View source for Problem: Refractive index for "light" and "heavy" water

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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.

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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

\(n=\sqrt{1-\frac{\lambda_0^2}{\pi} \left(6.4 \times 10^{-6}\right)} = \sqrt{1-\lambda_0^2\times 2.037 \times 10^{-6}}\)
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