Question

Derive an expression for de-Brogile wavelength of matter waves.

Answer 1

de Broglie's wavelength of matter waves: de Broglie equated the energy equations of Planck(wave) and Einstein(particle).
For a wave of frequency $\nu$, the energy associated with each photon is given by Planck's relation,

$E=h\nu$ ...............$\left(1\right)$
Where $h$ is Planck's constant.
According to Einstein's mass energy relation, a mass m is equivalent to energy,
$E=m{c}^2$ ............$\left(2\right)$
where $c$ is the velocity of light.
If, $h\nu =m{c}^2$
$\therefore \frac{hc}{\lambda }=m{c}^2$ (or) $\lambda =\frac{h}{mc}$ ...........$\left(3\right)$
(since $\nu =\frac{c}{\lambda }$)
For a particle moving with a velocity $v$, if $c=v$ from equation $\left(3\right)$
$\lambda =\frac{h}{mv}=\frac{h}{p}$
Where $p=mv$, the momentum of the particle. These hypothetical matter waves will have appreciable wavelength only for very light particles.