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	<title>Problem: Attenuation of the neutron beam - Revision history</title>
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	<updated>2026-04-17T08:35:13Z</updated>
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		<title>Wikiadmin: 1 revision imported</title>
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		<updated>2020-02-18T22:15:11Z</updated>

		<summary type="html">&lt;p&gt;1 revision imported&lt;/p&gt;
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				&lt;td colspan=&quot;1&quot; style=&quot;background-color: #fff; color: #202122; text-align: center;&quot;&gt;Revision as of 22:15, 18 February 2020&lt;/td&gt;
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		<id>https://e-learning.pan-training.eu/wiki/index.php?title=Problem:_Attenuation_of_the_neutron_beam&amp;diff=978&amp;oldid=prev</id>
		<title>ucph&gt;Tommy: Created page with &quot;&lt;!--\label{prob:attenuation}--&gt; Consider a neutron beam penetrating a sample. For simplicity we can assume the sample to be shaped like a slab with the large area perpendicula...&quot;</title>
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		<updated>2019-07-14T21:22:29Z</updated>

		<summary type="html">&lt;p&gt;Created page with &amp;quot;&amp;lt;!--\label{prob:attenuation}--&amp;gt; Consider a neutron beam penetrating a sample. For simplicity we can assume the sample to be shaped like a slab with the large area perpendicula...&amp;quot;&lt;/p&gt;
&lt;p&gt;&lt;b&gt;New page&lt;/b&gt;&lt;/p&gt;&lt;div&gt;&amp;lt;!--\label{prob:attenuation}--&amp;gt;&lt;br /&gt;
Consider a neutron beam penetrating a sample. For simplicity we can assume the sample to be shaped like a slab with the large area perpendicular to the neutron beam. &lt;br /&gt;
&lt;br /&gt;
=====Question 1=====&lt;br /&gt;
Derive the exponential decay of the beam inside the sample, [[Basics of neutron scattering#label-eq:attenuation|this equation]] from the [[Basics of neutron scattering]] page&amp;lt;!--(\ref{eq:attenuation})--&amp;gt;, \(\Psi(z) = \Psi(0) \exp(-\mu z)\).&lt;br /&gt;
&lt;br /&gt;
{{hidden begin|toggle=right|title=Solution|titlestyle=background:#ccccff}}&lt;br /&gt;
Consider a thin slab of material with thickness \(dz\) and area \(A(z)\). The number of neutrons scattered by this piece of material is \(\Psi(z) \Sigma A(z) dz\). The number of neutrons after passage of the slab will have decreased by this amount, i.e.&lt;br /&gt;
&lt;br /&gt;
:\( dN = - \Psi(z) \Sigma A(z) dz \, . \)&lt;br /&gt;
&lt;br /&gt;
Flux is neutrons pr. second hitting an area divided by the area whence&lt;br /&gt;
&lt;br /&gt;
:\( d\Psi = -\Psi \Sigma dz  \quad\Rightarrow\quad  \dfrac{d\Psi}{dz} = -\Psi \Sigma , \)&lt;br /&gt;
&lt;br /&gt;
and the solution to this first order differential equation is&lt;br /&gt;
&lt;br /&gt;
:\( \Psi(z) = \Psi_0 e^{-\Sigma z} . \,\)&lt;br /&gt;
&lt;br /&gt;
The constant \(\Sigma = \mu\) is also called the attenuation coefficient.&lt;br /&gt;
{{hidden end}}&lt;br /&gt;
&lt;br /&gt;
=====Question 2=====&lt;br /&gt;
Prove that when there is both absorption and scattering, the total attenuation coefficient is the sum of the individual coefficients.&lt;br /&gt;
&lt;br /&gt;
{{hidden begin|toggle=right|title=Solution|titlestyle=background:#ccccff}}&lt;br /&gt;
Similarly the decrease of neutrons in the beam after \(dz\) due to absorption is&lt;br /&gt;
&lt;br /&gt;
:\( d N_a = -\Psi(z)\mu_a A(z)dz .\, \)&lt;br /&gt;
&lt;br /&gt;
The total decrease is \(d N_\text{tot} = dN + d N_a\) giving decrease in flux&lt;br /&gt;
&lt;br /&gt;
:\( d \Psi_{\rm tot} = -\Psi_{\rm tot}(z)(\mu+\mu_a) dz \,\)&lt;br /&gt;
&lt;br /&gt;
with solution&lt;br /&gt;
&lt;br /&gt;
:\( \Psi_{\rm tot}(z) = \Psi_0 e^{-(\mu+\mu_a) z} . \)&lt;br /&gt;
{{hidden end}}&lt;/div&gt;</summary>
		<author><name>ucph&gt;Tommy</name></author>
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