Neutron Attenuation and Mean Free Path

Neutron Attenuation

Consider a monoenergetic neutron beam with intensity $I_0$ incident on a target of thickness $X$, with a neutron detector placed at some distance behind the target. Assume both the target and detector are very small, and the detector has a small solid angle that can only detect a portion of the neutrons passing through the target. In this scenario, all neutrons colliding with the target will either be absorbed or scattered in different directions, so only neutrons that do not interact with the target will reach the detector.

Let $I(x)$ be the intensity of the neutron beam that remains uncollided after traveling a distance $x$ within the target. When the neutron beam passes through a very thin target of thickness $\tau$, the number of collisions per unit area is $\Delta I = \sigma_t I\tau N = \Sigma_t I\tau \ \text{[neutrons/cm}^2\cdot\text{s]}$ (refer to equations (1) and (4) in Neutron Interactions and Cross-sections). Therefore, the decrease in neutron beam intensity as it travels a distance $dx$ within the target is:

\[-dI = \sigma_t IN dx = \Sigma_t I dx \tag{1}\]

Integrating this equation yields the following result:

\[\frac{dI}{I} = -\Sigma_t dx\] \[I(x) = I_0e^{-\Sigma_t x} \tag{2}\]

Thus, we can see that the neutron beam intensity decreases exponentially as the distance traveled through the target increases.

Mean Free Path

• The average distance a neutron travels between successive collisions with nuclei
• In other words, the average distance a neutron travels without collision
• Denoted by the symbol $\lambda$

$I(x)/I_0=e^{-\Sigma_t x}$ represents the probability that a neutron will not collide with a nucleus while traveling a distance $x$ within the medium. Therefore, the probability $p(x)dx$ that a neutron will travel a distance $x$ without collision and then collide within a distance $dx$ is:

\[\begin{align*} p(x)dx &= \frac{I(x)}{I_0} \Sigma_t dx \\ &= e^{-\Sigma_t x}\times \Sigma_t dx \\ &= \Sigma_t e^{-\Sigma_t x}dx \end{align*}\]

From this, we can derive the mean free path $\lambda$ as follows:

\[\begin{align*} \lambda &= \int_0^\infty xp(x)dx \\ &= \Sigma_t \int_0^\infty xe^{-\Sigma_t x}dx \\ &= \Sigma_t \left(\left[-\frac{1}{\Sigma_t}xe^{-\Sigma_t x} \right]_0^\infty +\int_0^\infty \frac{1}{\Sigma_t}e^{-\Sigma_t x} \right) \\ &= \left[-\frac{1}{\Sigma_t}e^{-\Sigma_t x} \right]_0^\infty \\ &= 1/\Sigma_t \tag{3} \end{align*}\]