**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=mc^{2} – – (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 mv^{2}=1/2 m^{2}v^{2}=p^{2}/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 = p^{2}/2m

Comparing above two equations, we get:

p^{2}/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 = p^{2}/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= p^{2}/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.

**Brain Teaser:**

If you have understood the article, then please answer the following question:

1. If an electron is accelerated under the potential of 100 volts, then what should be the wavelength of electron?

a) 12.27 Å

b) 1.27 Å

c) 10 Å

d) 100 Å

Please submit your answer in comments section.

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Ans= 1.27 Armstrong

I enjoy while reading thanks.

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1.27 Å

Ans= 1.27 Å

Ans= 1.27 Å

Thanks