Question

Consider a beam of electrons in a vacuum, passing through a very narrow slit of width 2.00μm. The electrons then head toward an array of detectors a distance 1.068 m away. These detectors indicate a diffraction pattern, with a broad maximum of electron intensity (i.e., the number of electrons received in a certain area over a certain period of time) with minima of electron intensity on either side, spaced 0.527 cm from the center of the pattern. What is the wavelength λ of one of the electrons in this beam? Recall that the location of the first intensity minima in a single slit diffraction pattern for light is y=Lλ/a, where L is the distance to the screen (detector) and a is the width of the slit. The derivation of this formula was based entirely upon the wave nature of light, so by de Broglie's hypothesis it will also apply to the case of electron waves.

Answer 1

Answer:

$9.9\cdot 10^{-9}m$

Explanation:

In a single-slit diffraction pattern, the location of the first minimum is given by

$y=\frac{L\lambda }{a}$

where

L is the distance between the slit and the screen

$\lambda$ is the wavelength

a is the width of the slit

In this problem, we have

$y=0.527 cm = 5.27\cdot 10^{-3} m$ (location of first minimum)

L = 1.068 m (distance of the screen)

$a=2.00\mu m=2.00\cdot 10^{-6}m$ (width of the slit)

Solving the equation for $\lambda$, we find the De Broglie wavelength of the electron:

$\lambda = \frac{ya}{L}=\frac{(5.27\cdot 10^{-3} m)(2.00\cdot 10^{-6} m)}{1.068 m}=9.9\cdot 10^{-9}m$