# Archived Relative error of a geiger counter?

1. Apr 1, 2014

### carnivalcougar

1. The problem statement, all variables and given/known data

A geiger counter measures 750 counts in 5 min
a) determine the average counting rate and the standard deviation of the counting rate.
b) find the relative error
c) How long would you want to count to get a relative error of 1%?

2. Relevant equations

standard deviation = $\sqrt{}navg$
relative error = δx/x

3. The attempt at a solution

The average counting rate is 750/5 which is 150 counts / min and the standard deviation is the square root of that which is 12.247. The relative error is 12.247/150 which is .0816 which is 8.16%

I'm not sure how to determine how long you would want to count to get a relative error of 1%

2. Feb 5, 2016

### gleem

There is an error in the attempt of the first part of the question i.e. standard deviation of the count rate.

The count rate = total counts/time = N/t

A change , error, or uncertainty in the count rate is related to those of the total counts and the time by taking the differential of the count rate equation.
.
dR = dN/t - N⋅dt/t2

where the terms on the right hand side are the contribution of the uncertainties of both the total number of counts and the time to that of the count rate.

Using the propagation of error to determine the average uncertainly in the count rate we can write

σR2 = √[(dN/t)2 + (N⋅dt/t2)2]

where dN = √N because nuclear decay follows a Gaussian distribution..For all practical purposes dt =0 so

σR =√N/t that is √750/5 = 5.48 cnts/min

with a relative error of σR/R = 5.48/150 =.0365

Now to address the OP's question on how to determine the count time to achieve a certain precision for the count rate.

The precision sought is 1% thus the relative error is .01 so

σR/R = .01 =√N/t / N/t =1 /√N Thus N = 10000

So one must count until the total accumulated counts is equal 10000, Notice that this is true for any count rate! If you need to know the time say for scheduling purposes then using the estimate of the count rate as 150 cnts/min we expect it to take about 66.7 minutes to reach 10000 counts.

Considering the uncertainly in the count rate of 3.65% we can estimate that the uncertainly in the time to accumulate 10000 counts will be about 66.7×.0365 = 2.4 minutes.

Or more exactly οt = (N/R)×√[ (σR/R)2 + 1/N]

3. Feb 5, 2016

### haruspex

Just to note that the original method, taking the square root of the rate, makes no sense dimensionally. It produces a s.d. of dimension $T^{-\frac 12}$. This might have been a clue.

4. Feb 5, 2016

### gleem

Good point thanks.