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Q : 11.17 (b) Also find the de Broglie wavelength of a neutron, in thermal equilibrium with matter, having an average kinetic energy of  (3/2)\hspace{1mm}kT at  300\hspace{1mm}K.

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The kinetic energy of the neutron is

\\K=\frac{3}{2}kT\\ K=\frac{3}{2}\times 1.38\times 10^{-23}\times 300\\ K=6.21\times 10^{-21} J

Where k Boltzmann's Constant is 1.38\times10-23 J/K

The momentum of the neutron will be p

\\p=\sqrt{2m_{N}K}\\ p=\sqrt{2\times 1.675\times 10^{-27}\times 6.21\times 10^{-21}}\\ p=4.56\times 10^{-24}kg\ m\ s^{-1}

Associated De Broglie wavelength is

\\\lambda =\frac{h}{p}\\ \lambda =\frac{6.62\times 10^{-34}}{4.56\times 10^{-24}}\\ \lambda =1.45\times 10^{-10} m

De Broglie wavelength of the neutron is 0.145 nm.

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