Question

Write down de Broglie’s relation and explain the terms therein.

Answer 1

Hint: Here, we will proceed by stating de Broglie’s relation along with its some features. Then, we will use planck's relation and Einstein’s energy relation to derive the relation between wavelength and momentum which is de Broglie’s relation.

Complete Step-by-Step solution:
De Broglie's equation states that a matter can act as waves such as light and radiation and can also act as waves and particles. The equation further describes the probability of diffracting a beam of electrons much like a beam of light. The de Broglie equation essentially helps us to understand the idea that matter has a wavelength.
The wave properties of matter are defined using this equation. In essence it explains the electron's wave existence. Therefore, if we look at each moving particle, whether it is microscopic or macroscopic, the wavelength will also be granted.
Massless photons as well as massive particles must satisfy one common relationship that link the energy E with the frequency f and the linear momentum p with the wavelength $\lambda $, according to de Broglie 's hypothesis.
Any particle that has energy and momentum is a de Broglie wave of frequency $\nu $ and wavelength $\lambda $.
According to Plank’s quantum theory, the energy of an electromagnetic wave is given by
\[{\text{E}} = {\text{h}}\nu \] where h is Planck’s constant (h = $6.63 \times {10^{ - 34}}$ Joule second) and $\nu $ is frequency
As we know that \[\nu = \dfrac{c}{\lambda }\] where $\nu $ is frequency, c is the speed of light (c = $3 \times {10^8}$ m/s) and \[\lambda \] is wavelength
Using the above relation, we get
\[
   \Rightarrow {\text{E}} = {\text{h}}\left( {\dfrac{c}{\lambda }} \right) \\
   \Rightarrow {\text{E}} = \dfrac{{{\text{h}}c}}{\lambda }{\text{ }} \to {\text{(1)}} \\
 \]
According to Einstein, energy of any particle having mass m is given by
${\text{E}} = {\text{m}}{{\text{c}}^2}{\text{ }} \to {\text{(2)}}$
As the particle exhibits dual nature (i.e., particle type and wave type) and energy being the same according to de Broglie, so by equating both the energy relations for the particle given by equations (1) and (2), we get
\[
   \Rightarrow {\text{m}}{{\text{c}}^2} = \dfrac{{{\text{h}}c}}{\lambda } \\
   \Rightarrow \lambda = \dfrac{{{\text{h}}c}}{{{\text{m}}{{\text{c}}^2}}} \\
   \Rightarrow \lambda = \dfrac{{\text{h}}}{{{\text{mc}}}} \\
 \]
The above relation for the particle of mass m and moving with velocity v is given by
\[ \Rightarrow \lambda = \dfrac{{\text{h}}}{{{\text{mv}}}}\]
Since, momentum (p) of any particle having mass m and velocity v is given by p = mv
Therefore, Wavelength \[\lambda = \dfrac{{\text{h}}}{{{\text{mv}}}} = \dfrac{{\text{h}}}{{\text{p}}} = \dfrac{{{\text{Planck's constant}}}}{{{\text{momentum}}}}\]
The above equation represents de Broglie’s relation.

Note- In the construction of electron microscopes and in the study of the surface structure of solids by electron diffraction, de Broglie 's idea of dual existence of matter finds application. The principle of de-Broglie can be extended not only to electrons but to other small particles such as neutrons , protons, atoms, molecules etc.