Question

Define particle velocity group velocity and phase velocity and also derive the relation between them

Answer 1

Answer:

REFER THE ATTACHMENT GIVEN ABOVE

Answer 2

Explanation:

‘Group velocity’ and ‘phase velocity’ are two 'important concepts' in physics. They both play an important role in fields such as optics, mechanics, sound engineering and quantum mechanics.

Definition of phase velocity and group velocity:

‘Phase velocity’ and ‘group velocity’ are the two velocities of the waves with higher and lower frequencies respectively.

The relation between group and ‘particle velocity’:

Group velocity is

$v_{g}=\frac{d \omega}{d k}$

or

$v_{g}=\frac{d(2 \pi \nu)}{d\left(\frac{2 \pi}{\lambda}\right)}=\frac{d(\nu)}{d\left(\frac{1}{\lambda}\right)}$

Where $\omega=2 \pi \nu$  and

$k=\frac{2 \pi}{\lambda}$

Or

$\frac{1}{v_{g}}=\frac{d\left(\frac{1}{\lambda}\right)}{d \nu}_{\rightarrow(1.12)}$

Where E= Total energy

V= 'Potential energy' of the particle.

We have,

$\frac{1}{2} m v^{2}=E-V$  

Therefore

$v=\left(\frac{2(E-V)}{m}\right)^{\frac{1}{2}}$

By using the de-Broglie formula we have

$\frac{1}{\lambda}=\frac{m v}{h}=\frac{m}{h}\left(\frac{2(E-V)}{m}\right)^{\frac{1}{2}}$

$\frac{1}{\lambda}=\frac{m}{h}\left(\frac{2(h \nu-V)}{m}\right)^{\frac{1}{2}}$

Substituting the value 1/λ  of in equation (1.12), then group velocity is

$\frac{1}{v_{g}}=\frac{d}{d \nu}\left[\frac{m}{h}\left(\frac{2(h \nu-V)}{m}\right)^{1 / 2}\right]$

$\frac{1}{v_{g}}=\frac{2 m}{h}\left(\frac{2(h \nu-V)}{m}\right)^{-1 / 2} \frac{2 h}{m}$  

$\frac{1}{v_{g}}=\left(\frac{2(h \nu-V)}{m}\right)^{-1 / 2}=\frac{1}{v}$  

$\frac{1}{v_{g}}=\frac{1}{v}$