Wave Nature of Matter

DE-BROGLIE WAVES

In 1925, before the discovery of electron diffraction, de Broglie proposed that the wavelength (l) of waves associated with particles (like Electron, photons) of momentum 'p' is given by

$\lambda =\frac{h}{p}=\frac{h}{mv}$

The wavelength associated with an electron accelerated through a potential difference of V volt is given by

$\frac{1}{2}{m}_e{v}^2=eVorv=\sqrt{\frac{2eV}{m_e}}$

$\therefore \lambda =\frac{h}{m_{e}v}=\frac{h}{\sqrt{2eV{m}_e}}$

Illustration : A particle of mass m is confined to a narrow tube of length L. Find

(a) the wavelengths of the de-Broglie wave which will resonate in the tube,

(b) the corresponding particle momenta, and

(c) the corresponding energies.

Solution:

(a) The de Broglie waves will resonate with a node at each end of the tube.

A few of the possible resonance forms are as follows:

${\lambda }_n=\frac{2L}{n};n=1,2,3,......$

(b) Since de-Broglie wavelengths are

${\lambda }_n=\frac{h}{p_n}$

$\therefore p_n=\frac{h}{{\lambda }_n}=\frac{nh}{2L}n=1,2,3,......$

(c) ${\left(KE\right)}_n=\frac{p_n^2}{2m}=\frac{n^2{h}^2}{8L^{2}m},n=1,2,3,......$