Problem:The moderator temperature - Revision history

Wikiadmin: Wikiadmin moved page Problem: The moderator temperature to Problem:The moderator temperature

2020-09-20T15:25:33Z

Wikiadmin moved page Problem: The moderator temperature to Problem:The moderator temperature

← Older revision	Revision as of 15:25, 20 September 2020
(No difference)

Wikiadmin at 16:50, 20 April 2020

2020-04-20T16:50:09Z

← Older revision		Revision as of 16:50, 20 April 2020
Line 28:		Line 28:

	{{hidden begin\|toggle=right\|title=Solution\|titlestyle=background:#ccccff}}		{{hidden begin\|toggle=right\|title=Solution\|titlestyle=background:#ccccff}}
	<figtable id="tab:specs_moderated_neutrons">[[File:~~Specs_moderated_neutrons1~~.png\|frame\|<caption> Specifics for neutrons moderated at 30 K and 300 K.</caption>]]</figtable>		<figtable id="tab:specs_moderated_neutrons">[[File:Specs_moderated_neutrons.png\|frame\|<caption> Specifics for neutrons moderated at 30 K and 300 K.</caption>]]</figtable>
	The values shown are shown in <xr id="tab:specs_moderated_neutrons">Table %i</xr>.		The values shown are shown in <xr id="tab:specs_moderated_neutrons">Table %i</xr>.

Wikiadmin: /* Question 2 */

2020-04-20T12:46:16Z

Question 2

← Older revision		Revision as of 12:46, 20 April 2020
Line 28:		Line 28:

	{{hidden begin\|toggle=right\|title=Solution\|titlestyle=background:#ccccff}}		{{hidden begin\|toggle=right\|title=Solution\|titlestyle=background:#ccccff}}
	<figtable id="tab:specs_moderated_neutrons">[[File:~~specs_moderated_neutrons~~.png\|frame\|<caption> Specifics for neutrons moderated at 30 K and 300 K.</caption>]]</figtable>		<figtable id="tab:specs_moderated_neutrons">[[File:Specs_moderated_neutrons1.png\|frame\|<caption> Specifics for neutrons moderated at 30 K and 300 K.</caption>]]</figtable>
	The values shown are shown in <xr id="tab:specs_moderated_neutrons">Table %i</xr>.		The values shown are shown in <xr id="tab:specs_moderated_neutrons">Table %i</xr>.

Wikiadmin at 12:14, 16 April 2020

2020-04-16T12:14:49Z

← Older revision		Revision as of 12:14, 16 April 2020
Line 55:		Line 55:
	\exp\big( {-C(T)(\lambda_1^{-2} -\lambda_2^{-2} )} \big)		\exp\big( {-C(T)(\lambda_1^{-2} -\lambda_2^{-2} )} \big)
	\end{equation*}		\end{equation*}
	where $ C(T) = h^2 / (2 m_n k_B T) $, hence $C(30\text{K}) = 3.17\times 10^{-19} \text{m}^2$ and $C(300\text{K}) = 3.17\times 10^{-20}\text{m}^2$.		where $ C(T) = h^2 / (2 m_n k_B T) $, hence $C(30\text{K}) = 3.17\times 10^{-19} \text{m}^2$ and $C(300\text{K}) = 3.17\times 10^{-20}\text{m}^2$.

	So for $T=30 \text{K}$		So for $T=30 \text{K}$

	\begin{equation*}		\begin{equation*}
	\frac{I(4\tt{~~\AA~~})}{I(20\tt{~~\AA~~})}		\frac{I(4\tt{Å})}{I(20\tt{Å})}
	= 5^5 \cdot \exp\big( -3.17\times 10^{-19}\cdot 6\times 10^{18} \big)		= 5^5 \cdot \exp\big( -3.17\times 10^{-19}\cdot 6\times 10^{18} \big)
	= 466		= 466
	\end{equation*}		\end{equation*}

	since $(4 \times 10^{-10} \text{m})^{-2} - (20 \times 10^{-10} \text{m})^{-2} = 6 \times 10^{18} \text{m}^{-2}$.		since $(4 \times 10^{-10} \text{m})^{-2} - (20 \times 10^{-10} \text{m})^{-2} = 6 \times 10^{18} \text{m}^{-2}$.

	For $T=300 \text{K}$		For $T=300 \text{K}$

	\begin{equation*}		\begin{equation*}
	\frac{I(4\tt{~~\AA~~})}{I(20\tt{~~\AA~~})}		\frac{I(4\tt{Å})}{I(20\tt{Å})}
	= 5^5 \cdot \exp\big( {-3.17\times 10^{-20}\cdot 6\times 10^{18}} \big)		= 5^5 \cdot \exp\big( {-3.17\times 10^{-20}\cdot 6\times 10^{18}} \big)
	= 2584		= 2584
Line 85:		Line 85:

	According to the calculation above apparantly there is several orders of magnitude more intensity of the neutrons at 4 Å than at 20 Å. If we however take a look at a small window of energies around 4 Å and 20 Å, respectively and integrate over these the ratios are not so extreme:		According to the calculation above apparantly there is several orders of magnitude more intensity of the neutrons at 4 Å than at 20 Å. If we however take a look at a small window of energies around 4 Å and 20 Å, respectively and integrate over these the ratios are not so extreme:
	Consider the intensity constant within the small wavelength bands $[\lambda_1-\Delta \lambda_1; \lambda_1+\Delta \lambda_1]$ and $[\lambda_2-\Delta \lambda_2; \lambda_2+\Delta \lambda_2]$. The relative intensity between wavelength bands is		Consider the intensity constant within the small wavelength bands $\lambda_1-\Delta \lambda_1; \lambda_1+\Delta \lambda_1$ and $\lambda_2-\Delta \lambda_2; \lambda_2+\Delta \lambda_2$. The relative intensity between wavelength bands is

	\begin{equation*}		\begin{equation*}
Line 97:		Line 97:

	The approximation is shown in <xr id="fig:moderator_spectrum">Figure %i</xr>.		The approximation is shown in <xr id="fig:moderator_spectrum">Figure %i</xr>.
	If ${\Delta \lambda}/{\lambda}=10\%$ and $\lambda_1=4$ Å and $\lambda_2=20$ Å then $\Delta\lambda_1=0.4$ Å and $\Delta\lambda_2=2$ Å so at 30 K where $I_1 / I_2 = I(~~4\AA{}~~) / I(~~20\AA{}~~) = 466$		If ${\Delta \lambda}/{\lambda}=10\%$ and $\lambda_1=4$ Å and $\lambda_2=20$ Å then $\Delta\lambda_1=0.4$ Å and $\Delta\lambda_2=2$ Å so at 30 K where $I_1 / I_2 = I(4Å) / I(20Å) = 466$

	\begin{equation*}		\begin{equation*}
Line 103:		Line 103:
	\end{equation*}		\end{equation*}

	and at 300 K where ${I_1} / {I_2}={I(~~4\AA{}~~)} / {I(~~20\AA{}~~)} = 2584$		and at 300 K where ${I_1} / {I_2}={I(4Å)} / {I(20Å)} = 2584$

	\begin{equation*}		\begin{equation*}

Wikiadmin: 1 revision imported

2020-02-18T22:15:12Z

1 revision imported

← Older revision	Revision as of 22:15, 18 February 2020
(No difference)

ucph>Tommy: Created page with "In typical sources for neutron scattering purposes, the neutrons are moderated by water. These "thermal" neutrons will (almost) reach thermal equilibrium with the moderator. C..."

2019-07-14T21:26:31Z

Created page with "In typical sources for neutron scattering purposes, the neutrons are moderated by water. These "thermal" neutrons will (almost) reach thermal equilibrium with the moderator. C..."

New page

In typical sources for neutron scattering purposes, the neutrons are moderated by water. These "thermal" neutrons will (almost) reach thermal equilibrium with the moderator. Calculate the equivalent energy, $E$, and velocity, $v$, of a thermal neutron, moderated by water at $T_{{\rm H}_2 {\rm O}} = 300\,\text{K}$.

=====Question 1=====
Calculate the corresponding de Broglie wavelength, $\lambda$, and wavenumber, $k$.

{{hidden begin|toggle=right|title=Hint|titlestyle=background:#ccccff}}
The energy at a specific temperature is $ E = k_B T $ [meV] and the neutron velocity at a specific temperature is

\begin{equation*}
v = \sqrt{\frac{2k_BT}{m_n}}
\end{equation*}

with unit [m/s].

Use the constants $ m_n = 1.675 \times 10^{-27} \text{kg}$, $1 \text{eV} = 1.602 \times 10^{-19} \text{J}$, $ k_B = 1.38 \times 10^{-23} \text{J/K} = 8.61 \times 10^{-5} \text{eV/K}$.
{{hidden end}}

{{hidden begin|toggle=right|title=Hint|titlestyle=background:#ccccff}}
The de Broglie wavelength can be calculated from $\lambda = h / (m_n v) $ [Å].
{{hidden end}}

{{hidden begin|toggle=right|title=Solution|titlestyle=background:#ccccff}}
The calculated values can be found in the table of the solution of the next question.
{{hidden end}}

=====Question 2=====
Perform the same calculations for neutrons thermalised by liquid H$_2$ at $T_H = 30 \text{K}$.

{{hidden begin|toggle=right|title=Solution|titlestyle=background:#ccccff}}
<figtable id="tab:specs_moderated_neutrons">[[File:specs_moderated_neutrons.png|frame|<caption> Specifics for neutrons moderated at 30 K and 300 K.</caption>]]</figtable>
The values shown are shown in <xr id="tab:specs_moderated_neutrons">Table %i</xr>.

{{hidden end}}

=====Question 3=====
For each of the two types of moderators above, calculate the ratio of intensities: $I$(4 Å)/$I$(20 Å).

{{hidden begin|toggle=right|title=Solution|titlestyle=background:#ccccff}}

\begin{equation*}
\frac{I(\lambda_1)}{I(\lambda_2)}
= \left(\frac{\lambda_2}{\lambda_1}\right)^5
\exp\big( {-C(T)(\lambda_1^{-2} -\lambda_2^{-2} )} \big)
\end{equation*}
where $ C(T) = h^2 / (2 m_n k_B T) $, hence $C(30\text{K}) = 3.17\times 10^{-19} \text{m}^2$ and $C(300\text{K}) = 3.17\times 10^{-20}\text{m}^2$.

So for $T=30 \text{K}$

\begin{equation*}
\frac{I(4\tt{\AA})}{I(20\tt{\AA})}
= 5^5 \cdot \exp\big( -3.17\times 10^{-19}\cdot 6\times 10^{18} \big)
= 466
\end{equation*}

since $(4 \times 10^{-10} \text{m})^{-2} - (20 \times 10^{-10} \text{m})^{-2} = 6 \times 10^{18} \text{m}^{-2}$.

For $T=300 \text{K}$

\begin{equation*}
\frac{I(4\tt{\AA})}{I(20\tt{\AA})}
= 5^5 \cdot \exp\big( {-3.17\times 10^{-20}\cdot 6\times 10^{18}} \big)
= 2584
\end{equation*}

{{hidden end}}

=====Question 4=====
Many neutron instruments utilize a band of incident wavelength, $\Delta \lambda$. For many instruments, $\Delta \lambda / \lambda$ is almost constant and is of the order 0.1%-10%, depending on instrument type. For instruments with these bandwidths, calculate the ratio of the neutron fluxes at the sample: $\Psi$(4 Å)/$\Psi$(20 Å).

{{hidden begin|toggle=right|title=Solution|titlestyle=background:#ccccff}}

<figure id="fig:moderator_spectrum"> [[File:Moderator spectrum.png| thumb | 400px | <caption> A sketch of a moderator spectrum and the intensity in a particular bandwidth at 4 Å and 20 Å respectively </caption>]] </figure>

According to the calculation above apparantly there is several orders of magnitude more intensity of the neutrons at 4 Å than at 20 Å. If we however take a look at a small window of energies around 4 Å and 20 Å, respectively and integrate over these the ratios are not so extreme:
Consider the intensity constant within the small wavelength bands $[\lambda_1-\Delta \lambda_1; \lambda_1+\Delta \lambda_1]$ and $[\lambda_2-\Delta \lambda_2; \lambda_2+\Delta \lambda_2]$. The relative intensity between wavelength bands is

\begin{equation*}
\frac{I(\lambda_1\pm\Delta\lambda_1)}{I(\lambda_2\pm\Delta\lambda_2)}
= \frac{\displaystyle\int_{\lambda_1-\Delta \lambda_1}^{\lambda_1+\Delta \lambda_1}
I(\lambda_1)d\lambda_1}
{\displaystyle\int_{\lambda_2-\Delta \lambda_2}^{\lambda_2+\Delta \lambda_2}
I(\lambda_2)d\lambda_2}
\approx \frac{I_1\cdot 2\Delta \lambda_1}{I_2\cdot 2\Delta \lambda_2}
\end{equation*}

The approximation is shown in <xr id="fig:moderator_spectrum">Figure %i</xr>.
If ${\Delta \lambda}/{\lambda}=10\%$ and $\lambda_1=4$ Å and $\lambda_2=20$ Å then $\Delta\lambda_1=0.4$ Å and $\Delta\lambda_2=2$ Å so at 30 K where $I_1 / I_2 = I(4\AA{}) / I(20\AA{}) = 466$

\begin{equation*}
\frac{I(4\pm0.4)}{I(20\pm2)} = 466\frac{0.8}{4} = 93.2
\end{equation*}

and at 300 K where ${I_1} / {I_2}={I(4\AA{})} / {I(20\AA{})} = 2584$

\begin{equation*}
\frac{I(4\pm0.4)}{I(20\pm2)} = 2584\frac{0.8}{4} = 517
\end{equation*}

{{hidden end}}