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

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

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

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

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.

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

Given that, the uniform spacing $1.60\mathrm{μ}m$, the speed of the electrons is $1.26×{10}^{4}m/s$and the distance x between the photographic film and the CD-ROM is .

The angles of diffraction are:

$$\begin{gathered} \sin {\mathrm{θ}}_1=\frac{\mathrm{λ}}{d} \\ \sin {\mathrm{θ}}_1=\frac{\left(h/p\right)}{d} \\ \sin {\mathrm{θ}}_1=\frac{\left[\left(6.526×{10}^{-34}\right)/\left\{\left(9.11×{10}^{-31}\right)\left(1.26×{10}^4\right)\right\}\right]}{\left(1.60×{10}^{-6}\right)} \\ \sin {\mathrm{θ}}_1=0.0361 \\ {\mathrm{θ}}_1={\sin }^{-1}\left(0.0361\right) \\ {\mathrm{θ}}_1=2.07° \end{gathered}$$

Andlocalid="1664010386168" $\sin {\mathrm{θ}}_2=\frac{2\mathrm{λ}}{d}$

$$\begin{gathered} \sin {\mathrm{θ}}_2=2\sin {\mathrm{θ}}_1 \\ {\mathrm{θ}}_2=2\left(2.07°\right) \\ {\mathrm{θ}}_2=4.14° \end{gathered}$$

Hence, if the speed of electron is$1.26×{10}^{4}m/s$at$\mathrm{θ}=2.07°\&\mathrm{θ}=4.14°$will the m = 1 and m = 2 intensity maxima appear.

If be the distance between the slit and the screen, the vertical distance between the center of the pattern from the$m^{th}$ dark band is:

$y_m=x\tan {\mathrm{θ}}_m$

Now, smaller angles $y_m=x{\mathrm{θ}}_m$

As, for smaller angle $\tan \mathrm{θ}=\mathrm{θ}$.

The distance between the two maxima is:

$$\begin{gathered} y=y_2-y_1 \\ =x\left({\mathrm{θ}}_2-{\mathrm{θ}}_1\right) \\ =\left(0.500\right)\left(0.0722-0.361\right) \\ =18.0mm \end{gathered}$$

Hence, the spacing on the film between given maxima is 18.0 mm