De Broglie Wavelength Of A Thermal Neutron

A particle in thermal equilibrium has average translational kinetic energy $K=\frac{3}{2}{k}_{B}T$, so its momentum is $p=\sqrt{2mK}=\sqrt{3m{k}_{B}T}$ and its de Broglie wavelength is $\lambda =\frac{h}{\sqrt{3m{k}_{B}T}}$. Substituting $m=1.67\times 10^{-27}\text{kg}$, $k_B=1.38\times 10^{-23}{\text{J K}}^{-1}$ and $T=300\text{K}$, the quantity under the root is $3\times 1.67\times 10^{-27}\times 1.38\times 10^{-23}\times 300\approx 2.07\times 10^{-47}$, whose square root is about $4.55\times 10^{-24}{\text{kg m s}}^{-1}$. Then $\lambda =\frac{6.6\times 10^{-34}}{4.55\times 10^{-24}}\approx 1.45\times 10^{-10}\text{m}=1.45\text{A˚}$. The value $0.29\text{A˚}$ wrongly inflates the temperature or momentum. The value $14.5\text{A˚}$ misplaces a power of ten. The value $0.91\text{A˚}$ omits the factor of three from the kinetic-energy expression. Crucially, this wavelength is comparable to interatomic spacings, which is precisely why thermal neutrons are superb probes for crystal-structure determination by neutron diffraction, building on the same de Broglie idea highlighted in NCERT. A magnitude check confirms a near-angstrom value, consistent with the slow speeds of thermalised neutrons. A subtle point is that the average thermal energy used here is a statistical mean, so individual neutrons have a spread of speeds and therefore a distribution of wavelengths about this central value. Nonetheless, because the typical wavelength lands near one angstrom, beams of thermal neutrons rival X-rays for probing crystal structures, with the added advantage of being scattered strongly by light atoms such as hydrogen.