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Madeline Horn edited The_equation_found_for_Single__.tex
over 8 years ago
Commit id: 4c87d2c23f52ddd60aeedfeee2435108e11f46b9
deletions | additions
diff --git a/The_equation_found_for_Single__.tex b/The_equation_found_for_Single__.tex
index 874964c..af6e2b2 100644
--- a/The_equation_found_for_Single__.tex
+++ b/The_equation_found_for_Single__.tex
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\begin{equation}
\begin{split}
E_n [eV] (fit)
& = (0.136\pm0.03)n +
(4.35\pm0.16) (4.35\pm0.16)\\
E_a [eV] (Peaks) &=(0.136\pm0.04)(0.03) + (4.35\pm0.16)\\
E_a [eV] (Peaks) &=4.418\pm0.20
\end{split}
\end{equation}
\end{equation}
Using the above equation, the intercept was found to be a value of $E_a = 4.418 \textrm{eV}$ at $n=0.5$. This value has been compared to the accepted value of $4.6674ev$ and $4.8865ev$ (cite), and the value extracted from the fit has a percent uncertainty of $5.34 \%$ and $9.59 \%$. The method used (Figure 6) proved to () values much closes to the accepted values. Thus it was concluded that although the value of $n=2$ is not on the line of best fit, using separate linear fits to approximate $\Delta E$ versus $n$ is a better method.