Question

The electron interference pattern of Figure 38.12 was made by shooting electrons with $50\mathrm{keV}$of kinetic energy through two slits spaced role="math" localid="1650737433408" $1.0\mathrm{\mu m}$apart. The fringes were recorded on a detector $1.0\mathrm{m}$behind the slits.

a. What was the speed of the electrons? (The speed is large enough to justify using relativity, but for simplicity do this as a nonrelativistic calculation.)

b. Figure 38.12 is greatly magnified. What was the actual spacing on the detector between adjacent bright fringes?

Short answer

(a) The speed of the electron is$\mathrm{v}=1.33\times {10}^8\frac{\mathrm{m}}{\mathrm{s}}$

(b) The actual distance between consecutive light fringes on the detector is$\mathrm{\Delta y}=5.47\mathrm{\mu m}$

Step-by-step solution

Given information

The kinetic energy of electrons is 50 KeV and a detector captured the fringes is 1.0 m

Part (a) Step 1: Given information

The electron starts at rest (or close to it), thus the kinetic energy obtained is$\frac{1}{2}{\mathrm{mv}}^2$

Part (a) Step 2: Finding the electron's speed using kinetic energy 

We can begin by calculating the electron's speed using kinetic energy,

$$\begin{gathered} {\mathrm{E}}_{\mathrm{k}}=\frac{{\mathrm{mv}}^2}{2} \\ \mathrm{v}=\sqrt{\frac{2{\mathrm{E}}_{\mathrm{k}}}{\mathrm{m}}}\left(\mathrm{express}\mathrm{v}\right) \\ \mathrm{v}=\sqrt{\frac{\left(2\cdot 50\times {10}^3\right)\times \left(1.6\times {10}^{-19}\right)}{9.11\times {10}^{-31}}} \\ \mathrm{v}=1.33\cdot {10}^8\frac{\mathrm{m}}{\mathrm{s}} \end{gathered}$$

Part (b) Step 1: Given information

On a detector 1.0 m behind the slits, the fringes were recorded

Part (b) Step 2: Determine the double slit experiment, fringe spacing 

In a double slit experiment, we can now utilise an expression for fringe spacing.

$$\begin{gathered} \mathrm{\Delta y}=\frac{\mathrm{L\lambda }}{\mathrm{d}} \\ \mathrm{\lambda }=\frac{\mathrm{h}}{\mathrm{mv}}\text{(de Broglie wavelength)} \\ \mathrm{\Delta y}=\frac{\mathrm{Lh}}{\mathrm{mvd}}\text{(substitute}\mathrm{\lambda }\text{)} \\ \mathrm{\Delta y}=\frac{\left(1\cdot 6.63\times {10}^{-34}\right)}{\left(9.11\times {10}^{-31}\right)\times \left(1.33\times {10}^8\right)\times \left(1\times {10}^{-6}\right)}(\mathrm{substitute}\text{)} \\ \mathrm{\Delta y}=5.47\mathrm{\mu m} \end{gathered}$$