Comparing With Whitney Rates

$\textbf{Navigation}$

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Using the values from Whitney:

For a 5cm target of LH2:

$\rho_{target}\times l_{target}=\frac{70.85 kg}{1 m^3}\times \frac{1 mole}{2.02 g} \times \frac{1000g}{1 kg} \times \frac{6.022\times10^{23} molecules\ LH_2}{1 mole} \frac{2 atoms}{1\ molecule\ LH_2} \times \frac{1m^3}{(100 cm)^3} \times \frac{5 cm}{ }=\frac{2.11\times 10^{23}}{cm^2} \times \frac{1^{-24} cm^{2}}{barn}=0.211 barns^{-1}$

Using the beam current of 100nA,

$\Phi_{beam}=\frac{100\times 10^{-9}\ A}{1} \times \frac{1\ C}{1\ A} \times \frac{1\ e^{-}}{1\ s} \times \frac{1}{1.602\times 10^{-19}C}\Rightarrow 6.2422\times 10^{11}\ \frac{1}{s}=\Phi_{e^{-}}$

Given the beam Luminosity of:

$\mathcal {L}=\frac{1.32\times 10^{35}}{cm^2 \cdot s}\frac{10^{-24} cm^{2}}{barn}=\frac{1.32\times 10^{11}}{barn\cdot s}=\Phi_{beam}\ \rho\ l_{target}$

We can check to make sure the density makes sense

$\Rightarrow \rho_{target}=\frac{1.32\times 10^{35}}{ \Phi_{beam}\ l_{target}cm^2 \cdot s}=\frac{1.32\times 10^{35}}{ 6.24\times 10^{11}\ \cdot 5cm^3}=\frac{4.23\times 10^{22}}{cm^3}$

$\Rightarrow \rho_{target}=\frac{70.85 kg}{1 m^3}\times \frac{1 mole}{2.02 g} \times \frac{1000g}{1 kg} \times \frac{6.022\times10^{23} molecules\ LH_2}{1 mole} \times \frac{1m^3}{(100 cm)^3}\frac{2 atoms}{1\ molecule\ LH_2}=\frac{4.23\times 10^{22}}{cm^3}$

Using the values from Whitney :

https://wiki.iac.isu.edu/index.php/CLAS12_RateEst_byWA

Rates

Energy	6 GeV	11 GeV
Process	(nb)	(nb)
Moller	22773001	75008636
DIS + radiative tail	128	83
Elastic e-p	5511220	3670740
Elastic radiative tail	24705	12944
π0 electro-production	14802	17908
π0 photo-production	569	852
π+ electro-production	4032	5536
π+ photo-production	282	487
π− electro-production	2806	3843
π− photo-production	199	342
Total	2.83317E7	7.87214E7

$R_{events}= \sigma_{events} \mathcal{L}=7.87\times 10^{-2}\ barn\ \cdot \frac{1.33\times 10^{11}}{barn\cdot s}=\frac{1.05\times 10^{10}events}{s}$

$R_{Moller}= \sigma_{Moller} \mathcal{L}=7.50\times 10^{-2}\ barn\ \cdot \frac{1.33\times 10^{11}}{barn\cdot s}=\frac{9.97\times 10^{9}Moller}{s}$

File:CLAS12ExpWS WA.pdf

$\sigma=\frac{R}{\mathcal{L}}=\frac{dN}{dt}\cdot \frac{1}{\mathcal{L}}$

$\Rightarrow \int\limits_{0}^{t}\, dt=\int\limits_{0}^{N}\frac{1}{\sigma \mathcal{L}}\, dN$

$t_{simulated}=\frac{N_{events}}{\sigma_{events} \Phi \rho \ell}=\frac{1000000\ barn \cdot s}{7.87\times 10^{-2} \cdot 1.33\times 10^{11}\ barn}=9.54\times 10^{-5} s$

$N_0=\Delta t \cdot R_{events}=\Delta t \cdot \frac{N_{events}}{t_{simulated}}=500\times 10^{-9}\ s \cdot \frac{1\times 10^6}{0.0954\times 10^{-3}\ s}=5240$

$Occupancy=\frac{N_{hits}}{N_0}=\frac{N_{hits}}{\Delta t \cdot R_{events}}=\frac{t_{simulated}\cdot N_{hits}}{N_{events}\cdot \Delta t}=$

Similarly,

$N_{events} =R_{events}\cdot t_{simulated}=\frac{1.05\times 10^{10}events}{s} \cdot 9.54\times 10^{-5} s=1001700\ events$

$N_{Moller} =R_{Moller}\cdot t_{simulated}=\frac{9.97\times 10^{9}Moller}{s} \cdot 9.54\times 10^{-5} s=951138\ Moller$

Using just the scattered particles, for the total time run,

$\sigma=\frac{ N_{scattered}}{ N_{incident}}\frac{1}{\rho \ell}\Rightarrow N_{incident}=\frac{ N_{scattered}}{\sigma \rho \ell}=\frac{ 1000000}{ .078\cdot 2.11\times 10^{-1}}\approx 6\times 10^7\ incident$

If we view the window for each region as how many events can be detected per second

$\sigma=\frac{ t\ N_{scattered}}{\Delta t N_{incident}}\frac{1}{\rho \ell}\Rightarrow N_{incident}=\frac{t\ N_{scattered}}{\Delta t\sigma \rho \ell}=\frac{0.0954\times 10^{-3}\ s\ \times 1000000}{250 \times 10^{-9}\ s \times .078\cdot 2.11\times 10^{-1}}\approx 6\times 10^7\ incident \times 380 \approx 2\times 10^{10}$

$\left ( \frac{Number\ of\ hits}{Moller\ electron}\right ) \left (\frac{Moller\ electrons}{incidents\ electron} \right) \left (\frac{incident\ electrons}{sec} \right )$