Spread the love

de Broglie Equation

de Broglie Equation :- Because radiation has a dual nature, that is, radiation exhibits both wave and particle properties, and because the universe is made up of radiation and matter, de-Broglie concluded that a moving particle must also have a dual nature because nature seeks symmetry.

De Broglie’s hypothesis(de Broglie Equation) :-

According to de Broglie hypothesis, a moving material particle sometimes behaves like a wave and sometimes as a particle or in other words a wave is associated with a moving material particle, this wave is called a matter wave or de Broglie wave, the wavelength of which is called the de Broglie wavelength and it is given by the following formula…

$\displaystyle \lambda =\frac{h}{mv}$

where m and v are the mass and velocity of the particle and h is the Planck’s constant.

De Broglie wavelength formula

According to Planck’s quantum theory, the energy of a photon is given by the following formula

E = hν …..(1)

Where h is Planck’s constant and ν is the frequency of light.

If the mass of a particle is m and its velocity is c, then its energy from Einstein’s energy-mass relation is given by the following formula

E = mc² …..(2)

From equation (1) and (2)

hν = mc²

⇒ hν/c = mc …..(3)

Since each photon moves with velocity c, the momentum of the photon, p = mc

So from equation (3)

p = (hν/c) = (hν/νλ) = (h/λ)

λ = h/p …..(4)

De Broglie assumed that equation (4) is applicable for both photons and matter particles.

From equation (4)

λ ∝ 1/p

de Broglie wavelength is inversely proportional to the momentum, which is why the wavelength of small particles like electrons appears but not of a moving ball or car, because they have more momentum.

λ = h/p = h/mv

Note :- De Broglie waves are not mechanical waves nor electromagnetic waves (because electromagnetic waves are produced by accelerating electric charges).

Relationship between de Broglie wavelength and temperature

(de Broglie Equation)

According to kinetic theory, the average kinetic energy of a particle at T Kelvin temperature is given by the following formula

$\displaystyle K=\frac{3}{2}{{k}_{B}}T$

where k_B is Boltzmann constant.

If a particle of mass m is moving with a velocity v, then its kinetic energy…

$\displaystyle K=\frac{1}{2}m{{v}^{2}}$

Momentum of particle, $\displaystyle p=mv=\sqrt{2mK}=\sqrt{2m\times \frac{3}{2}{{k}_{B}}T}=\sqrt{3m{{k}_{B}}T}$

Hence the de Broglie wavelength…

$\displaystyle \lambda =\frac{h}{p}=\frac{h}{\sqrt{3m{{k}_{B}}T}}$

Example 1.

We may state that the energy E of a photon of frequency v is E = hv where h is a Planck’s constant. The momentum p of a photon is p = h/λ where λ is the wavelength of the photon. From the above statement one may conclude that the wave velocity of light is equal to :-

(A) $\displaystyle 3\times {{10}^{8}}m{{s}^{-1}}$

(B) $\displaystyle \frac{E}{p}$

(D) $\displaystyle {{\left( \frac{E}{p} \right)}^{2}}$

Solution.

Given that E = hv and p = h/λ

$\displaystyle \Rightarrow \frac{E}{p}=\frac{h\nu }{h/\lambda }=\nu \lambda =c$

Hence (B) is correct option.

Example 2.

If the momentum of an electron is changed by P, then the de-Broglie wavelength associated with it changes by 0.5 %. The initial momentum of the electron will be :

(A) 400P

(B) P/200

(D) 200P

Solution :-

Let initial de – Broglie wavelength of electron is given by

$\displaystyle {{\lambda }_{i}}=\frac{h}{{{p}_{i}}}$ …..(1)

Let momentum is increased by P, then new de – Broglie wavelength(decreased) of electron,

$\displaystyle {{\lambda }_{f}}=\frac{h}{{{p}_{i}}+P}$ …..(2)

Equation (1) – (2) gives

$\displaystyle {{\lambda }_{i}}-{{\lambda }_{f}}=\frac{h}{{{p}_{i}}}-\frac{h}{{{p}_{i}}+P}=\frac{hP}{({{p}_{i}}+P){{p}_{i}}}$ …..(3)

Dividing (3) by (1) we get,

$\displaystyle \frac{{{\lambda }_{i}}-{{\lambda }_{f}}}{{{\lambda }_{i}}}=\frac{hP}{({{p}_{i}}+P){{p}_{i}}}\times \frac{{{p}_{i}}}{h}=\frac{P}{{{p}_{i}}+P}$