### de-Broglie Concept of Matter wave

Louis de-Broglie thought that similar to the dual nature of light, material particles must also possess the dual character of particle and wave. This means that material particles sometimes behave as particle nature and sometimes behave like a wave nature.

According to de-Broglie –
A moving particle is always associated with a wave, called as de-Broglie matter-wave, whose wavelengths depend upon the mass of the particle and its velocity.

According to Planck’s theory of radiation–

$E=h\nu \qquad(1)$

Where
h – Planck’s constant
$\nu$ - frequency

According to Einstein’s mass-energy relation –

$E=mc^ {2} \qquad (2)$

According to de Broglie's hypothesis equation $(1)$ and equation $(2)$ can be written as –

$mc^ {2} = h \nu$

$mc^ {2} = \frac{hc}{\lambda }$

$\lambda =\frac{h}{mc}\qquad(3)$

$\lambda =\frac{h}{P}$

Where $P$ –Momentum of Photon

Similarly from equation $(3)$ the expression for matter waves can be written as

 $\lambda=\frac{h}{mv}=\frac{h}{P}\qquad(4)$

Here $P$ is the momentum of the moving particle.

1.) de-Broglie Wavelength in terms of Kinetic Energy

$K=\frac{1}{2} mv ^{2}$

$K=\frac{m^{2}v^{2}}{2m}$

$K=\frac{P^{2}}{2m}$

$P=\sqrt{2mK}$

Now substitute the value of $P$ in equation $(4)$ so

 $\lambda =\frac{h}{\sqrt{2mK}} \qquad (5)$

2.) de-Broglie Wavelength for a Charged particle

The kinetic energy of a charged particle is $K = qv$

Now substitute the value of $K$ in equation$(5)$ so

 $\lambda =\frac{h}{\sqrt{2mqv}}$

3.) de-Broglie Wavelength for an Electron

The kinetic energy of an electron

$K=ev$

If the relativistic variation of mass with a velocity of the electron is ignored then $m=m_{0}$ wavelength

 $\lambda =\frac{h}{\sqrt{2m_{0}ev}}$
So wavelength of de-Broglie wave associated with the electron in non-relativistic cases

4.) de-Broglie wavelength for a particle in Thermal Equilibrium