In the experiment by Jönsson, 50-keV electrons impinged on slits of width 500 nm separated by a distance of 2000 nm. The observation screen was located 350 mm beyond the slits. What was the distance between the first two maxima? Strategy The equation specifying the orders of maxima and the angle θ

Question

In the experiment by Jönsson, 50-keV electrons impinged on slits of width 500 nm separated by a distance of 2000 nm. The observation screen was located 350 mm beyond the slits. What was the distance between the first two maxima?

Strategy The equation specifying the orders of maxima and the angle θ from incidence is (see Figure 5.18) given by

[latex]d \sin heta=n \lambda[/latex] (5.36)

The first maximum is of order n = 0 and occurs for θ = 0. The next maximum, at angle θ, occurs for n = 1:

[latex]\sin heta=\frac{\lambda}{d}=\frac{\lambda}{2000 nm }[/latex]

We need to determine the wavelength λ to find the angle θ. Once we know θ, we can find the distance between the two maxima on the observation screen.

Accepted Answer

We have already calculated the wavelength for electrons of energy [latex]e V_{0}[/latex] Equation (5.7).

[latex]\lambda=\frac{1.226 nm \cdot V ^{1 / 2}}{\sqrt{V_{0}}}[/latex] (5.7)

[latex]\lambda=\frac{1.226 nm \cdot V ^{1 / 2}}{\sqrt{50 imes 10^{3} V }}=5.48 imes 10^{-3} nm[/latex]

Because 50 keV may be too high an energy for a nonrelativistic calculation such as that done in Equation (5.7), we should perform a relativistic calculation to be certain. We first find the momentum and insert that into [latex]\lambda=h / p[/latex].

[latex]\begin{aligned}(p c)^{2}=E^{2}-E_{0}^{2}=&\left(K+E_{0} ight)^{2}-E_{0}^{2} \=&\left(50 imes 10^{3} eV +0.511 imes 10^{6} eV ight)^{2} \&-\left(0.511 imes 10^{6} eV ight)^{2} \=&\left(0.231 imes 10^{6} eV ight)^{2}\end{aligned}[/latex]

Now we can determine the wavelength.

[latex]\lambda=\frac{h}{p}=\frac{h c}{p c}=\frac{1240 eV \cdot nm }{0.231 imes 10^{6} eV }=5.36 imes 10^{-3} nm[/latex]

Therefore, we find the more accurate relativistic value to be somewhat less (2%) than the nonrelativistic value. Now we can determine the angle.

[latex]\sin heta=\frac{5.36 imes 10^{-3} nm }{2000 nm }=2.68 imes 10^{-6}[/latex]

The distance of the first maximum along the screen is [latex]y=\ell an heta[/latex], but for such a small angle, [latex]\sin heta \approx an heta[/latex].

[latex]\begin{aligned}y &=\ell an heta \approx \ell \sin heta=350 mm \left(2.68 imes 10^{-6} ight) \&=9.38 imes 10^{-4} mm \frac{10^{6} nm }{ mm }=938 nm\end{aligned}[/latex]

Such a diffraction pattern is too small to be viewed by the naked eye. Jönsson magnified the pattern by a series of electronic lenses and then observed a fluorescent screen with a 10-power optical microscope to see the diffraction pattern as expected.

Question 5.7: In the experiment by Jönsson, 50-keV electrons impinged on s...

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