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===== Table of Important Numbers =====

{| class="wikitable"
|+ style="caption-side:right; text-align:left;"|''
:Sources:
:b\(_c\): V. F. Sears, Neutron News 3(3) (1992) 26
:Relative mass and density: http://www.webelements.com
:M: C. Kittel, Introduction to Solid State Physics 7th ed. (1996) Wiley, New York
''
|-
! scope="col"|
! scope="col"| b\(_c\)
! scope="col"| Relative mass
! scope="col"| density (g/cm\(^3\))
! scope="col"| M (\(\mu_B\))
|-
| \(^1\)H
| -3.7406
| 1.00794
|
|
|-
| \(^2\)H
| 6.671
| 2.0141
|
|
|-
| O
| 5.803
| 15.9994
|
|
|-
| \(^1\)H\(_2\)O
| -1.6482
| 18.01
| 1.0
|
|-
| Si
| 4.1491
| 28.0855
| 2.33
|
|-
| SiO\(_2\)
| 15.7551
| 60.0843
| 2.2
|
|-
| Co
| 2.49
| 58.933
| 8.9
| 1.6
|-
| Ru
| 7.02
| 101.07
| 12.37
|
|-
|}

Calculate \(q_c\) for the following interfaces (left to right moves from the first medium to the second):

=====Question 1=====
air / silicon

{{hidden begin|toggle=right|title=Solution|titlestyle=background:#ccccff}}
The refractive index is given by [[Neutron reflectivity#label-reflectivityeq:Critqc|this equation]], \(q_c=\sqrt{16\pi\left(\rho_2\overline{b}_2-\rho_1\overline{b}_1\right)}\).

The exercise is therefore about calculating \(\rho\overline{b}_c\) for the different media.
It is convenient to express \(q_c\) in Å\(^{-1}\). The numbers in the table above can be used to calculate \(\rho\overline{b}_c\) so long as they are expressed in Å.

Note that cm\(^{–3}\) = 10\(^{–24}\) Å\(^{–3}\)and \(\rho\) is a ''formula unit number density''.

The conversion between mass density and number density is performed by:

\begin{align}
\text{number density} = \frac{\text{mass density} \cdot \text{Avogadro's number}}{\text{atomic mass for the formula unit}}
\end{align}
For air: \(\rho\overline{b}_c\) = 0

For silicon:
\begin{align}
\rho\overline{b}_c &= \frac{2.33 \cdot 10^{-24}}{28.0855} \cdot 6.023 \cdot 10^{23} \cdot 4.1491 \cdot 10^{-5}\\

&= 2.0732 \cdot 10^{-6} \text{Å}^{-2}
\end{align}

Thereby \(q_c\) for the air/silicon interface is:

\begin{align}
q_c = \sqrt{16\pi \cdot 2.0732 \cdot 10^{-6} \text{Å}^{-2}} = 0.001 \text{Å}^{-1}
\end{align}
{{hidden end}}

=====Question 2=====
silicon / heavy water (\(^2\)H\(_2\)O)

{{hidden begin|toggle=right|title=Solution|titlestyle=background:#ccccff}}
For deuterated water (note that the number density for light and heavy water
will be the same) \(\rho\overline{b}_c\) is:

\begin{align}
\rho\overline{b}_c &= \frac{1 \cdot 10^{-24}}{18.01} \cdot 6.023 \cdot 10^{23} \cdot (5.803+2\cdot6.667) \cdot 10^{-5}\\

&= 3.576 \cdot 10^{-6} \text{Å}^{-2}
\end{align}

Thereby \(q_c\) for the silicon/heavy water (\(^2\)H\(_2\)O) interface is:

\begin{align}
q_c = \sqrt{16\pi \cdot (3.576-2.0732) \cdot 10^{-6} \text{Å}^{-2}} = 0.0087 \text{Å}^{-1}
\end{align}

{{hidden end}}

=====Question 3=====
silicon / light water (\(^1\)H\(_2\)O)

{{hidden begin|toggle=right|title=Solution|titlestyle=background:#ccccff}}
For protonated water (note that this is negative) \(\rho\overline{b}_c\) is:

\begin{align}
\rho\overline{b}_c &= \frac{1 \cdot 10^{-24}}{18.01} \cdot 6.023 \cdot 10^{23} \cdot (5.803-2\cdot3.7406) \cdot 10^{-5}\\

&= -2.9719 \cdot 10^{-7} \text{Å}^{-2}
\end{align}

Thereby \(q_c\) for the silicon/heavy water (\(^2\)H\(_2\)O) interface is:

\begin{align}
q_c = \sqrt{16\pi \cdot (-2.029719-2.0732) \cdot 10^{-6} \text{Å}^{-2}}
\end{align}

This is imaginary, i.e. there is no critical edge for a silicon / light water interface. This also comes from Snell's Law - the radiation is moving from a low refractive index to a high refractive index, so at the interface the wave is bent towards the surface normal.
{{hidden end}}

=====Question 4=====
(38.8% \(^1\)H\(_2\)O + 61.2% \(^2\)H\(_2\)O) / silicon

{{hidden begin|toggle=right|title=Solution|titlestyle=background:#ccccff}}
For (38.8% \(^1\)H\(_2\)O + 61.2% \(^2\)H\(_2\)O) \(\rho\overline{b}_c\) is:
\begin{align}
\rho\overline{b}_c &= (0.612 \cdot 3.576 - 0.388 \cdot 0.029719)\cdot 10^{-6}\\

&= -2.0734 \cdot 10^{-6} \text{Å}^{-2}\\
&\approx \text{the value for Si}
\end{align}

Thereby \(q_c\) for the (38.8% \(^1\)H\(_2\)O + 61.2% \(^2\)H\(_2\)O)/siliconinterface is:

\begin{align}
q_c = \sqrt{16\pi \cdot (2.0734-2.0732) \cdot 10^{-6}} \approx 0
\end{align}

{{hidden end}}

=====Question 5=====
Will there be reflectivity in the case of question 3 or 4?

{{hidden begin|toggle=right|title=Solution|titlestyle=background:#ccccff}}
There is reflectivity in case silicon/light water (\(^1\)H\(_2\)O), as the two media have different refractive indices. The radiation sees a contrast between them. There is no critical edge, but there is still some reflection.

There is no reflectivity in case (38.8% \(^1\)H\(_2\)O + 61.2% \(^2\)H\(_2\)O)/silicon. The refractive indices for the two media are the same. As far as the radiation is concerned, it is still travelling in a homogeneous medium and it will not experience an interface.

{{hidden end}}

Problem: Critical edge - Revision history

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