Question

A neutron beam of energy $E$ scatters from atoms on a surface with a spacing $d=0.1nm$. The first maximum of intensity in the reflected beam occurs at $\theta ={30}^{\circ }$. What is the kinetic energy $E$ of the beam in $eV$?

Answer 1

Given, $d=0.1nm$

$\theta ={30}^{\circ }$

$n=1$

According to Breagg's law,

$2d\sin \theta =n\lambda$

$\Rightarrow 2\times 0.1\times \sin {30}^{\circ }=\lambda$

$\lambda =0.1nm={10}^{-10}m$

Momentum $P=\frac{h}{\lambda }$

$P=\frac{6.62\times {10}^{-34}}{{10}^{-10}}=6.62\times {10}^{-24}kgm/s$

Kinetic energy

$K.E.=\frac{1}{2}\frac{p^2}{m}=\frac{1}{2}m{v}^2$

$=\frac{1}{2}\frac{(6.62\times {10}^{-24}{)}^2}{1.67\times {10}^{-27}}J$

$=\frac{1}{2}\frac{(6.62\times {10}^{-24}{)}^2}{1.67\times {10}^{-27}\times 1.6\times {10}^{-19}}J$

$K.E.=0.082eV$

$\text{Final Answer}:0.082eV$