(a) Obtain the de Broglie wavelength of a neutron of kinetic energy 150 eV. As you have seen in Exercise 11.31, an electron beam of this energy is suitable for crystal diffraction experiments. Would a neutron beam of the same energy be equally suitable? Explain. (mn = 1.675 × 10⁻²⁷ kg) (b) Obtain the de Broglie wavelength associated with thermal neutrons at room temperature (27 ºC). Hence explain why a fast neutron beam needs to be thermalised with the environment before it can be used for neutron diffraction experiments.

Ans (a).

De Broglie wavelength = 2.327 x 10^-12 m; neutron is not suitable for the diffraction experiment

Kinetic energy of the neutron, K = 150 eV = 150 x 1.6 x 10^-19 = 2.4 x 10^-17 J

Mass of a neutron ,m_n=1.675 x 10^-27 kg

The kinetic energy of the neutron is given by the relation:

K = 1/2 m_nv²

=> m_nv = √(2Km_n)

Where,v = Velocity of the neutron

m_nv = Momentum of the neutron

De-Broglie wavelength of the neutron is given as:

λ =h/m_nv   = h/√(2Km_n)

It is clear that wavelength is inversely proportional to the square root of mass.

Hence. wavelength decreases with increase in mass and vice versa.

λ = (6.6 x 10⁻³⁴)/√ [2(2.4 x 10^-17)(1.675 x 10^-27)]

= 2.327 x 10⁻¹² m

It is given in the previous problem that the inter-atomic spacing of a crystal is about 1 Aº, i.e., 10^-10 m. Hence, the inter-atomic spacing is about a hundred times greater. Hence, a neutron beam of energy 150 eV is not suitable for diffraction experiments.

Ans (b).

De Broglie wavelength = 1.447 x 10^-10 m

Room temperature, T = 27°C = 27 + 273 = 300 K

The average kinetic energy of the neutron is given as:

E= 3/2 kT

Where, k = Boltzmann constant = 1.38 x 10⁻²³ J mol^-1 K⁻¹ The wavelength of the neutron is given as:

λ = h/√(2Em_n) = h/√(3Km_nT )

=   (6.6 x 10⁻³⁴)/√[3 x  (1.675 x  10⁻²⁷ )x (1.38 x  10⁻²³ )x 300]

= 1.447 x 10⁻¹⁰ m

This wavelength is comparable to the inter-atomic spacing of a crystal. Hence, the high energy neutron beam should first be thermalised, before using it for diffraction.