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General
MODULE 1. Magnetism
MODULE 2. Particle Physics
MODULE 3. Modern Physics
MODULE 4. Semicoductor Physics
MODULE 5. Superconductivty
MODULE 6. Optics
LESSON 6. Theory of Wave and particle
DeBroglie wavelength
The expression of the DeBroglie wave associated with a material particle can be derived on comparison of radiation.
Suppose, according to Plank theory of radiation the energy of photon (quantum) is
E = hv
But ν = \[{c \over \lambda }\] c  velocity of light
E = \[{{hc} \over \lambda }\]…… (1) \[\lambda\]  wavelength of light
But according to energy and mass relation E = mc^{2}…… (2)
Equating (1) and (2)
mc^{2} = \[{{hc} \over \lambda }\]
Hence \[\lambda\] = \[{h \over {mc}}\]……. (3)
We know that c velocity of light (photon) so
mc = P ……. (4) Associated with photon (radiation)
Now suppose a material particle of mass –m and material particle is moving with velocity ν
Hence, the momentum of particle P = mν
The wavelength (\[\lambda\]) is associated with particle (material particle)
\[\lambda\] = \[{h \over {mc}}\] = \[{h \over P}\]…….(5)
Let the kinetic energy of moving material particle is
E = \[{{m{v^2}} \over 2}\]
E = \[{{m{v^2}} \over 2}\] × \[{m \over m}\]
E = \[{{{m^2}{v^2}} \over {2m}}\]
E = \[{{{P^2}} \over {2m}}\] where, P = mν
E = \[{{m{v^2}} \over 2}\] = \[{{{m^2}{v^2}} \over {2m}}\] = \[{{{P^2}} \over {2m}}\]
P = \[\sqrt {2mE}\]…….(6)
DeBroglie wavelength
\[\lambda\] = \[{h \over {\sqrt {2mE} }}\]……(7)
DeBroglie wavelength associated with electron
Suppose
Electron is at rest
It has having mass m_{0}
Charge of electron e
Now electron associated by potential voltage and obtained the velocity ν from its rest position.
KE = \[{{{m_0}{v^2}} \over 2}\] = \[eV\]
ν^{2} = \[{{2eV} \over {{m_0}}}\]
ν = \[\sqrt {{{2eV} \over {{m_0}}}\]
DeBroglie wavelength is associated with electron
\[\lambda\] = \[{h \over {{m_0}v}}\]
\[\lambda\] = \[{h \over {\sqrt {2{m_0}eV} }}\]
h  Plank’s constant 6.625 × 10^{34} Joule.sec
m_{0 } Rest mass of electron 9.11 × 10^{31} kg
E  Charge of electron 1.62 × 10^{19}C
V  100 Volts
\[\lambda\] = 1.226 Å ……(8)
The wavelength is associated with an electron accelerated to 100 volts is
1.226 Å
Properties of matter wave

Lighter in the particle, greater is the wavelength associated with it

\[\lambda\] = \[{h \over {mv}}\] = \[{h \over P}\]

Smaller is the velocity of the particle, greater is the wavelength associated with it.

Suppose,ν = 0, then \[\lambda\] = ∞ ; wave becomes indeterminate

Suppose ν = ∞ then \[\lambda\] = 0; wave becomes a zero. These conditions suggest, the matter waves are generated by the motion of particles.

These waves ( \[\lambda\] = ∞ and \[\lambda\] = 0) are produced whether the particles are charged particles or the particles are unchanged.

These waves are not electromagnetic waves but they are a new kind of waves (electromagnetic waves are produced only by motion of charged particle). The velocity of material particle is not a constant while the velocity of electromagnetic wave is constant.

The velocity of matter waves is greater than the velocity of light. This can be provide as under
The wave velocity ω is ν × \[\lambda\]
So, ω = \[{{{c^2}} \over v}\]
As a particle velocity ν can’t exceed c hence ω is greater than velocity of light.
 The wave and particle aspect of moving bodies can never appear together in the same experiment. What we can say is that waves have particle like properties and particle have wave like properties and the concepts are inseparably linked. Matter wave representation is only a symbolic representation.
 The wave nature of matter introduced an uncertainty in the location of the position of the particle because a wave can’t be said exactly at this point or exactly at that point. However, where the wave is large(strong) there is a good chance of finding the particle while, where the wave is small there is very small(weak) chance of finding the particle