Question

Show that wavelength of electromagnetic radiation is equal to the de-Broglie wavelength of its quantum (photon).

Answer 1

Hint: Electromagnetic radiations have both electric and magnetic fields and they travel at the speed of light through space in the form of waves. The electromagnetic radiation does not have mass or charge but it travels in the form of photons or quanta. The photons or quanta refers to the packets of energy. The electromagnetic radiations are natural and man-made. The examples of electromagnetic radiations include radio waves, microwave, infrared light, gamma rays, x rays and ultraviolet radiations.


Complete step by step solution: The distance between two successive crests or troughs of a wave is known as a wavelength. The de- Broglie wavelength is the wavelength associated with an object, in relation with the mass and momentum of that object.
A photon is a quantum of electromagnetic radiation and it has no mass. But, it has momentum and it is carried in the direction of their motion. Photon momentum is very small.
The Photon momentum is given by Compton Effect, which is based on the scattering of photons by an electron. According to the Compton Effect, the energy and momentum are conserved hence, the energy and momentum of a scattered photon decreases. The Compton Effect states that the photons carry momentum.
The momentum of an object is the product of the mass of that object and the velocity of that object.
The formula of momentum is,
$P = mv$……………………(1)
Here, P represents the momentum of an object in Kg.m/s. The mass of the object will be ‘m’ and ‘v’ is the velocity of that object.
Based on Compton Effect, the momentum of photon is given by:
$P = \dfrac{h}{\lambda }$…………………….(2)
Here, $\lambda $ is the wavelength of the electromagnetic radiation and is expressed in metres. The letter ‘h’ is Planck’s constant, equals $6.63 \times {10^{ - 34}}J.s$.
The photon have energy equals to $(h\nu )$ and hence, the momentum of that photon can be given as:
$P = \dfrac{{h\nu }}{c} = \dfrac{h}{\lambda }$ $\left( {\because \lambda = \dfrac{c}{\nu }} \right)$
$\therefore \lambda = \dfrac{h}{P}$ ……………………………..(3)
The idea of de-Broglie wavelength originated from de-Broglie hypothesis which was proposed in 1924-25 by French scientists, Louis de Broglie. He stated that electrons and other particles of matter possess wave-like characteristics and can behave like waves.
The de-Broglie wavelength of the photon is expressed as
$\lambda = \dfrac{h}{{mv}}$………………………(4)
Here, $\lambda $ is the wavelength of the electromagnetic radiation carrying photons and m is the mass of the photon and v is its velocity.
As shown in equation (1), $P = mv$
On equating equations (1) and (4), we get
$\lambda = \dfrac{h}{P}$……………………….(5)
From equations (2) and (5), we can conclude that the wavelength of the electromagnetic radiation equals that of the de-Broglie wavelength of the photon.

Note: There is an alternate way to prove that the wavelength of electromagnetic radiation is equal to the de-Broglie wavelength of its quantum.
The formula for wavelength of an electromagnetic radiation is given by,
$\lambda = \dfrac{c}{v}$…………………..(1)
Here, $\lambda$ is the wavelength of an electromagnetic radiation in metres, c is the velocity of light equals to that of $3 \times {10^8}m/s$ and v is the frequency of the wave in hertz.
The momentum of photon with energy $h\nu $equals to,
$P = \dfrac{{h\nu }}{c}$………………….(2)
Here, h is Planck’s constant.
The de-Broglie wavelength of photon is given as,
$\lambda = \dfrac{h}{{mv}}$ ………………(3)
Here, $\lambda $ is the wavelength of an electromagnetic radiation in metres, h is the Planck’s constant, m is the mass of a photon and v is the velocity of that photon.
As $P = mv$, the equation (3) can be written as,
$\lambda = \dfrac{h}{P}$……………….(4)
Now, substitute the value of momentum of photon is de-Broglie equation numbered as (3), from equation (2), we get,
$\lambda = \dfrac{h}{{\dfrac{{hv}}{c}}}$
$\Rightarrow \lambda = \dfrac{c}{v}$…………….(5)
From equations (1) and (5), we can infer that the wavelength of the electromagnetic radiation equals that of the de-Broglie wavelength of the photon.