Expression for De-Broglie wavelength :

Expression for De-Broglie wavelength :

Expression for de-Broglie wavelength : 

According to quantum theory, the energy of a photon is given by

$E = h\nu \qquad \dots (i)$

According to Einstein's mass-energy equivalence, the energy of the photon is

$E = mc^2 \qquad \dots (ii)$

From equations (i) and (ii), we have,

$mc^2 = h\nu$

$mc^2 = h\frac{c}{\lambda}( \because c = \nu \lambda)$

where, $\lambda$ is de-Broglie wavelength.

$mc = \frac{h}{\lambda}$ 

$ \lambda = \frac{h}{mc} \qquad \dots (iii)$

But $mc = p$, the momentum of photon, therefore,

$\lambda = \frac{h}{p} \qquad \dots (iv)$

If instead of a photon, we have a material particle of mass $m$ moving with velocity $v$, then eqn. (iii) becomes,

$\lambda = \frac{h}{mv} \qquad \dots (v)$

which is the expression for de-Broglie wavelength.

Conclusions. From eqn. (v), it is concluded that,

(i) de-Broglie wavelength is inversely proportional to the velocity of the particle, i.e. $\lambda \propto \frac{1}{v}$. If particle moves faster, the wavelength of the wave associated with it is smaller and vice-versa.

(ii) The particle at rest ($v=0$), has uniformed de-Broglie wavelength. Such a wave cannot be observed.

(iii) de-Broglie wavelength is inversely proportional to the mass of the particle, i.e. $\lambda \propto \frac{1}{m}$. Thus, the wavelength of a matter wave due to a heavier particle is smaller than that due to a lighter particle.

(iv) de-Broglie or matter wave is independent of the charge on the material particle. It means, matter wave or de-Broglie wave will result due to every moving particle (whether charged or uncharged).

(v) de-Broglie waves are not electromagnetic in nature because electromagnetic waves are produced by accelerated charged particles, whereas de-Broglie waves are produced by charged as well as uncharged particles.

(vi) Macroscopic objects like ball, stone, cars etc. do not show wave-like properties. This is because their mass is large and speed is small and the value of '$\lambda$' is extremely small. Hence, wavelength associated with macroscopic objects is extremely small and insignificant.

Note : If the material particle is moving with a velocity comparable to the velocity of light, then the mass of the particle changes according to the relation.

$m = \frac{m_0}{\sqrt{1 - v^2/c^2}}$

So de-Broglie wavelength is given by,

$\lambda= \frac{h}{mv} = \frac{h}{\frac{m_0}{\sqrt{1 - v^2/c^2}} v} = \frac{h\sqrt{1 - v^2/c^2}}{m_0 v}$

Previous Post Next Post