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${\mathrm{I}}_{\mathrm{s}}+\mathrm{X}→$fluorescence intensity from the unknown + standard addition $V→$total volume of (unknown + standard)${\mathrm{V}}_0→$ fluorescence intensity from unknown initial concentration of Unknown initial concentration of standard ${\mathrm{V}}_S→$volume of standard.

${\mathrm{I}}_{\mathrm{s}+\mathrm{x}}\left(\frac{\mathrm{V}}{{\mathrm{V}}_0}\right)={\mathrm{I}}_{\mathrm{x}}+\frac{{\mathrm{I}}_{\mathrm{x}}}{[\mathrm{X}{]}_i}â‹\dots [\mathrm{S}{]}_iâ‹\dots \left(\frac{{\mathrm{V}}_{\mathrm{s}}}{{\mathrm{V}}_0}\right)$

The graph${\mathrm{I}}_{\mathrm{S}}+\mathrm{x}\left(\frac{\mathrm{V}}{{\mathrm{V}}_0}\right)$

Versus is,$[\mathrm{S}{]}_i\left(\frac{{\mathrm{V}}_{\mathrm{s}}}{{\mathrm{V}}_0}\right)$

The x-intercept is the concentration of Se. We calculate x-intercept by dividing b / a. So, the concentration of Se is$0.03842\mathrm{μ²\mu }/\mathrm{mL}$

$\begin{array}{r}\mathrm{w}\left(\mathrm{Se}\right)=\frac{\mathrm{m}\left(\mathrm{Se}\right)}{\mathrm{m}\left(\text{nuts}\right)}\\ \text{Where Mass of Se is,}\\ \mathrm{m}\left(\mathrm{Se}\right)=\mathrm{V}â‹\dots \mathrm{γ}\left(\mathrm{Se}\right)\\ \mathrm{m}\left(\mathrm{Se}\right)=10\mathrm{mL}â‹\dots 0.03842\mathrm{μ}\mathrm{g}/\mathrm{mL}\\ \mathrm{m}\left(\mathrm{Se}\right)=0.3842\mathrm{μ}\mathrm{g}\\ \mathrm{w}\left(\mathrm{Se}\right)=\frac{0.3842â‹\dots {10}^{−6}\mathrm{g}}{0.108\mathrm{g}}\\ \mathrm{w}\left(\mathrm{Se}\right)=3.56â‹\dots {10}^{−4}\mathrm{\%}\end{array}$

The standard deviation of x-intercept is calculated by formula:

$\frac{s_y}{\left|a\right|}â‹\dots \sqrt{\frac{1}{n}+\frac{{\stackrel{¯}{y}}^2}{a^2â‹\dots \mathrm{Î\pounds }{\left(x_i−\stackrel{¯}{x}\right)}^2}}$

n is a number of data, y is a average of y,$x_i$are individual values of x, and is average of x.

$∑{\left({\mathrm{x}}_{\mathrm{i}}−\stackrel{¯}{\mathrm{X}}\right)}^2$

The relative uncertainty in the intercept is:

$\frac{0.000732}{0.03842}=1.91\mathrm{\%}$

Uncertainty of :

$\begin{array}{r}0.0192â‹\dots 3.56â‹\dots {10}^{−4}\mathrm{\%}=6.84â‹\dots {10}^{−6}\mathrm{\%}\\ 3.56(Â\pm 0.068)â‹\dots {10}^{−4}\mathrm{\%}\end{array}$

95% confidence interval is calculated:

$Â\pm$t. standard deviation

t is a Student's t. We read value from Table 4-4. The degrees of freedom are:

n - 2 = 5 - 2 =3

For 95% confidence level and 3 degrees of freedom, the value is 3.182.

$\begin{array}{r}3.182â‹\dots 0.068â‹\dots {10}^{−4\mathrm{\%}}=0.22â‹\dots {10}^{−4\mathrm{\%}}\\ 3.56(Â\pm 0.22)â‹\dots {10}^{−4}\mathrm{\%}\end{array}$