Question

De-Broglie wavelength associated with thermal neutrons at room temperature of ${27}^0$C is :

• A. $1.452\times {10}^{-10}$m
• B. $2.57\times {10}^{-10}$
• C. $0.452\times {10}^{-10}$m
• D. $3.452\times {10}^{-10}$m

Answer 1

The correct option is A $1.452\times {10}^{-10}$m

Given,

$T=27+273=300K$

$\text{Boltzmann's constant.}$

$k=1.38\times {10}^{-23}Jmo{l}^{-1}{k}^{-1}$

Kinetic energy of neutron at absolute temperature

$\text{T is given by,}E=\frac{3}{2}kT$

$\lambda =\frac{h}{\sqrt{2mE}}$$=\frac{h}{\sqrt{3mkT}}$

$\lambda =\frac{6.63\times {10}^{-34}}{\sqrt{3\times 1.675\times {10}^{-27}\times 1.38\times {10}^{-23}\times 300}}$$=1.45\times {10}^{-10}m$

Hence, De Broglie wavelength is $1.45\times {10}^{-10}m$