Question
What is the average number of interactions per bunch crossing for a luminosity of 2.5E31 cms^-1, given a total cross-section of 80 mb?
Answers
GPT-4o mini
To calculate the average number of interactions per bunch crossing for a given luminosity and total cross-section, you can use the relation:
\[
\text{Average number of interactions per bunch crossing} = \mathcal{L} \times \sigma
\]
where:
- \(\mathcal{L}\) is the luminosity in \(\text{cm}^{-2}\text{s}^{-1}\)
- \(\sigma\) is the total cross-section in \(\text{cm}^2\)
Given that:
- \(\mathcal{L} = 2.5 \times 10^{31} \, \text{cm}^{-2} \text{s}^{-1}\)
- \(\sigma = 80 \, \text{mb} = 80 \times 10^{-27} \, \text{cm}^2 = 8.0 \times 10^{-26} \, \text{cm}^2\) (since 1 mb = \(10^{-27} \, \text{cm}^2\))
Now, substituting these values into the formula:
\[
\text{Average number of interactions per bunch crossing} = (2.5 \times 10^{31} \, \text{cm}^{-2} \text{s}^{-1}) \times (8.0 \times 10^{-26} \, \text{cm}^2)
\]
Calculating this gives:
\[
\text{Average number of interactions per bunch crossing} = 2.5 \times 8.0 \times 10^{31 - 26}
\]
\[
= 20.0 \times 10^{5}
\]
\[
= 2.0 \times 10^{6}
\]
Thus, the average number of interactions per bunch crossing at a luminosity of \(2.5 \times 10^{31} \, \text{cm}^{-2} \text{s}^{-1}\) and a total cross-section of \(80 \, \text{mb}\) is:
\[
\boxed{2.0 \times 10^{6}}
\]
\[
\text{Average number of interactions per bunch crossing} = \mathcal{L} \times \sigma
\]
where:
- \(\mathcal{L}\) is the luminosity in \(\text{cm}^{-2}\text{s}^{-1}\)
- \(\sigma\) is the total cross-section in \(\text{cm}^2\)
Given that:
- \(\mathcal{L} = 2.5 \times 10^{31} \, \text{cm}^{-2} \text{s}^{-1}\)
- \(\sigma = 80 \, \text{mb} = 80 \times 10^{-27} \, \text{cm}^2 = 8.0 \times 10^{-26} \, \text{cm}^2\) (since 1 mb = \(10^{-27} \, \text{cm}^2\))
Now, substituting these values into the formula:
\[
\text{Average number of interactions per bunch crossing} = (2.5 \times 10^{31} \, \text{cm}^{-2} \text{s}^{-1}) \times (8.0 \times 10^{-26} \, \text{cm}^2)
\]
Calculating this gives:
\[
\text{Average number of interactions per bunch crossing} = 2.5 \times 8.0 \times 10^{31 - 26}
\]
\[
= 20.0 \times 10^{5}
\]
\[
= 2.0 \times 10^{6}
\]
Thus, the average number of interactions per bunch crossing at a luminosity of \(2.5 \times 10^{31} \, \text{cm}^{-2} \text{s}^{-1}\) and a total cross-section of \(80 \, \text{mb}\) is:
\[
\boxed{2.0 \times 10^{6}}
\]