De – Broglie’s wavelength | E=hf & E=mc2

De – Broglie’s hypothesis mainly proposes the dual nature of light. According to Louis De-Broglie, a light may act as a particle as well as a wave. He observed the relationship of De – Broglie’s wavelength with the momentum. This relation is given by Albert Einstien.

Planck’s in 1901 studied the energy of light of different frequencies radiated from the black body radiations. On the basis of these studies he put forward his theory known as Planck’s Quantum Theory According to this theory: A black body did not emit or absorb energy continuously, but in whole numbers of small packets of energy called a quantum. The energy of each quantum is equal to Planck’s constant (h) and the frequency (v) of radiations.

E = hv

In 1905 Albert Einstein extended the quantum theory by postulating that energy, like light, is not emitted or absorbed in packets but is also propagated in space in packets. Each packet is called Photon. According to this extension, light has wave as well as particle characteristics. The Energy is given by the product of mass in kg and the square of the speed of light.

E = mc²

Planck’s Quantum Theory:

The electromagnetic wave theory of radiations cannot explain the energy distribution in intensity wavelength curves. Planck’s Quantum theory is given by Max Plancks. Max Plancks proposed the quantum theory in 1900 to explain the emission and absorption of radiations. He also proved that electron has a wave nature. The main points of the theory are:

• Energy is not emitted or absorbed continuously. Rather, it is emitted or absorbed in a discontinuous manner and in the form of wave packets called photons. Each wave packet or quantum is associated with a definite amount of energy. Hence, in the case of light, the quantum of energy is often called a photon.
• The amount of energy that is associated with the quantum of radiation is directly proportional to the frequency of radiation. Therefore, the relation becomes:

E ∝ v

E = hv

In this equation, “h” is Planck’s constant and has the value of 6.626×10^-24 Js.

• A body can emit or absorb energy in the form of quanta.

E = hv

v = c / λ

E = hc / λ

Einstein’s extensions & E = mc²:

De – Broglie used the concept of Einstein’s mass-energy equivalence. According to the mass-energy equivalence relation:

“When an object is in a stationary position it possesses certain energy. It does not contain any Kinetic energy (K.E) but possesses Potential energy (P.E) in it. It is most probable that they also contain some of the chemical and thermal energy”.

So, the field of applied mechanics suggested that:

“Sum of all energies present in an object is equal to the product of the mass of object & square of the speed of light (c)”

Einstein extended his idea and postulated that packets or tiny bundles or tiny bundles of energy are an integral part of all Electromegnatic radiations and they could not be subdivided. These indivisible tiny bundles of energy he called photons.

He also gives the formula for calculating the momentum of a photon by relating Planck’s formula by E=pc.

P = h/λ

Deriving Energy-mass equivalance (E=mc²)

Mass-energy equivalence satisfied the meaning of interconversion of mass into energy and vice versa. Albert Einstein described a conceptual relation between mass and energy using the THEORY OF RELATIVITY.

In the equation E = mc² [E represents the energy in any form, m represents the mass in kg, and c is the speed of light having the value 3 × 10⁸ ms^-1].

Consider an object moving at the speed approximate to light (c). So, it possesses some of the force. This force will allow inducing energy and momentum.

So, ENERGY GAINED is given by:

E = Force × Distance by which force act (Velocity)

E = F × c ……………(1)

MOMENTUM is given by:

Momentum = Mass × Velocity

P = m × c ……………(2)

According to newton’s 2nd law, (Change in momentum will lead to force applied). So,

Force = Momentum

F = m × c ………….(3)

Put the value of F in equation (1)

E = m × c × c

E = mc²

Einstein’s mass-energy equivalence has the following Applications:

1. Used to understand the concept of nuclear fission and nuclear fusion reaction.
2. Involve in finding binding energy of atom and nucleus.
3. Used to calculate released energy during a nuclear reaction.
4. Used to calculate the change in mass during a chemical reaction.
5. Used to specify the effect of gravity on stars, moons, planets, etc.
6. This equation is used to estimate the constituents and age of planets.

De – Broglie’s wavelength & Hypothesis:

In 1924, Louis De – Broglie in his thesis for his Ph.D. proposed mathematically the dual nature of light. He uses the concept of Einstein’s mass-energy equivalence and Planck’s quantum theory of radiation. For proving this, Loius De-Broglie is awarded the noble prize.

EINSTEIN’S MASS-ENERGY EQUIVALENCE:

E = mc²………….(a)

PLANCK’S QUANTUM THEORY:

E = hf ……….(b)

Comparing both equations:

mc² = hf

As we know that by wave equation [v = fλ]

So,

f = v/λ

Put it in the equation:

mc² = h (c/λ)

mc² = h (c/λ)

mc = h/ λ

By rearranging it:

λ = h / mc (Formula for De – Broglie’s wavelength)

Numerical Problems Regarding De – Broglie’s wavelength:

How you can determine the wavelength of protons by using De – Broglie’s wavelength formula?

First of all, you have to know the value of Planck’s constant, the speed of light, and the mass of a proton.

Planck’s constant = h = 6.626 × 10^-34 Js

mass of proton = m = 1.6726219 × 10^-27 kilograms

Speed of light = c = 3 × 10⁸ ms^-1

λ = h / mc

λ = 6.626 × 10^-34 / 1.6726219 × 10^-27( 3 × 10⁸ )

 λ = 1.2086 × 10^-11 m

TEST YOURSELF?????????

1. Calculate the De – Broglie wavelength for the electron? (2.42×10−12 m)
2. De – Broglie wavelength is valid for macroscopic bodies. So, calculate the wavelength having the mass of 100 kg and the object is not moving with the speed of light. So, its velocity given to you is 100ms^-1. (6.63×10⁻³⁰m)

Hence, we will drive a conclusion that Light did not has dual nature but it contains number of particles that move in wave like pattern.