Problem: Scattering length density for "light" and "heavy" water
The average coherent scattering length for elemental oxygen is 5.803 fm.
Question 1
Calculate the refractive index for "light" water, \(^1\)H\(_2\)O, and for "heavy" water, $^2$H$_2$O.
The neutron refractive index for a non-magnetic medium (and water is non-magnetic!) is given by $n^2=1-\frac{\lambda_0^2}{\pi}\rho\overline{b}$, where $\rho$ is the molecular number density for water and $\overline{b}$ is the mean coherent scattering length.
The coherent scattering length for:
$^1$H is $-3.7406$ fm,
$^2$H is 6.671 fm,
O is 5.803 fm.
Therefore, the mean coherent scattering length, $\overline{b}$, for:
$^1$H$_2$O$ = 2\times-3.7406 + 5.803 = -1.6782$ fm,
$^2$H$_2$O$ = 2\times6.671 + 5.803 = 19.145$ fm.
The relative atomic mass for:
$^1$H is 1.00794, $^2$H is 2.0141, O is 15.9994.
The density of natural water is 1 g$\cdot$cm$^{-3}$. Natural water is almost 100\% $^1$H$_2$O. The molecular number density is therefore:
$\rho\left(\textrm{H}_2\textrm{O}\right) = \frac{1}{2 \times 1.00794 + 15.9994}\cdot 6.023 \times 10^{23} = 3.343 \times 10^{22} ~\textrm{g}\cdot\textrm{cm}^{-3}$
Note that the molecular number density will be the same for both $^1$H$_2$O and $^2$H$_2$O.
The wavelength, $\lambda_0$, is often given in Ångströms, [Å]. Note that 1 fm = 10$^{-5}$ Å and 1 cm$^{-3}$ = 10$^{-24}$ Å$^{-3}$.
The scattering length density, $\rho\overline{b}$, for $^1$H$_2$O is thus:
$3.343\times 10^{22}\times 10^{-24} \cdot -1.6782\times 10^{-5} = -0.561 \times 10^{-6}$ Å$^{-2}$
and the refractive index is
$n=\sqrt{1-\frac{\lambda_0^2}{\pi} \left(-0.561 \times 10^{-6}\right)} = \sqrt{1+\lambda_0^2\times 1.786 \times 10^{-7}}$
The scattering length density, $\rho\overline{b}$, for $^2$H$_2$O is thus:
$3.343\times 10^{22}\times 10^{-24} \cdot 19.145\times 10^{-5} = 6.400 \times 10^{-6}$ Å$^{-2}$
and the refractive index is
$n=\sqrt{1-\frac{\lambda_0^2}{\pi} \left(6.4 \times 10^{-6}\right)} = \sqrt{1-\lambda_0^2\times 2.037 \times 10^{-6}}$