Compute the average number of interactions per bunch crossing for a luminosity L=2.5E31 cms-1,if the total cross section is sigma =20mb and bunch crossing occur every T=4.What is the probability of having zero interactions in a bunch?
What is a bunch crossing? Bunch of what?
What are mb units? Millbarnes?
cm/s is not a unit of luminosity
What is T? What are its units?
this is jibberish
Trying to sort out the units.
20 mb = 20 millibarns (mbarn)?
the luminosity is presumably cm2 s^-1
because by definition
event rate = luminosity * cross section
Not sure of units of T ns?
To compute the average number of interactions per bunch crossing, we can use the formula:
Average number of interactions = L * sigma * T
Given:
L = 2.5E31 cm⁻²s⁻¹ (Luminosity)
sigma = 20 mb (Total cross section)
T = 4 (Bunch crossing time)
Substituting the given values into the formula:
Average number of interactions = (2.5E31 cm⁻²s⁻¹) * (20 mb) * (4)
To simplify the calculation, let's convert the units:
1 mb = 10⁻¹⁸ cm²
Average number of interactions = (2.5E31) * (20) * (4E-18 cm²)
Calculate the result:
Average number of interactions = 2E14 interactions per bunch crossing
Now, to calculate the probability of having zero interactions in a bunch, we can use the Poisson distribution formula:
Probability of having zero interactions = e^(-N) * (N^0 / 0!)
where N is the average number of interactions calculated above.
Substituting N = 2E14 into the formula:
Probability of having zero interactions = e^(-2E14) * (2E14^0 / 0!)
Since 0! = 1, the formula simplifies to:
Probability of having zero interactions = e^(-2E14)
Calculating the result:
Probability of having zero interactions = approximately 1.831E-86908575234534
The probability of having zero interactions in a bunch is extremely low.