De Broglie Hypothesis

Till Neil Bohr’s model, we have considered the wave nature of light. However, when Max Planck introduced Quantum theory, he said that light is made up of tiny packets of energy, and these packets of energy are quanta or photons.

Therefore, light has a particle nature, and each packet emits a fixed or quantized energy. So, the energy of a photon is:

E = hf = h * c/λ

Where,

h = Planck’s constant

f = frequency of a given wave

λ = Wavelength

This means light has wave and particle duality.

Inspired by his theory, one of the students of Neil Bohr named Louis de Broglie gave his hypothesis that electron has a very small mass of 9.1 x 10\[^{-31}\]kg, and a speed of 218 x 10\[^{6}\] m/s

This means if an electron moves with this speed, it appears like a wave. Therefore, an electron has wave-particle duality.

This hypothesis was proven true by his student, Davisson-Germer in his diffraction experiment on an electron.

De Broglie Equation

According to de Broglie, every moving particle sometimes acts as a wave and sometimes as a particle and vice versa. The wave associated with moving particles is the matter-wave or de Broglie wave whose wavelength is called the de Broglie wavelength.

For an electron, de Broglie wavelength equation is:

λ = \[\frac{h}{mv}\]

Where

λ = Wavelength of the electron

m and v = mass, and velocity of an electron, respectively.

mv = momentum

This equation turned out to be true for all the particles.

This means that all the moving particles in the universe, be it a car, a human being, or anything, have wave-particle duality.

So,

λ = \[\frac{h}{mv}\]

Significance of De Broglie Equation

de Broglie says that all the objects that are in motion have a particle nature. However, if we look at a moving ball or a moving car, they don’t seem to have particle nature.

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To make this clear, de Broglie derived the wavelengths of electron and a cricket ball. Now, let’s understand how he did this.

De Broglie Wavelength

1. De Broglie Wavelength for a Cricket Ball

Let’s say,

Mass of the ball = 150 g (150 x 10⁻³ kg),

Velocity = 35 m/s, and

h = 6.626 x 10⁻³⁴ Js

Now, putting these values in the equation λ = \[\frac{h}{mv}\]

λ = \[\frac{6.626 \times 10^{-34}}{150 \times 10^{-3} \times 35}\]

On solving, we get,

λBALL = 1.2621 x 10\[^{-34}\]m = 1.2621 x 10\[^{-24}\]Å

We know that Å is a very small unit, and therefore the value is in the power of 10\[^{-24}\], which is a very small value. From here, we see that the moving cricket ball is a particle.

Now, the question arises if this ball has a wave nature or not. Your answer will be a big no because the value of λBALL is immeasurable.

This proves that de Broglie’s theory of wave-particle duality is valid for the moving objects ‘up to’ the size (not equal to the size) of the electrons.

Let’s understand how this theory applies to electrons and find the de Broglie wavelength of an electron.

1. De Broglie Wavelength for an Electron

We know that me = 9.1 x 10\[^{-31}\]kg, and ve = 218 x 10\[^{6}\]m/s

Now, putting these values in the equation λ = \[\frac{h}{mv}\]

= \[\frac{9.1 \times 10^{-31}}{218 \times 10^{6}}\]

λ = 3.2 Å

This value is measurable. Therefore, we can say that electrons have wave-particle duality.

Thus all the big objects have a wave nature and microscopic objects like electrons have wave-particle nature.

The conclusion of de Broglie hypothesis

From de Broglie equation for a material particle, i.e., λ = \[\frac{h}{p}\] = \[\frac{h}{mv}\] , we conclude the following:

i. If v = 0, then λ = ∞, and

If v = ∞, then λ = 0

It means that waves are associated with the moving material particles only. This implies these waves are independent of their charge.

De Broglie Wavelength of an Electron Derivation

Let’s say an electron accelerated from rest position through a voltage difference of V volts. Then,

Gain in KE = \[\frac{1}{2}\] mv\[^{2}\]

Work done on the electron = eV

So, \[\frac{1}{2}\] mv\[^{2}\] = eV or v = \[\sqrt{\frac{2eV}{m}}\]

We know that λ = \[\frac{h}{mv}\] = \[\frac{h}{\sqrt{2meV}}\]

On substituting the values:

h = 6.626 x 10\[^{-34}\] Js

m = 9.1 x 10\[^{-31}\]kg

e = 1.6 x 10\[^{-19}\] C

= \[\frac{6.626 \times 10^{-34}}{\sqrt{2 \times 9 \times 10^{-31} \times 1.6 \times 10^{-19} \times V}}\]

On solving, we get

λ = \[\frac{12.27}{\sqrt{V}}\]Å

FAQ (Frequently Asked Questions)

Q1: Derive the Relation Between De Broglie Wavelength and Temperature.

Ans: We know that the average KE of a particle is:

K = 3/2 k_{b}T

Where k_{b} is Boltzmann’s constant, and

T = temperature in Kelvin

The kinetic energy of a particle is ½ mv²

The momentum of a particle, p = mv = √2mK

= √2m(3/2)KbT = √2mKbT

de Broglie wavelength, λ = h/p = h√2mkbT

Q2: If the KE of an Electron Increases By 21%, Find the Percentage Change in its De Broglie Wavelength.

Ans: We know that λ = h/√2mK

So, λ_{i} = h/√(2m x 100) , and λ_{f} = h/√(2m x 121)

% change in λ is:

Change in wavelength/Original x 100 = (λ_{fi} - λ_{f})/λ_{i} = ((h/√2m)(1/10 - 1/21))/(h/√2m)(1/10)

On solving, we get

% change in λ = 5.238 %

Q3: Find λ Associated with an H_{2} of Mass 3 a.m.u Moving with a Velocity of 4 km/s.

Ans: Here, v = 4 x 10³ m/s

Mass of hydrogen = 3 a.m.u = 3 x 1.67 x 10⁻²⁷kg = 5 x 10⁻²⁷kg

On putting these values in the equation λ = h/mv we get

λ = (6.626 x 10⁻³⁴)/(4 x 10³ x 5 x 10⁻²⁷) = 3 x 10⁻¹¹ m

Q4: If an Electron Behaves Like a Wave, What Should Determine its Wavelength and Frequency?

Ans: Momentum and energy determine the wavelength and frequency of an electron.