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NCERT Class XII Chapter
Dual Nature of Radiation and Matter
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Question : 32 of 37
Marks: +1, -0
(a) Obtain the de Broglie wavelength of a neutron of kinetic energy 150 eV. An electron beam of this energy is suitable for crystal diffraction experiments. Would a neutron beam of the same energy be equally suitable?
Explain. (mnm_n = 1.675 × 10−2710^{-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 thermalised with the environment before it can be used for neutron diffraction experiments.
Solution:  
(a) Let us first calculate the wavelength of matter wave associated with neutron of kinetic energy 150 eV.
K = p22m\frac{p^2}{2m}
So, momentum p = 2mK\sqrt{2mK}
Wavelength λ = hp\frac{h}{p} = h2mK\frac{h}{\sqrt{2mK}}
λ =
6.63×10−342×1.675×10−27×150×1.6×10−19\frac{6.63 \times 10^{-34}}{\sqrt{2 \times 1.675 \times 10^{-27} \times 150 \times 1.6 \times 10^{-19}}}
m = 2.33 × 10−1210^{-12} m
As the interatomic spacing (1 Å = 10−1010^{-10} m) is about hundred times greater than this wavelength, so a neutron beam of 150 eV energy is not suitable for diffraction experiments.
(b) Average kinetic energy of a neutron at absolute temperature T is
12mv2\frac{1}{2}mv^2 = 32\frac{3}{2} kT or p22m\frac{p^2}{2m} = 32\frac{3}{2} kT or p = 3mkT\sqrt{3mkT}
∴ de Broglie wavelength , λ = hp\frac{h}{p} = h3mkT\frac{h}{\sqrt{3mkT}}
Given mnm_n = 1.675 × 10−2710^{-27} kg, k = 1.38 × 10−23 J K−110^{-23} \,\text{J K}^{-1}
T = 27 + 273 = 300 K, h = 6.63 × 10−3410^{-34} J s
λ =
6.63×10−343×1.675×10−27×1.38×10−23×300\frac{6.63 \times 10^{-34}}{\sqrt{3 \times 1.675 \times 10^{-27} \times 1.38 \times 10^{-23} \times 300}}
= 6.63×10−104.56\frac{6.63 \times 10^{-10}}{4.56} m = 1.45 × 10−1010^{-10} m = 1.45 Å
As this wavelength is comparable to interatomic spacing (≈ 1 Å) in a crystal, so thermal neutrons can be used for diffraction experiments. So high energy neutron beam should be first thermalised before using it for diffraction.
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