Question

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 aaa. The electron beam was generated using a potential difference of magnitude VVV. After passing through the slit, the diffracted electrons are collected on a screen that is a distance LLL away from the slit. Assume that VVV is small enough so that the electrons are nonrelativistic. Ultimately, you will find the width of the central maximum for the diffraction pattern.

Answer 1

The question involves calculating the wavelength and diffraction patterns of electrons in a single slit experiment. The student would use the de Broglie wavelength formula and the principles of single slit diffraction to find the dimensions of the central maximum. It demonstrates the wave-particle duality exhibited by electrons.

Step-by-step explanation:

The student is seeking to understand how to find the wavelength and diffraction patterns of electrons. The context involves an electron beam passing through a single slit, creating a diffraction pattern on a distant screen. From the given potential difference, one could calculate the de Broglie wavelength of the electrons, as their velocity can be determined under the assumption that they are nonrelativistic. The central maximum's width on the diffraction pattern can be deduced using the slit width and the wavelength.

The phenomenon of diffraction and interference highlighted in the question is a demonstration of the wave-like properties of electrons, referred to as wave-particle duality. The experimental setup often includes narrow slits whose sizes are comparable to the wavelength of electrons, resulting in observable wave effects such as constructive and destructive interference.

The angular positions of the minima and maxima in the diffraction pattern are crucial for determining the dimensions of the pattern. The de Broglie wavelength plays a significant role in these calculations, linking the microscopic quantum world to observable macroscopic patterns.

Answer 2

y = L h / a √√ (2q V m)

Step-by-step explanation:

This is a diffraction exercise, so we must use the D'Broglie relation to encode the wavelength of the electron beam.

p = h / λ

λ= h / p

the moment is

p = m v

λ = h / mv

Let's use energy conservation

E = K

q ΔV = ½ m v²

v = √ 2qΔv / m

λ = h / (m √2q ΔV / m)

λ = h / √ (2q ΔV m)

Having the wavelength of the electrons we can use the diffraction ratio

a sin θ = m λ

First minimum occurs for m = 1

sin θ = λ / a

let's use trigonometry for the angles

tan θ = y / x

as in these experiments the angles are very small

tan θ = sin θ / cos θ = sin θ

sin θ = y / x

we substitute

y / x = λ / a

y = x λ / a

we replace the terms

y = L h / a √√ (2q V m)