De Broglie concept of matter waves: dual
nature of matter
MATTER WAVES : DE-BROGLIE
CONCEPT
In 1924, Lewis de-Broglie proposed that matter has dual
characteristic just like radiation. His concept about the dual nature of matter
was based on the following observations:-
(a) The whole universe is composed of matter
and electromagnetic radiations. Since both are forms of energy so can be
transformed into each other.
(b) The matter loves symmetry. As the
radiation has dual nature, matter should also possess dual character.
According to the de Broglie concept of matter waves, the matter
has dual nature. It means when the matter is moving it shows the wave
properties (like interference, diffraction etc.) are associated with it and
when it is in the state of rest then it shows particle properties. Thus the
matter has dual nature. The waves associated with moving particles are matter
waves or de-Broglie waves.
WAVELENGTH
OF DE-BROGLIE WAVES
Consider
a photon whose energy is given by
E=hυ=hc/λ
– – (1)
If
a photon possesses mass (rest mass is zero), then according to the theory of
relatively ,its energy is given by
E=mc2 – – (2)
From
(1) and (2) ,we have
Mass
of photon m= h/cλ
Therefore
Momentum of photon
P=mc=hc/cλ=h/λ
– – (3)
Or
λ = h/p
If
instead of a photon, we consider a material particle of mass m moving
with velocity v,then the momentum of the particle ,p=mv. Therefore, the
wavelength of the wave associated with this moving particle is given by:
h/mv
–
Or
λ = h/p (But here
p = mv) (4)
This wavelength is called DE-Broglie
wavelength.
Special Cases:
1. dE-Broglie wavelength for material
particle:
If
E is the kinetic energy of the material particle of mass m moving with velocity
v,then
E=1/2
mv2=1/2 m2v2=p2/2m
Or
p=√2mE
Therefore
the by putting above equation in equation (4), we get de-Broglie wavelength
equation for material particle as:
λ
= h/√2mE – – (5)
2. dE-Broglie wavelength for particle in
gaseous state:
According
to kinetic theory of gases , the average kinetic energy of the material
particle is given by
E=(3/2)
kT
Where
k=1.38 x 10-23 J/K is the Boltzmann’s constant and T is the absolute
temperature of the particle.
Also
E = p2/2m
Comparing
above two equations, we get:
p2/2m =
(3/2) kT
or
p = /√3mKT
Therefore
Equation (4) becomes
λ=h/√3mKT
This is the dE-Broglie wavelength for particle
in gaseous state:
3. dE-Broglie wavelength for an accelerated
electron:
Suppose
an electron accelerates through a potential difference of V volt. The work done
by electric field on the electron appears as the gain in its kinetic energy
That
is E = eV
Also
E = p2/2m
Where
e is the charge on the electron, m is the mass of electron and v is the
velocity of electron, then
Comparing
above two equations, we get:
eV=
p2/2m
or
p = √2meV
Thus
by putting this equation in equation (4), we get the the de-Broglie wavelength
of the electron as
λ
= h/√2meV 6.63 x 10-34/√2 x 9.1
x 10-31 x1.6 x 10-19 V
λ=12.27/√V
Å
This is the de-Broglie wavelength for electron moving in a potential
difference of V volt.
Comments
Post a Comment