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Ω=2π(1−cosθa)Ω=2π(1−cosθa) { $\Omega = 2\pi(1 - cos\theta_{a})$ θa=atan(1.2733.7)θa=atan(1.2733.7) { $\theta_{a} = atan(\frac{1.27}{33.7})$  = 0.0377rad. Therefor

Ωa=4.4547e-3 Sr

Ω=2π(1−cosθb)Ω=2π(1−cosθb) { $\Omega = 2\pi(1 - cos\theta_{b})$ θb=atan(1.5536.16)θb=atan(1.5536.16) { $\theta_{b} = atan(\frac{1.55}{36.16})$  = 0.0428rb Therefor

Ωb=5.7615e-3 Sr

Source 1 : 32.5mCi on 05/19/69 decayed to 9.93mCi on 10/14/2020.

Source 2 : 30.0mCi on 07/11/69 decayed to 9.20mCi on 10/14/2020.

Total source strength = 19.13mCi = 7.078e8 dps.  Assuming 1mCi = 3.7e7 dps (disintegrations per sec)

Since 93.5% of decays produce a 0.662MeV gamma that rate is 6.618e8dps.

Number of source emissions on scatterer should be;

Ωa4π∗Source=(4.4547e−3Sr4π)6.618e8dps=2.347e5dpsΩa4π∗Source=(4.4547e−3Sr4π)6.618e8dps=2.347e5dps { $\frac{\Omega_{a}}{4\pi}*Source = (\frac{4.4547e-3Sr}{4\pi})6.618e8dps=2.347e5dps$ . is

According to Wolfram Alpha https://demonstrations.wolfram.com/KleinNishinaFormulaForComptonEffect/ the total cross section for 0.662MeV gammas is:

3.226re23.226r2e { $3.226 r^{2}_{e}$   where re is the classical electron radius and re2=7.94×10−26cm−2r2e=7.94×10−26cm−2 { $r^{2}_{e} = 7.94x10^{-26} cm^{-2}$ .

So we get the fraction of the incident flux which scatters in the rod as:

Φ=Φ0e−nσxΦ=Φ0e−nσx { $\Phi = \Phi_{0}e^{-n\sigma x}$ where n is the electron density of Al, σ is the total cross section and x is the path length through the rod which we take as an average 1.27cm to account for the fact that it is round.

The number density of atoms in Al is 6.0e28 m-3.  Since each atom contains 13 electrons n = 7.8e29 m-3 = 7.8e23 cm-3.

So,

ΦΦ0=exp(−7.8e23∗3.226∗7.94e−26∗1.27)ΦΦ0=exp(−7.8e23∗3.226∗7.94e−26∗1.27) { $\frac{\Phi}{\Phi_{0}}=exp(-7.8e23 * 3.226 * 7.94e-26 * 1.27)$  = 0.7759 for the flux that passes through the rod.  Therefor, 0.2241∗2.347x105dps=5.26x104dps0.2241∗2.347x105dps=5.26x104dps { $0.2241 * 2.347 x 10^{5}dps=5.26 x 10^{4}dps$  total scattered flux into 4π.

This plot shows the intrinsic efficiency of NaI.

Note, the values are calculated.

Also note that the calculations are for a 3“ x 3” crystal with the source at a distance of 20“.  I am guessing out much larger crystal with only the central area exposed is more efficient.

Intrinsic efficiency means the fraction of gammas which will produce a pulse from the PMT.  In other words, anything in the compton shelf and the photopeak.

{ ${/download/attachments/256644197/Screen%20Shot%202020-10-13%20at%2011.12.08%20AM.png?version=1&modificationDate=1602705935000&api=v2}$

The differential cross section can be found from:

dσdΩ=1ntΔΩFoutFindσdΩ=1ntΔΩFoutFin { $\frac{d\sigma}{d\Omega}=\frac{1}{nt\Delta\Omega}\frac{F_{out}}{F_{in}}$ where n is the electron density of Al, t is the path length through the rod which we take as an average 1.27cm to account for the fact that it is round, and ΔΩ is the solid angle subtended bu the detector.  Fin is the flux incident on the detector and Fout is the flux incident on the detector given as,

Fin=RθϵθFin=Rθϵθ { $F_{in} = \frac{R_{\theta}}{\epsilon_{\theta}}$ where Rθ is the observed rate at scattering angle θ and εθ is the crystal detection efficiency at the scattering angle θ.

Therefor,

dσdΩ=1.7521×10−22RθϵθFin=7.4654e−28RθϵθdσdΩ=1.7521×10−22RθϵθFin=7.4654e−28Rθϵθ { $\frac{d\sigma}{d\Omega}=1.7521x10^{-22}\frac{R_{\theta}}{\epsilon_{\theta}F_{in}}=7.4654e-28\frac{R_{\theta}}{\epsilon_{\theta}}$ { ${/download/attachments/256644197/download.png?version=1&modificationDate=1602705238000&api=v2}$