for what does "68%" and "confidence level/interval" mean.

4. How to see my comments:

1) Click the "Grade Mark" when you open your report.

2) Find the blue bubbles.

3) Move the cursor to the blue bubbles, then it will expand.

Fig 1. Go to the "GradeMark".

Fig 2. Find the blue bubbles.