Physics and Engineering of Radiation Detection

700 Chapter 12. Radiation Spectroscopy

and meters to angstroms conversion factors. This gives

En =

1. 3106 × 10 −^40

λ^2

1

(1. 602 × 10 −^19 )(10−^20 )

=

81. 8128

λ^2 n

meV, (12.3.5)

wheremeV stands formillielectron volts andλnis in ̊A.Nowthatwehavea
relation between neutron energy and wavelength we can estimate the wavelength of
neutrons of any kinetic energy. Let us see what we get for thermal neutrons, which
typically have an energy of 25meV. Using above equation we get

λn =

[

81. 8128

En

] 1 / 2

=

[

81. 8128

25

] 1 / 2

=1. 81 ̊A, (12.3.6)

which corresponds to the typical interatomic distance in most materials. This makes
slow neutrons an excellent tool for studying structure of materials. Note that 25
meV approximately corresponds to the thermal excitations at room temperature
(see example below).

Example:

Determine the thermal excitation energy corresponding to the room temper-

ature of 300K.

Solution:

The energy of thermal excitations is given by

E=kBT,

wherekBis the Boltzmann’s constant andTis the absolute temperature. For

T= 300Kthis equation gives

E =

(

1. 3806 × 10 −^23

)

(300)

=4. 1418 × 10 −^21 J

=

4. 1418 × 10 −^21

1. 602 × 10 −^19

=0. 0258 eV

=25. 8 meV.

In neutron spectroscopic analyses, most experimenters prefer to use the term
wave vector instead of wavelength. Wave vector was previously defined for x-rays
ask=2π/λ. Same definition applies to neutron wavelength as well. Hence the
expression 12.3.5 can also be written as

En =2. 0723 kn^2 , (12.3.7)