**De Broglie’s Equation**

In 1924 Louis de Broglie theorized that only light possesses both wave and particle properties, but rather particles with mass, such as electrons do as well. The wavelength of these material waves also known as the de Broglie wavelength.

**What is De Broglie’s Equation?**

The De Broglie equation is an equation used to describe the wave properties of matter, specifically, the wave nature of the electron. De Broglie suggested that particles can exhibit properties of waves.

λ = h/ mv.

Where,

λ = Wavelength,

h = Planck’s Constant,

m = Mass of the Particle,

V = Velocity of moving particle.

**Derivation of De – Broglie Equation: **The wavelength of the wave associated with any material was calculated by analogy with photon.

In case of photon, if it is assumed to have wave character, its energy is given by:

E = hυ … (1)

If the photon is supposed to have particle character, its energy is given by:

E = mc² … (2)

Now, by equating (1) and (2), we get:

hv = mc²

\(h\left( \frac{c}{\lambda } \right)\,\,=\,\,m{{c}^{2}}\)\(\left( \because \,v\,=\,\frac{c}{\lambda } \right)\)

\(\Rightarrow \,\,\,\,\lambda \,\,=\,\,\frac{h}{mc}.\)

The above equation is applicable to material particle. If mass and velocity of photon is replaced by the mass and velocity of material particle. Thus, for any material particle like electron.

\(\lambda \,\,=\,\,\frac{h}{mv}\),

Where,

mv = p Where, p is the momentum of the particle.

**How to find the De Broglie’s Equation?**

**Problem: **What is the wavelength of an electron moving at 5.31 x 10⁶ m/sec?

**Solution: **Given,

Mass of an electron (m) = 9.1 x 10⁻³¹ kg

Planck’s constant (h) = 6.626 x 10⁻³⁴ J sec

We know that:

De Broglie’s Equation:

\(\lambda \,\,=\,\,\frac{h}{mv}\),

\(\lambda \,\,=\,\,\frac{6.626\times {{10}^{-34}}J.\sec }{9.1\times {{10}^{-31}}kg\times 5.31\times {{10}^{6}}m/\sec }\,\,=\,\,1.37\times {{10}^{-10}}m\).

Wavelength (λ) = 1.37 x 10⁻⁸ m = 1.37°A.

∴ Wavelength (λ) = 1.37°A.