Dose estimation numerical factor

In Sec. 2.9, we arrived at Eq. 5 for estimating the dose received by a specimen. We show here the numerical factors from that equation in meter-kilogram-second units:

$\displaystyle {\cal D} (\mbox{Gray})$	$\textstyle =$	$\displaystyle 2 N_{A} r_{e} hc \frac{1}{\eta \Delta^{2} A} \sum_{i} \phi_{i} \tau_{i} f_{2,i}$
	$\textstyle =$	$\displaystyle 2\cdot (6.022\times 10^{23} \frac{\mbox{atoms}}{\mbox{mole}})\cdot (2.818\times 10^{-15}\mbox{ m})$
		$\displaystyle \cdot (1239.852\mbox{ eV}\cdot\mbox{nm}) \cdot (1.6022\times 10... ...9} \frac{\mbox{Joule}}{\mbox{eV}}) \cdot (10^{-9} \frac{\mbox{m}}{\mbox{nm}})$
		$\displaystyle \cdot \frac{1}{\eta \cdot [\Delta (\mbox{nm})]^{2}\cdot A (\mbox{g/mole})} \cdot \frac{1}{(10^{-9}\mbox{ m/nm})^{2}\cdot (10^{-3}\mbox{ kg/g})}$
		$\displaystyle \cdot \sum_{i} \phi_{i} (10^{3}\mbox{ photons/sec})\cdot \tau_{i} (10^{-3}\mbox{sec})\cdot f_{2,i}$
	$\textstyle =$	$\displaystyle \frac{6.74\times 10^{5}} {\eta \cdot [\Delta (\mbox{nm})]^{2}\cd... ...i} (10^{3}\mbox{ photons/sec})\cdot \tau_{i} (10^{-3}\mbox{sec})\cdot f_{2,i}$	(9)