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doing delta-X calculations

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caes 2017-01-27 13:50:05 -05:00
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alpha_spectroscopy/data/spectroscope_measurements.ods
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alpha_spectroscopy/report/report.tex
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@ -87,7 +87,7 @@ When an absorber is present between the alpha radiation source and the detector,
\subsection{Interference from a Thin Film}
\label{subsec:film}
A thin film placed between the source and the detector causes interference such that a loss of energy should be observed, proportional to the penetration depth of the foil. In order to build a rough model of this response function, 3 thicknesses of Nickel foil are placed between the detector and the source, and the average kinetic energy of the alpha particles emitted from the Americium-241 sample is computed using the Gaussian fit as outlined above for each condition. The thicknesses and computed average kinetic energies are tabulated below, with the error for the kinetic energy computed as the standard deviation of the mean, i.e., $\frac{\sigma}{\sqrt{n}}$. The density of Nickel near room temperature is extremly well-known to be approximately $\rho_{lead} = 8.908 g/cm^3$. The absorption depth is therefore computed as $\Delta x = \rho_{lead} T$ where T is the thickness of the absorber in centimeters.
A thin film placed between the source and the detector causes interference such that a loss of energy should be observed, proportional to the penetration depth of the foil. In order to build a rough model of this response function, 3 thicknesses of Nickel foil are placed between the detector and the source, and the average kinetic energy of the alpha particles emitted from the Americium-241 sample is computed using the Gaussian fit as outlined above for each condition. The thicknesses and computed average kinetic energies are tabulated below, with the error for the kinetic energy computed as the standard deviation of the mean, i.e., $\frac{\sigma}{\sqrt{n}}$. The density of Nickel near room temperature is extremly well-known to be approximately $\rho_{nickel} = 8908 mg/cm^3$. The absorption depth is therefore computed as $\Delta x = \rho_{nickel} T$ where T is the thickness of the absorber in centimeters.
\begin{table}[]
\label{tab:foil}
@ -105,7 +105,7 @@ When an absorber is present between the alpha radiation source and the detector,
\subsection{Interference from a Gas}
\label{subsec:gas}
A gas can act as an attenuator similarly to a film. An alpha spectrum is collected under increasing pressure (8 settings total). The gas occupies the entire space between the source and the detecter, so the depth is exactly that distance, but at each step the density changes, dependent on the pressure. Here, air is used as the absorbant gas, and the pressure is reported by a simple spin gauge connected to the vacuum chamber. The density of air at room temperature can be estimated as $\rho_A = 1.28 mg/cm^2$, and while this number is presented without uncertainty, it is an estimate only and can be considered accurate to the precision it is quoted here. Then, assuming any changes to equilibrium between gasses inside the chamber and outside occur quasistatically, the density of the gas at a given pressure can be related to its density at atmospheric pressure as
A gas can act as an attenuator similarly to a film. An alpha spectrum is collected under increasing pressure (8 settings total). The gas occupies the entire space between the source and the detecter, so the depth is exactly that distance, but at each step the density changes, dependent on the pressure. Here, air is used as the absorbant gas, and the pressure is reported by a simple spin gauge connected to the vacuum chamber. The density of air at room temperature can be estimated as $\rho_A = 1.28 mg/cm^3$, \cite{EngToolbox:AirDensity} and while this number is presented without uncertainty, it is an estimate only and can be considered accurate to the precision it is quoted here. Then, assuming any changes to equilibrium between gasses inside the chamber and outside occur quasistatically, the density of the gas at a given pressure can be related to its density at atmospheric pressure as
\begin{center}
$\rho_P = \frac{P}{P_A} \rho_A$.