JEE Main & Advanced Physics Photo Electric Effect, X- Rays & Matter Waves Matter Waves (de-Broglie 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 \[h=\] Plank's constant, \[m=\] Mass of the particle, \[\upsilon =\] 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, neutron, \[\alpha -\]particle etc. is of the order of \[{{10}^{-10}}m\]. 

(2) de-Broglie wavelength associated with the charged particles : The energy of a charged particle accelerated through potential difference V is \[E=\frac{1}{2}m{{v}^{2}}=qV\]

Hence de-Broglie wavelength \[\lambda =\frac{h}{p}=\frac{h}{\sqrt{2mE}}=\frac{h}{\sqrt{2mqV}}\]

\[{{\lambda }_{Electron}}=\frac{12.27}{\sqrt{V}}\]\[\overset{\text{o}}{\mathop{\text{A}}}\,\],

\[{{\lambda }_{\Pr oton}}=\frac{0.286}{\sqrt{V}}\]\[\overset{\text{o}}{\mathop{\text{A}}}\,\],

\[{{\lambda }_{Deutron}}=\frac{0.202}{\sqrt{V}}\]\[\overset{\text{o}}{\mathop{\text{A}}}\,\],  

\[{{\lambda }_{\alpha -particle}}=\frac{0.101}{\sqrt{V}}\]\[\overset{\text{o}}{\mathop{\text{A}}}\,\] 

(3) de-Broglie wavelength associated with uncharged particles : For Neutron de-Broglie wavelength is given as

\[{{\lambda }_{Neutron}}=\frac{0.286\times {{10}^{-10}}}{\sqrt{E\,(\text{in}\,eV)}}m=\frac{0.286}{\sqrt{E\,(\text{in}\,eV})}{\AA}\]

Energy of thermal neutrons at ordinary temperature

\[\because \] \[E=kT\Rightarrow \lambda =\frac{h}{\sqrt{2mkT}}\]; where \[T=\] Absolute temperature,

\[k=\] Boltzman's constant \[=1.38\times {{10}^{-23}}\]Joule/kelvin,

So, \[{{\lambda }_{Thermal\ neutron}}=\frac{6.62\times {{10}^{-34}}}{\sqrt{2\times 1.67\times {{10}^{-27}}\times 1.38\times {{10}^{-23}}T}}=\frac{30.83}{\sqrt{T}}{\AA}\]

(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 \[\frac{{{\lambda }_{ph}}}{{{\lambda }_{e}}}=\frac{c}{E}\sqrt{2mK}=\sqrt{\frac{2m{{c}^{2}}K}{{{E}^{2}}}}\]