The de-Broglie wavelength of neutron at 127°C is: (Given Boltzmann constant, \( k = 1.38 \times 10^{-23}\,\mathrm{J\,K^{-1}} \), \( h = 6.63 \times 10^{-34}\,\mathrm{J\,s} \), mass of neutron = \( 1.66 \times 10^{-27}\,\mathrm{kg} \))
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 1.264\,\mathring{\mathrm{A}} \)
The de Broglie wavelength of a particle is \( \lambda = h/p \). A neutron in thermal equilibrium at temperature \( T \) has an average kinetic energy of \( \frac{3}{2}kT \), so its momentum is found from that energy.
\( \displaystyle \frac{p^{2}}{2m} = \frac{3}{2}kT \;\Rightarrow\; p = \sqrt{3mkT} \;\Rightarrow\; \lambda = \frac{h}{\sqrt{3mkT}} \)
\( T = 127 + 273 = 400\,\mathrm{K} \)
\( \displaystyle 3mkT = 3 \times 1.66\times10^{-27} \times 1.38\times10^{-23} \times 400 = 2.749\times10^{-47}\,\mathrm{kg^{2}\,m^{2}\,s^{-2}} \)
\( \displaystyle \sqrt{3mkT} = 5.243\times10^{-24}\,\mathrm{kg\,m\,s^{-1}} \)
\( \displaystyle \lambda = \frac{6.63\times10^{-34}}{5.243\times10^{-24}}\,\mathrm{m} = 1.264\times10^{-10}\,\mathrm{m} = 1.264\,\mathring{\mathrm{A}} \)