Set 57 Problem #4

Set 57 Problem number 4

Problem

A beam consisting of 60 eV electrons (electron mass 9.11 * 10^-31 kg) is incident on a thin wafer of a crystal with layer spacing 3.7 Angstroms. Surprisingly we find that the electrons, which are particles, are scattered in such a way as to form an interference pattern identical to that of a wave. What will be the distance between central interference maxima at a distance of 13 cm from the wafer?

Solution

A particle with momentum p behaves in many situations as a wave with wavelength

deBroglie wavelength: `lambda = h / p,

where h is Planck's constant. This wavelength is called the deBroglie wavelength after Louis deBroglie, who postulated this model in the 1920's.

To find the momentum of the electron from the given information we first find the KE in Joules:

The KE of the given electron in Joules is 60 eV * (1.6 * 10^-19 J / eV) = .96 * 10^-17 J.

We can find the momentum m v of the electron by multiplying its KE, which is .5 m v^2, by 2 m to obtain m^2 v^2, then taking the square root:

momentum of electron: p = m v = `sqrt( m^2 v^2) = `sqrt ( 2 * m * .5 mv^2) = `sqrt( 2 * m * KE), so
momentum = `sqrt[ (2 * 9.11 * 10^-31 kg) * ( .96 * 10^-17 J ) ] = 4.156922 * 10^-24 kg m/s.

The deBroglie wavelength of the electron is therefore

wavelength: `lambda = h / p = 6.63 * 10^-34 J s / ( 4.156922 * 10^-24 kg m/s ) = 1.59493 * 10^-10 m = 1.59493 * 10^-10 Angstroms.

The layers of the crystal effectively form slits through which the beam of electrons will diffract. Interference maxima will occur with path differences of 0, 1, 2, ... wavelengths. The central maximum will occur along the direction of the original beam, provided the crystal is oriented perpendicular to the beam. The next maximum will occur at an angle for which the path difference between paths from adjacent slits is one wavelength. If a stands for slit separation and `lambda for wavelength, this occurs at angle `theta such that

sin(`theta) = `lambda / a.

Thus the angle at which the maximum occurs is

`theta = sin^-1 ( 1.59493 Angstroms / 3.7 Angstroms) = 25.54794 degrees.

At a distance of 13 cm the separation of this angle from the central beam will be

separation of adjacent maxima = 13 cm * sin( 25.54794 degrees) = 6.210429 cm.

General Solution

An electron with kinetic energy E will have momentum m v = `sqrt( 2 m * (1/2 mv^2) ) = `sqrt( 2 m E ).

Its deBroglie wavelength will therefore be

deBroglie wavelength: `lambda = h / p = h / `sqrt( 2 m E ).

A beam of such electrons incident on a crystal with layer spacing a will exhibit interference maxima at angles `theta such that

sin(`theta) = n `lambda / a, n = 0, 1, 2, ... .

The first-order maximum (n=1) will thus occur at angle

`theta1 = sin^-1( `lambda / a) = sin^-1 ( h / [ a*( `sqrt( 2 m E) ].

We will observe at least the first-order maximum as long as h / [ a `sqrt( 2 m E) )]< 1. If E is too small, we will not observe a maximum.