Question

(a) For what kinetic energy of a neutron will the associated de Broglie wavelength be $1.40\times {10}^{-10}m$?

(b) Find the de Broglie wavelength of a neutron in thermal equilibrium with matter having an average kinetic energy of $\left(3/2\right)kT$ at $300K$.

Answer 1

(a) The wavelength associated with neutron, $\lambda =1.4\times {10}^{-10}m$

Mass of neutron, $m=1.66\times {10}^{-27}kg$

The kinetic energy is given by$K=\frac{1}{2}m{v}^2$

For de broglie wavelength and velocity,

$\lambda =\frac{h}{mv}$

So,

$K=\frac{h^2}{2{\lambda }^{2}m}=6.75\times {10}^{21}J=4.219\times {10}^{2}eV$

(b) The temperature of surrounding $T=300K$, $m$ is the mass of neutron.

Boltzmann constant, $K=1.38\times {10}^{-23}kg{m}^2{s}^{-2}{K}^{-1}$

Average K.E. of neutron is,

$K^{{}^{\prime }}=\frac{3}{2}KT\Rightarrow \frac{3}{2}(1.38\times {10}^{-23}\times 300)=6.21\times {10}^{-21}J$

From de broglie equation,

${\lambda }^{{}^{\prime }}=\frac{h}{\sqrt{2K^{{}^{\prime }}m}}\Rightarrow \frac{6.6\times {10}^{-34}}{\sqrt{2\times 6.21\times {10}^{-21}\times 1.66\times {10}^{-27}}}=0.146$