Question

What is the de Broglie wavelength associated with an electron accelerated?

Answer 1

Hint: Electric flux measures the number of electric field lines passing through a point. The electric flux through a surface can be defined as the product of the electric field and the area of the surface through which it passes normally. Gauss’s law is useful in finding the electric charge in a closed surface.

Complete answer:
As we know, matter and energy are related by:
$E = mc^{2}$
E is the energy, m is the mass and c is the speed of light.
Using planck's theory, energy is related with frequency:
$E = h \nu$
We will equate both equations.
$ mc^{2} = h \nu$
But the electron will not move with the speed of light so, we will replace c with v.
$ mv^{2} = h \nu$
Wavelength and frequency is related by:
$\nu = \dfrac{v}{\lambda}$
Therefore,
$ mv^{2} = h \dfrac{v}{\lambda}$
$\implies mv = \dfrac{h}{\lambda}$
$\implies \lambda = \dfrac{h}{mv}$ ……($1$)
And, we know,
$mv = \sqrt{2mqV}$……($2$)
From (1) & (2),
$\lambda = \dfrac{h}{\sqrt{2mqV}}$
By putting all the values for electron,
We get, the De Broglie wavelength connected with an electron accelerated by a potential difference V is provided by the equation:
$\lambda = \dfrac{1.227}{\sqrt{V}} nm$.

Note: De Broglie wavelength is an essential concept while examining quantum mechanics. The wavelength associated with an object concerning its momentum and mass is de Broglie wavelength. A particle’s de Broglie wavelength usually is inversely proportional to its force.