Group velocity is equal to particle velocity

Prove that: Group velocity is equal to Particle Velocity

Solution:

We know that group velocity

$V_{g}=\frac{d\omega}{dk}$

$V_{g}=\frac{d(2\pi\nu )}{d(\frac{2\pi }{\lambda })} \qquad \left(k=\frac{2\pi}{\lambda} \right)$

$V_{g}=\frac{d\nu}{d(\frac{1}{\lambda })}$

$\frac{1}{V_{g}}=\frac{d( \frac{1}{\lambda })}{d\nu}\qquad(1)$

We know that the total energy of the particle is equal to the sum of kinetic energy and potential energy. i.e

$E=K+V$

Where

$K$ – kinetic energy
$V$ – Potential energy

$E=\frac{1}{2} mv^{2}+V$

$E-V=\frac{1}{2}\frac{(mv)^2}{m}$

$E-V=\frac{1}{2m }(mv)^2$

$2m(E-V)=(mv)^2$

$mv=\sqrt{2m(E-V)}$

According to de-Broglie wavelength-

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

$\lambda =\frac{h}{\sqrt{2m(E-V)}}$

$\frac{1}{\lambda} =\frac{\sqrt{2m(E-V)}}{h}\qquad(3)$

Now put the value of $\frac{1}{\lambda }$ in equation$(1)$

$\frac{1}{V_{g}} =\frac{d}{dv}[\frac{{2m(E-V)}^\tfrac{1}{2}}{h}]$

$\frac{1}{V_{g}} =\frac{d}{dv}[\frac{{2m(h\nu -V)}^\tfrac{1}{2}}{h}]$

$\frac{1}{V_{g}} =\frac{1}{2h}[{2m(h\nu -V)}]^{\tfrac{-1}{2}}{2m.h}$

$\frac{1}{V_{g}} =\frac{1}{2h}[{2m(E -V)}]^{\tfrac{-1}{2}}{2m.h} \qquad \left(\because E=h\nu \right) $

$\frac{1}{V_{g}} =\frac{m}{mv}$     {from equation $(2)$}

$V_{g}=V$

Thus, the above equation shows that group velocity is equal to particle velocity.