Chapter 4.3, Problem 20E

A physicist wants to estimate the rate of emissions of alpha particles from a certain source. He makes two counts. First, he measures the background rate by counting the number of particles in 100 seconds in the absence of the source. He counts 36 background emissions. Then, with the source present, he counts 324 emissions in 100 seconds. This represents the sum of source emissions plus background emissions.

1. a. Estimate the background rate, in emissions per second, and find the uncertainty in the estimate.
2. b. Estimate the sum of the source plus background rate, in emissions per second, and find the uncertainty in the estimate.
3. c. Estimate the rate of source emissions in particles per second, and find the uncertainty in the estimate.
4. d. Which will provide the smaller uncertainty in estimating the rate of emissions from the source: (1) counting the background only for 150 seconds and the background plus the source for 150 seconds, or (2) counting the background for 100 seconds and the source plus the background for 200 seconds? Compute the uncertainty in each case.
5. e. Is it possible to reduce the uncertainty to 0.03 particles per second if the background rate is measured for only 100 seconds? If so, for how long must the source plus background be measured? If not, explain why not.

To determine

Estimate the background rate in emissions per second.

Find the uncertainty in the estimate.

Answer to Problem 20E

The estimate of the background rate in emissions per second is 0.36

The uncertainty in the estimate is 0.06.

Explanation of Solution

Given info:

There are 36 background emissions in 100 seconds in the absence of source and 324 emissions in the presence of source.

Calculation:

The random variables X is defined as the number of background events counted in 100 seconds. The background rate is denoted as ${\lambda }_B$. Then X follows Poisson with parameter, $100{\lambda }_B$. The observed value is $X=36$.

The estimate ${\lambda }_B$ can be obtained with the sample concentration ${\widehat{\lambda }}_B$.

The sample concentration ${\widehat{\lambda }}_B$ can be obtained as follows:

${\widehat{\lambda }}_B=\frac{X}{t}$

Substitute $X=36\text{ and }t=100$ in the formula.

$\begin{array}{c}{\widehat{\lambda }}_B=\frac{36}{100}\\ =0.36\end{array}$

Thus, the estimate is 0.36.

The uncertainty in ${\widehat{\lambda }}_B$ can be obtained as follows:

${\sigma }_{{\widehat{\lambda }}_B}=\sqrt{\frac{{\lambda }_B}{t}}$

Replacing $\lambda$ with ${\widehat{\lambda }}_B=0.36\text{ and }t=100$ in the formula

$\begin{array}{c}{\sigma }_{{\widehat{\lambda }}_B}=\sqrt{\frac{0.36}{100}}\\ =\sqrt{0.0036}\\ =0.06\end{array}$

Thus, the uncertainty in the estimate is 0.06.

To determine

Estimate the sum of the source plus background rate in emissions per second.

Find the uncertainty in the estimate.

Answer to Problem 20E

The estimate of the sum of the source plus background rate in emissions per second is 3.24.

The uncertainty in the estimate is 0.18.

Explanation of Solution

Calculation:

The random variable Y is defined as the numbers of events both background and source counted in 100 seconds.

The estimate ${\lambda }_T$ can be obtained with the sample concentration ${\widehat{\lambda }}_T$.

Then Y follows Poisson with parameter, $100{\lambda }_T$. The observed values is $Y=324$

The sample concentration ${\widehat{\lambda }}_T$ can be obtained as follows:

${\widehat{\lambda }}_T=\frac{Y}{t}$

Substitute $t=100\text{ and }Y=324$ in the formula.

$\begin{array}{c}{\widehat{\lambda }}_T=\frac{324}{100}\\ =3.24\end{array}$

Thus, the estimate is 3.24.

The uncertainty in ${\widehat{\lambda }}_T$ can be obtained as follows:

${\sigma }_{{\widehat{\lambda }}_T}=\sqrt{\frac{{\widehat{\lambda }}_T}{t}}$

Replacing ${\lambda }_T$ with ${\widehat{\lambda }}_T=3.24\text{and }t=100$ in the formula

$\begin{array}{c}{\sigma }_{{\widehat{\lambda }}_T}=\sqrt{\frac{3.24}{100}}\\ =\sqrt{0.0324}\\ =0.18\end{array}$

Thus, the uncertainty in the estimate is 0.18.

To determine

Estimate the rate of source emissions in particles per second and find the uncertainty in the estimate.

Answer to Problem 20E

The estimate of rate of source emissions in particles per second is 2.88.

The uncertainty in the estimate is 0.19

Explanation of Solution

Calculation:

The source rate is denoted as as ${\lambda }_S$. The total rate is denoted as ${\lambda }_T={\lambda }_B+{\lambda }_S$.

The rate of source emissions in particles per second is,

${\lambda }_S={\lambda }_T-{\lambda }_B$

The sample concentration ${\widehat{\lambda }}_S$ can be obtained as follows:

${\widehat{\lambda }}_S={\widehat{\lambda }}_T-{\widehat{\lambda }}_B$

Substitute ${\widehat{\lambda }}_B\text{=0}\text{.36 and }{\widehat{\lambda }}_T=3.24$ in the formula.

$\begin{array}{c}{\widehat{\lambda }}_S=3.24-0.36\\ =2.88\end{array}$

Thus, the estimate is 2.88.

Both ${\widehat{\lambda }}_T$ and ${\widehat{\lambda }}_B$ are independent, since they are based on the two independent events.

The uncertainty in ${\lambda }_S={\lambda }_T-{\lambda }_B$ can be obtained as follows:

${\sigma }_{{\widehat{\lambda }}_S}=\sqrt{{\sigma }^2{}_{{\widehat{\lambda }}_T}+{\sigma }^2{}_{{\widehat{\lambda }}_B}}$

Substitute  ${\sigma }_{{\widehat{\lambda }}_T}=0.06\text{ and }{\sigma }_{{\widehat{\lambda }}_B}=0.18$ in the formula

$\begin{array}{c}{\sigma }_{{\widehat{\lambda }}_S}=\sqrt{{\left(0.06\right)}^2+{\left(0.18\right)}^2}\\ =\sqrt{0.0036+0.0324}\\ =\sqrt{0.036}\\ =0.19\end{array}$

Thus, the uncertainty in the estimate is 0.19.

To determine

Identify which will provide the smaller uncertainty in estimating the rate of emissions from the source, counting the background only for 150 seconds and the background plus the source for 150 seconds or counting the background for 100 seconds and the source plus the background for 200 seconds.

Compute the uncertainty in each case.

Answer to Problem 20E

The counting the background only for 100 seconds and the background plus the source for 200 seconds has smaller uncertainty than counting the background only for 150 seconds and the background plus the source for 150 seconds.

The uncertainty in the case of counting the background only for 150 seconds and the background plus the source for 150 seconds is 0.15.

The uncertainty in the case of counting the background only for 100 seconds and the background plus the source for 200 seconds is 0.14.

Explanation of Solution

Calculation:

Case 1:

If counting the background only for 150 seconds and the background plus the source for 150 seconds.

The uncertainty in ${\widehat{\lambda }}_B$ can be obtained as follows:

${\sigma }_{{\widehat{\lambda }}_B}^2=\frac{{\lambda }_B}{t}$

Replacing ${\lambda }_B$ with ${\widehat{\lambda }}_B=0.36\text{and }t=150$ in the formula

$\begin{array}{c}{\sigma }_{{\widehat{\lambda }}_B}^2=\frac{0.36}{150}\\ =2.4\times {10}^{-3}\end{array}$

Thus, the uncertainty in the ${\widehat{\lambda }}_B$ is 0.0024.

The uncertainty in ${\widehat{\lambda }}_T$ can be obtained as follows:

${\sigma }_{{\widehat{\lambda }}_T}^2=\frac{{\lambda }_T}{t}$

Replacing ${\lambda }_T$ with ${\widehat{\lambda }}_T=3.24\text{and }t=150$ in the formula

$\begin{array}{c}{\sigma }_{{\widehat{\lambda }}_T}^2=\frac{3.24}{150}\\ =0.0216\end{array}$

Thus, the uncertainty in the ${\widehat{\lambda }}_T$ is 0.0216.

Then,

${\sigma }_{{\widehat{\lambda }}_S}=\sqrt{{\sigma }_{{\widehat{\lambda }}_B}^2+{\sigma }_{{\widehat{\lambda }}_T}^2}$

Substitute ${\sigma }_{{\widehat{\lambda }}_B}^2=0.0024$ and ${\sigma }_{{\widehat{\lambda }}_T}^2=0.0216$ in the above equation.

$\begin{array}{c}{\sigma }_{{\widehat{\lambda }}_S}=\sqrt{0.0024+0.0216}\\ =\sqrt{0.024}\\ =0.15\end{array}$

Thus, the uncertainty in this case is 0.15.

Case 2:

If counting the background only for 100 seconds and the background plus the source for 200 seconds.

The uncertainty in ${\widehat{\lambda }}_B$ can be obtained as follows:

${\sigma }_{{\widehat{\lambda }}_B}^2=\frac{{\lambda }_B}{t}$

Replacing ${\lambda }_B$ with ${\widehat{\lambda }}_B=0.36\text{and }t=100$ in the formula

$\begin{array}{c}{\sigma }_{{\widehat{\lambda }}_B}^2=\frac{0.36}{100}\\ =3.6\times {10}^{-3}\end{array}$

Thus, the uncertainty in the ${\widehat{\lambda }}_B$ is 0.0036.

The uncertainty in ${\widehat{\lambda }}_T$ can be obtained as follows:

${\sigma }_{{\widehat{\lambda }}_T}^2=\frac{{\lambda }_T}{t}$

Replacing ${\lambda }_T$ with ${\widehat{\lambda }}_T=3.24\text{and }t=200$ in the formula

$\begin{array}{c}{\sigma }_{{\widehat{\lambda }}_T}^2=\frac{3.24}{200}\\ =0.0162\end{array}$

Thus, the uncertainty in the ${\widehat{\lambda }}_T$ is 0.0162.

Then,

${\sigma }_{{\widehat{\lambda }}_S}=\sqrt{{\sigma }_{{\widehat{\lambda }}_B}^2+{\sigma }_{{\widehat{\lambda }}_T}^2}$

Substitute ${\sigma }_{{\widehat{\lambda }}_B}^2=0.0036$ and ${\sigma }_{{\widehat{\lambda }}_T}^2=0.0162$ in the above equation.

$\begin{array}{c}{\sigma }_{{\widehat{\lambda }}_S}=\sqrt{0.0036+0.0612}\\ =\sqrt{0.0198}\\ =0.14\end{array}$

Thus, the uncertainty in this case is 0.14.

Thus, it can be seen that the counting the background only for 100 seconds and the background plus the source for 200 seconds has smaller uncertainty than counting the background only for 150 seconds and the background plus the source for 150 seconds.

To determine

Check whether it is possible to reduce the uncertainty to 0.03 particles per second if the background rate is measured for only 100 seconds. If so find how long must be the source plus background be measured. If not, explain the reason.

Answer to Problem 20E

It is not possible to reduce the uncertainty to 0.03 particles per second if the background rate is measured for only 100 seconds.

Explanation of Solution

Justification:

If the background emissions are counted for 100 seconds and the source plus background is counted for N seconds.

Then uncertainty in the estimate of the source rate is, $\sqrt{\frac{0.36}{100}+\frac{3.24}{N}}$

If one increase the value of N then the estimate of the source rate must be greater than

$\sqrt{\frac{0.36}{100}}=0.06$.

Thus, an estimate of source rate must be always greater than 0.03.