# (a) Obtain the de Broglie wavelength of a neutron of kinetic energy 150 eV.

Question:

(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.

Solution:

(a) De Broglie wavelength $=2.327 \times 10^{-12} \mathrm{~m}$; neutron is not suitable for the diffraction experiment Kinetic energy of the neutron, $K=150 \mathrm{eV}$

= 150 × 1.6 × 10−19

= 2.4 × 10−17 J

Mass of a neutron, mn = 1.675 × 10−27 kg

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

$K=\frac{1}{2} m_{n} v^{2}$

$m_{n} v=\sqrt{2 K m_{n}}$

Where,

= Velocity of the neutron

mnv = Momentum of the neutron

De-Broglie wavelength of the neutron is given as:

$\lambda=\frac{h}{m_{n} v}=\frac{h}{\sqrt{2 K m_{n}}}$

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

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

$\therefore \lambda=\frac{6.6 \times 10^{-34}}{\sqrt{2 \times 2.4 \times 10^{-17} \times 1.675 \times 10^{-27}}}$

$=2.327 \times 10^{-12} \mathrm{~m}$

It is given in the previous problem that the inter-atomic spacing of a crystal is about 1 Å, 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.

(b) De Broglie wavelength $=1.447 \times 10^{-10} \mathrm{~m}$

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

The average kinetic energy of the neutron is given as:

$E=\frac{3}{2} k T$

Where,

k = Boltzmann constant = 1.38 × 10−23 J mol−1 K−1

The wavelength of the neutron is given as:

$\lambda=\frac{h}{\sqrt{2 m_{n} E}}=\frac{h}{\sqrt{3 m_{n} k T}}$

$=1.447 \times 10^{-10} \mathrm{~m}$

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.