Matter Waves

Matter Waves

According to de-Broglie a moving material particle sometimes acts as a wave and sometimes as a particle. The wave associated with moving particle is called matter wave or de-Broglie wave and it propagates in the form of wave packets with group velocity.

1) de-Broglie wavelength: According to de-Broglie theory, the wavelength of de-Broglie wave is given by, $$\lambda =\frac{h}{p}=\frac{h}{mv}=\frac{h}{\sqrt{2mE}}$$$$\Rightarrow \lambda \propto \frac{1}{p}\propto \frac{1}{v}\propto \frac{1}{\sqrt{E}}$$

Where, $$h=$$ Planck’s constant, $$m=$$ Mass of the particle, $$v=$$ Speed of the particle, $$E=$$ Energy of the particle. The smallest wavelength whose measurement is possible is that of $$\gamma$$ rays. The wavelength of matter waves associated with the microscopic particles like electron, proton and neutron and $$\alpha$$– particle etc. is of the order of$${{10}^{-10}}m$$.

2) De – Broglie Wavelength associated with the charged particle: The energy of a charged particle accelerated through potential difference $$V$$ is$$E=\frac{1}{2}m{{v}^{2}}=qV$$. Hence, de – Broglie wavelength, $$\lambda =\frac{h}{p}=\frac{h}{\sqrt{2mE}}=\frac{h}{\sqrt{2mqV}}$$

$${{\lambda }_{Electron}}=\frac{12.27}{\sqrt{V}}\overset{0}{\mathop{A}}\,$$ , $${{\lambda }_{Proton}}=\frac{0.286}{\sqrt{V}}\overset{0}{\mathop{A}}\,$$  ; $${{\lambda }_{Deuteron}}=\frac{0.202}{\sqrt{V}}\overset{0}{\mathop{A}}\,$$ , $${{\lambda }_{\alpha -Particle}}=\frac{0.101}{\sqrt{V}}\overset{0}{\mathop{A}}\,$$

3) De – Broglie wavelength associated with uncharged particles: For neutron de – Broglie wavelength is given as: $${{\lambda }_{Neutron}}=\frac{0.286\times {{10}^{10}}}{\sqrt{E\left( in\,eV \right)}}m=\frac{0.286}{\sqrt{E\left( in\,eV \right)}}\overset{0}{\mathop{A}}\,$$

Energy of thermal neutrons at ordinary temperature, $$E=kT$$$$\Rightarrow \lambda =\frac{h}{\sqrt{2mkT}}$$

Where, $$T=$$ Absolute temperature, $$k=$$ Boltzmann’s constant $$=1.38\times {{10}^{-23}}Joule/Kelvin$$

So, $${{\lambda }_{Thermal\,\,Neutron}}=\frac{6.626\times {{10}^{-34}}}{\sqrt{2\times 1.67\times {{10}^{-27}}\times 1.38\times 1{{-}^{-23}}T}}=\frac{30.83}{\sqrt{T}}\overset{0}{\mathop{A}}\,$$

4) Ratio of wavelength of photon and electron: The wavelength of a photon of energy $$E$$ is given by, $${{\lambda }_{ph}}=\frac{hc}{E}$$

While the wavelength of an electron of kinetic energy $$K$$ is given by, $${{\lambda }_{c}}=\frac{h}{\sqrt{2mK}}$$.

Therefore, for the same energy the ratio is, $$\frac{{{\lambda }_{ph}}}{{{\lambda }_{e}}}=\frac{c}{E}\sqrt{2mK}=\sqrt{\frac{2m{{c}^{2}}K}{{{E}^{2}}}}$$.