De Broglie Wavelength Formula

De Broglie Wavelength Formula

Light is electromagnetic energy that may operate as both a wave and a particle. De Broglie wavelengths explain the dual existence of light by describing the nature of the wave in respect to the particle. In other words, it relates a wave or particle’s wavelength with momentum. It claims that the wavelength of a particle is inversely related to its mass and velocity. The wavelength of a particle is represented by the symbol λ. The unit of measurement is metres (m), and the dimensional formula is [M⁰L¹T⁰]. Its formula is the Plank’s constant divided by the product of the particle’s mass and velocity.

Formula of De Broglie Wavelength

At the subatomic level, the quantum theory posits matter to be both a wave and a particle. According to the De Broglie equation, any moving particle can operate as both a wave and a particle. The wave associated with moving particles is known as the matter-wave, as well as the De Broglie wave. The De Broglie wavelength is the name given to this wavelength.

For an electron, the De Broglie Wavelength Formula   is:

The De Broglie Wavelength Formula

λ = h/mv

Where,

• λ is the De Broglie wavelength,
• h is Plank’s constant with the value of 6.62 × 10⁻³⁴ Js,
• m is the mass,

• v is the velocity of the particle.

The De Broglie Wavelength Formula is discovered to function and apply to all forms of matter in the cosmos, i.e., everything in the universe, from living beings to inanimate objects, has wave-particle duality.

Derivation For De Broglie Wavelength

The De Broglie wavelength of a particle is calculated using the formulae for its energy. Consider a photon with mass (m), energy (E), wavelength (λ), and velocity equal to the speed of light (c). A photon’s energy (E) can be expressed as-

E = hc/λ ⇢ (1)

Also we know that,

E = mc²  ⇢ (2)

Equating (1) and (2) we get,

hc/λ = mc²

h/λ = mc

λ = h/mc

For a particle with velocity v (less than c) the formula becomes,

λ = h/mv or λ = p

This derives the formula for De Broglie wavelength of a particle.

Solved Examples

Problem 1: Find the wavelength of an electron moving with a velocity of 200 m/s.

Solution:

We have,

m = 9.1 × 10^-31

v = 200

Using the formula we get,

λ = h/mv

= (6.62 × 10⁻³⁴) / (9.1 × 10^-31 × 200)

= 3645.5 nm

Problem 2: Find the wavelength of an electron moving with a velocity of 20 m/s.

Solution:

We have,

m = 9.1 × 10^-31

v = 20

Using the formula we get,

λ = h/mv

= (6.62 × 10⁻³⁴) / (9.1 × 10^-31 × 20)

= 91010 nm