91Ó°ÊÓ

If an electron with the mass m and momentum p , the de Broglie wavelength$\mathit{λ}$ of an electron is written as:

$\mathit{λ}\mathbf{=}\frac{\mathbf{h}}{\mathbf{p}}$

The value of Planck’s constant $\mathit{h}\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{626}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{34}}\mathit{J}\mathbf{.}\mathit{s}\mathbf{}\mathit{o}\mathit{r}\mathbf{}\mathbf{4}\mathbf{.}\mathbf{136}\mathbf{×}{\mathbf{10}}^{\mathbf{15}}\mathbf{}\mathit{e}\mathit{V}\mathbf{.}\mathit{s}$.

If the mass of particle is and speed is v , the relativistic kinetic energy is written as:

$\mathit{K}\mathbf{=}\left(\mathrm{γ}-1\right)\mathit{m}{\mathit{c}}^{\mathbf{2}}$

Where,$\mathrm{γ}=\frac{1}{\sqrt{1-{\displaystyle \frac{v^2}{c^2}}}}$

And the nonrelativistic kinetic energy is written as:

$\mathit{K}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathit{m}{\mathit{v}}^{\mathbf{2}}$

Dark fringes occurs when a monochromatic bean sent through a narrow slit of width produces a diffraction$\mathit{θ}$ pattern on a distant screen.

$\mathit{s}\mathit{i}\mathit{n}\mathit{θ}\mathbf{=}\frac{\mathbf{m}\mathbf{λ}}{\mathbf{d}}$where, m is the number of fringes.

Given that, spacing of crystal is$0.0910×{10}^{-9}m$ .

$$\begin{gathered} m=1 \\ \mathrm{θ}=28.6° \end{gathered}$$

The kinetic energy K is:

$$\begin{gathered} K=\frac{h^2}{2m{\mathrm{λ}}^2} \\ =\frac{h^2}{2m{\left(d\sin \mathrm{θ}\right)}^2} \\ =\frac{\left(6.626×{10}^{-34}\right)}{2\left(1.67×{10}^{-27}\right)\left[\left(0.0910×{10}^{-9}\right)\sin 28.6°\right]} \\ =\frac{{\left(6.626×{10}^{-34}\right)}^2}{2\left(1.67×{10}^{-27}\right){\left(4.36×{10}^{-11}\right)}^2} \\ =0.433eV \end{gathered}$$

Hence, the kinetic energy of each neutron in the beam is 0.433 eV