De-Broglie Hypothesis - Wave Particle Duality and Matter Waves

DE‒BROGLIE HYPOTHESIS ‒ WAVE PARTICLE DUALITY AND MATTER WAVES

de‒Broglie concept of Dual Nature

The universe is made of Radiation (light) and matter (particles). The light exhibits the dual nature (i.e.,) it can behave both as a wave (Interference, diffraction phenomenon) and as a particle (Compton effect, photo‒electric effect etc).

Since nature loves symmetry, in 1924 Louis de‒Broglie suggested that an electron (or) any other material particle must exhibit wave like properties in addition to particle nature.

The waves associated with a material particle are called Matter waves.

de‒Broglie Wavelength

From the theory of light, considering a photon as a particle the total energy of the photon is given by E=mc².                 …………(1)

where m→ Mass of the particle

c → Velocity of light

Considering the photon as a wave, the total energy is given by E=hv       ... (2)

where

h→ Planck's constant

v → Frequency of radiation

From equations (1) and (2) we can write E = mc² = hv           ... (3)

We know momentum = mass × velocity

p = mc

∴ Equation (3) becomes hv=pc

 p = hv / c

Since c/v = λ

we can write p= h/λ

 (or) The wavelength of a photon λ = h/p                …………(4)

 de‒Broglie suggested that equation (4) can be applied both for photons and material particles. If m is the mass of the particle and v is the velocity of the particle, then

Momentum p = mv.

de‒Broglie wavelength = λ = h / mv                …………(5)

Other forms of de‒Broglie Wavelength

(i) de‒Broglie wavelength interms of Energy

We know kinetic energy E = ½ mv²

Multiplying by 'm' on both sides we get

E_m =  ½ m²v²

(or) m²v² = 2Em

mv = √(2Em)

∴ de‒Broglie wavelength λ = h / √(2mE)        …………..(6)

(ii) de‒Broglie Wavelength interms of voltage

If a charged particle of charge 'e' is accelerated through a potential difference v.

Then the kinetic energy of the particle= ½ mv²             …………….(7)

Also we know energy = eV           …………….(8)

Equating equations (7) and (8) we get

 ½ mv² = eV

Multiplying by 'm' on both sides we get

m²v² = 2meV

(or) mv = √(2meV)     …………….(9)

Substituting equation (9) in (5), we get

de‒Broglie wavelength = λ = h / √(2meV)          ...(10)

(iii) de‒Broglie wavelength interms of Temperature

When a particle like neutron is in thermal equilibrium at temperature T, then they possess Maxwell distribution of velocities.

∴ Their kinetic energy E_k = ½ mv²_rms             ……………(11)

where v_rms is the Root mean square velocity of the particle.

Also, we know Energy = 3/2 K_BT           ……………(12)

where K_B is the Boltzmann constant.

∴ Equating equations (11) and (12) we get

½ mv² =  3/2 K_BT

 m²v² =  3m K_BT

mv = √(3m K_BT)

de‒Broglie wavelength λ = h / mv =  h / √(3m K_BT)   ………. (13)