Ncert Exercises & Solutions

Question 11.32:

(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. (m_n= 1.675 x 10^-27 kg)

(b) Obtain the de Broglie wavelength associated with thermal neutrons at room temperature (27 $°$ C). Hence explain why a fast neutron beam needs to be thermalized with the environment it can be used for neutron diffraction experiments.

(a)
Hint: $\lambda=\frac{h}{\sqrt{2mK.E}}$

Step 1: Find the wavelength of a neutron.
The wavelength of a neutron with kinetic energy K.E is given by:
$\lambda=\frac{h}{\sqrt{2mK.E}}$

\begin{aligned} ∴ λ & = \frac{6 .6 \times 10^{- 14}}{\sqrt{2 \times 2 .4 \times 10^{- 17} \times 1 .675 \times 10^{- 27}}} \\ = 2 .327 \times 10^{- 12} m \end{aligned}

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)
Hint: $\lambda=\frac{h}{\sqrt{2mE}}$
Step 1: Find the average kinetic energy of the neutron.

The average kinetic energy of the neutron is given as:

E = \frac{3}{2} kT

Where, k = Boltzmann Constant 1.38 x 10^-23 J mol^-1 K^-1

Step 2: Find the wavelength of the neutron.
The wavelength of the neutron is given as:

\begin{aligned} λ & = \frac{h}{\sqrt{2 m_{} E}} = \frac{h}{\sqrt{3 m_{} kT}} \\ = \frac{6 .6 \times 10^{- 34}}{\sqrt{3 \times 1 .675 \times 10^{- 27} \times 1 .38 \times 10^{- 23} \times 300}} \\ = 1 .447 \times 10^{- 10} m \end{aligned}

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

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