June, 7, 2007 Cf rate measurements using Photonics PMT

Cf source was placed 16.5 cm away from face of PHotonics PMT. The PMT output signal was descriminated by a Lecroy Model 821 descriminator set to pass any voltage below -0.5 mV. The descriminated output was then sent to a counter Model TC 531.

Configuration Definitions

1: Placed the Cf source 16.5 cm from face of Scintillator. neutrons passing through scintillator would also pass through PMT

2: Inserted a 2.5 cm thick steel shielding wall between Scintillator and Cf source. Right up against Scintillator

3: Replaced steel wall with 5 cm thick lead shielding brick

4: Replaced lead brick with 6.7 cm thick paraffin block ( Paraffin N.F. , Household wax made by Parowax. 4 cackes, total net weight 16 oz (454 g))

5: Increased thickness of paraffin block to 13.4 cm

6: No source backround

Measurements

Using Photonics PMT and BC

Configuration	Rate (counts/min)	Statistical Error in Rate (Counts/min)
1	2.7e3	0.09e3
2	2.2e3	0.01e3
3	0.8e3	0.02e3
4	2.2e3	0.08e3
5	1.8e3	0.02e3
6

Analysis

Properties of Materials used

Steel : Zeff = 26 and [math]\rho[/math]=8.02 [math]g/cm^3[/math]

Atomic composition for AISI 304) in %

Element	%
Si	1
Cr	19
Mn	2
Fe	68
Ni	10

Lead: Z =

Photon Mass Attenuation Lengh [math]\lambda = \frac{\rho}{\mu} (\frac{g}{cm^2})[/math]

Example Calculation

A 1 MeV photon has [math]\lambda \approx 18 \frac{g}{cm^2}[/math]

The photon intensity is attenuated such that [math]I = I_o e^{-\mu x}[/math]

The half length [math]\lambda_{\frac{1}{2}}[/math] is the distance needed to reduce the photon intensity in half

[math]I=\frac{I_o}{2}=I_o e^{-\mu \lambda_{\frac{1}{2}}}[/math]

Therefore [math]\lambda_{\frac{1}{2}} = -\frac{\ln(\frac{1}{2})}{\mu} = -\frac{\lambda \ln(\frac{1}{2})}{\rho} [/math]

In the case of steel: [math]\lambda_{\frac{1}{2}} = -\frac{18 \ln(\frac{1}{2})}{8} (cm) =1.5 cm = 0.61 inches [/math]

So if the Californium source were just 1 MeV photons then I would expect a 2.5 cm thick steel wall to drop the rate by at least a factor of 2. In reality the photons from Cf-252 are a distribution of energies so we would need to use a montecarlo to make a more accurate prediction. All that can be concluded at this time is that a drop in rate from 2700 to 2200 counts per minute corresponds to a 19% drop in rate which indicates that the particles causing light in the scintillator are not just 1 MeV photons. One can conjecture that because the photon distribution from CF-252 contains more photons at energies below 1 MeV than above that there must be other particles besides photons (probably neutrons) making a signal in the scintillator.