William edited After_removing_the_background_maxima__.tex  over 8 years ago

Commit id: b17182ddc9dcf80618047af464b4b690a6e1cfc6

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After removing the background, maxima and minima were found by fitting a quadratic function to a small range of data close to the perceived maximum and minimum values. This produces maxima and minima values with the smallest amount of error. Due to poor resolution, the last few peaks of the Franck-Hertz curve appear as points, and proved difficult to fit using a quadratic function.  Once these values were found, the measured spacings, ($\Delta E_{n}$), between the maxima and minima of the Franck-Hertz curve were plotted against the minimum order (n) of the peaks and dips and analyzed using a linear fit (Figure 6). The lowest excitation energy of Mercury was determined from the graph by finding the intercept of the linear fit at $n=0.5$.The value of the excitation energy found from the linear fit was then compared to the known value of the lowest excitation energy for Mercury I ($4.6674ev$ and $4.8865ev$) (REFERENCE) . For the dips, when $n=0.5$, $E_{a}=4.72eV$. For the peaks, when $n=0.5$, $E_{a}=4.63eV$ The values found from the linear fits of the data had a percent uncertainty of $0.80 \% -3.41 \%$. Thus, excitation value found for the lowest state of Mercury quantitatively agrees with the expected vale.\\  The fits for the peaks and dips are:  \begin{equation}  \begin{split}  E_n [eV] (Peaks) &  = (0.1040\pm0.04)n + (4.573\pm0.19) (4.573\pm0.19)\\  E_a [eV] (Peaks) &=(0.1040\pm0.04)(0.5) + (4.573\pm0.19)\\  E_a [eV] (Peaks) &=4.63\pm.23  \end{split}  \end{equation}  \begin{equation}  \begin{split}  E_n [eV] (Dips) = &=  (0.0660\pm0.05)n + (4.686\pm0.20) (4.686\pm0.20)\\  E_a [eV] (Dips) &= (0.0660\pm0.05)(0.5) + (4.686\pm0.20)\\  E_a [eV] (Dips) &= .72\pm.25\\  \end{split}  \end{equation} The value of the excitation energy found from the linear fit was then compared to the known value of the lowest excitation energy for Mercury I ($4.6674ev$ and $4.8865ev$) (citation needed) . For the dips, when $n=0.5$, $E_{a}=4.72eV$. For the peaks, when $n=0.5$, $E_{a}=4.63eV$ The values found from the linear fits of the data had a percent uncertainty of $1.68 \% -3.42 \%$. Thus, excitation value found for the lowest state of Mercury quantitatively agrees with the expected vale.\\  The fits for the peaks and dips are: