12.3. Neutron Spectroscopy 701
whereknhas units of ̊A−^1. This equation can be used to estimate the magnitude of
the wave vector for any given neutron energy (see example below). Note that here
the energy is in units ofmeV.
Example:
Determine the magnitude of the wave vector for a neutron having energy
equivalent to the thermal excitation energy at room temperature of 300K.Solution:
In the previous example we saw that the energy of thermal excitations at room
temperature of 300Kis 25.8meV. Substituting this value in equation 12.3.7
gives the required value of the wave vector.kn =[
En
2. 0723] 1 / 2
=
[
25. 8
2. 0723
] 1 / 2
=3. 5284 ̊A−^1
As mentioned earlier, the biggest advantage of neutrons as scattering probes is
that their wavelengths correspond to the thermal excitations. The other advantage
is their low cross section for most materials, which allows them to penetrate deep
into the materials. Now, of course, the wavelength and cross section also have energy
dependence but for spectroscopic purposes one mostly uses either thermal neutron
sources or cold neutron sources. The former have a energy range of up to about
100 meV while the latter can assume an energy of up to about 10meV.Atthese
energies, a diffraction pattern can reveal atomic structure of the material, which is
also evident from the computations performed earlier.
12.3.BNeutronSpectrometryTechniques
In this section we will look at some of the common neutron spectrometry techniques
and discuss the related instrumentation. But before we do that let us have a quick
look at he process of neutron scattering. Suppose a beam of neutrons interacts with
a sample and gets reflected as shown in Fig.12.3.1. The neutrons can undergo elastic
as well as inelastic scattering with the nuclei. These processes are described below.
Elastic Scattering:In an elastic scattering event the energy of the scattered
electrons is equal to that of the incident neutrons. That is
Ef = Ei
and kf = ki,where the subscriptsiandfrefer to the states before and after scattering. In
such as case we will haveEi−Ef = ω= 0 (12.3.8)
⇒ω =0. (12.3.9)