Modern Physics: Mastering Physics: Diffraction of an Electron Beam

Learning Goal: To understand how to find the wavelength and diffraction patterns of electrons.

An electron beam is incident on a single slit of width

. The electron beam was generated using a potential difference of magnitude

. After passing through the slit, the diffracted electrons are collected on a screen that is a distance

away from the slit. Assume that

is small enough so that the electrons are nonrelativistic. Ultimately, you will find the width of the central maximum for the diffraction pattern.

Part A
In any diffraction problem, the wavelength of the waves is important. To find the wavelength of electrons, you can use the de Broglie relation $\lambda=\frac{h}{p}$ , but you first must find the momentum of one of the electrons. The electrons are accelerated through a potential difference . Use this information to find the momentum of the electrons.

Express your answer in terms of the mass of an electron m_e

, the magnitude of the charge on an electron

, and

ANSWER:

$\sqrt{2m_{e}eV}$

Part B

What is the wavelength lambda

of the electron beam? Use the de Broglie relation and the momentum that you found in Part A.

Express your answer in terms of

, and

ANSWER:

$\frac{h}{\sqrt{2m_{e}eV}}$

Part C

The width of the central maximum is defined as the distance between the two minima closest to the center of the diffraction pattern. Since these are symmetric about the center of the pattern, you need to find only the distance to one of the minima, and then the width of the central maximum will be twice that distance. Find the angle theta

between the center of the diffraction pattern and the first minimum.
The equations for diffraction, which you have seen applied to light, are valid for any wave, including electron waves. Recall that the angle to a diffraction minimum for single-slit diffraction is given by the equation $\sin(\theta)=m\lambda/a$ , where

is the width of the slit and

is an integer. Recall that $m = \pm 1$ for the first minima on either side of the central maximum.
Do not make any approximations at this stage.

Express your answer in terms of

, and

ANSWER:

${\sin}^{-1}\left(\frac{h}{a\sqrt{2m_{e}eV}}\right)$

Part D
What is the width of the central maximum on the screen? Assume that is small enough that you can use the approximation $\sin(\theta) \approx \tan(\theta) \approx \theta$ , where is measured in radians.