Chapter 28, Problem 1CQ

To determine

Check whether the wavelength of the x-rays and the electron is same and also check whether the diffraction pattern will be same if their energies are the same.

Answer to Problem 1CQ

Diffraction pattern depends on the wavelength and since wavelength is same both electron and photon produce the same diffraction pattern.

Same energy photon and electron doesn’t produce same diffraction pattern.

Explanation of Solution

The angle of diffraction is same in both the electron diffraction and the x-ray diffraction experiments.

Write the condition for diffraction of x-rays

$n\lambda =2d\sin \theta$                                                              (I)

Here, $\lambda$ is the wavelength of the x-rays, $d$ is the interatomic distance in the sample, $\theta$ is the angle of diffraction and $n$ is the order of diffraction.

Since $\theta$,  and $d$ are fixed for the sample, the diffraction pattern only depends on $\lambda$.

Thus, the wavelength of the x-rays and electrons is the same given the diffraction maxima is same for both.

Consider the energy of the photon and the electron to be the same.

Write the expression for energy of the photon

$E_p=\frac{hc}{{\lambda }_p}$                                                           (II)

Here, $E_p$ is the energy of the photon, $h$ is the Planck’s constant and $c$ is the speed of light.

Write the relativistic expression for energy of the electron

$E^2=p^2{c}^2+{(E_0)}^2$                                                       (III)

Here, $p$ is the momentum of the electron, $E$ is the total energy of the electron and $E_0$ is its rest energy.

Rearranging (III) for $p$

$p=\frac{\sqrt{E^2-E_0^2}}{c}$                                                      (IV)

Write the expression for de Broglie wavelength of the electron

${\lambda }_e=\frac{hc}{\sqrt{E^2-E_0^2}}$                                                          (V)

Considering $E=E_p$ and substitute (II) in (V)

${\lambda }_e=\frac{hc}{\sqrt{{\left(\frac{hc}{{\lambda }_p}\right)}^2-E_0^2}}$                                                          (VI)

From (VI),

${\lambda }_e\ne {\lambda }_p$.

Since, the wavelengths are not the same they cannot produce the same diffraction pattern.