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Q:11.32 (a) Obtain the de Broglie wavelength of a neutron of kinetic energy 150\hspace{1mm}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\times 10^-^2^7\hspace{1mm}kg)

                              

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Kinetic energy of the neutron(K)=150eV

De Broglie wavelength associated with the neutron is

\\\lambda =\frac{h}{p}\\ \lambda =\frac{h}{\sqrt{2m_{N}K }}\\ \lambda =\frac{6.62\times 10^{-34}}{\sqrt{2\times 1.675\times 10^{-27}\times 150\times 1.6\times 10^{-19}}}\\ \lambda =2.327\times 10^{-12}m

Since an electron beam with the same energy has a wavelength much larger than the above-calculated wavelength of the neutron, a neutron beam of this energy is not suitable for crystal diffraction as the wavelength of the neutron is not of the order of the dimension of interatomic spacing.

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