Question

Write the expression for the de-Broglie wavelength associated with a charged particle having charge and mass when it is accelerated by a potential \[V\] .

Answer 1

Hint: In order to solve this question study the de Broglie hypothesis. The relationship between Kinetic Energy, potential difference and charge of the particle and also the relationship of kinetic energy with momentum is important.

Complete step by step answer:
Consider a charged particle carrying charge $q$ of mass $m$ accelerated from the rest through a potential $V$ .
The kinetic energy, $K$ of the particle equals the work done on it by the electric field. So,
$K = qV$
Here, $qV$ is the work done on the charged particle by the electric field.
Also, the kinetic energy of the charged particle is given by
$K = \dfrac{1}{2}m{v^2}$
Here, $m$ is the mass of the particle
$v$ is the velocity of the particle by which it is moving under the effect of an electric field.
Now, momentum of the particle is given by $p = mv$
Putting this in the expression of Kinetic Energy we get,

$K = \dfrac{1}{2}m{v^2}$
In terms of momentum it can be written as
$ \Rightarrow K = \dfrac{{{p^2}}}{{2m}}$
From here we get the expression for momentum as
$p = \sqrt {2mK} $
Putting the expression for kinetic energy it can be written as,
$ \Rightarrow p = \sqrt {2mqV} $
The de Broglie wavelength $\lambda $ of the charged particle is then
$\lambda = \dfrac{h}{p}$
Putting the expression for momentum we get
$ \Rightarrow \lambda = \dfrac{h}{{\sqrt {2mK} }}$
Putting the expression for Kinetic Energy we get
$ \Rightarrow \lambda = \dfrac{h}{{\sqrt {2mqV} }}$

Therefore, $\lambda = \dfrac{h}{{\sqrt {2mqV} }}$ is the expression for the de-Broglie wavelength associated with a charged particle having charge and mass when it is accelerated by a potential \[V\] .

Note: If the charged particle is considered as the electron then the de Broglie wavelength for electron would be given as $\lambda = \dfrac{h}{{\sqrt {2meV} }}$ .
After putting the respective values of $h$ , $m$ ,and $e$ we get,
$\lambda = \dfrac{{1.227}}{{\sqrt V }}nm$
This wavelength is of the same order as the spacing between the atomic planes in crystals. This suggests that matter waves associated with an electron could be verified by crystal diffraction experiments analogous to X-Ray diffraction.