De-Broglie’s Equation:-

According to de-Broglie a moving particle can be associated with a wave. So the waves associated with a moving material particles are known as de-Broglie waves or matter waves.

Expression For de-Broglie wavelength:-

According to quantum theory: the energy of the photon is

Given by              E = hν  ----(1)

According to Einstein’s mass energy equivalence, the energy of the photon is :

E = mc²               -----(2)

From equation (1) & (2)

hν = mc²

h(c/ λ) = mc²

λ = h/mc

If instead of a photon, we have a material particle of mass in moving with velocity v; then

  λ = h/mv      or        λ= h/p                        (p = mv)

• De-Broglie waves is independent of the charge of the material particle i.e. de-Broglie waves are associated with every moving particles.
• De-Broglie waves are not electromagnetic in nature E.M waves are produced only by accelerated charged particles.
• De-Broglie wavelength of an electromagnetic radiation is equal to the wavelength of the photon.

Q. What is the de-Broglie wavelength of a 2 kg object moving with a speed of 1 ms^-1?

Solution :             m = 2kg                v = 1m/sec

                                λ = h/mv = 6.625 x 10^-34 / 2 x 1

                                = 3.312 x 10^-34 m.

De-Broglie wavelength of an electron:

Let us consider a beam of electrons travelling through a potential difference of V volt.

The kinetic energy of an electron is given by

½ mv² = e  V = E

m²v² = 2eVm

mv = (2me)^1/2 V

According to de-Broglie wavelength

λ = (h/mv) = (h/(2me)^1/2V) or              h/(2mE)^1/2

is             h=6.625 x 10^-34 Js           m=9.1 x 10^-31kg              e = 1.6 x 10^-19 C

λ = 6.625 x 10^-34 / (2 x 9.1 x 10^-31 x 1.6 x 10^-19 x V)^1/2

 λ = 12.27 / (V)^1/2 A°

Q. A photon & electron have got same de-Broglie wavelength. Which has greater total energy? Explain.

Solution: De-Broglie wavelength is given by λ = h / (2mE)^1/2

Or           λ² = h² / (2mE)^1/2         or E = h²/2m λ²

Since λ is same for both the particles

E ∝ 1/m

As mass of the electron is less than that of a photon, so energy of electron is greater than that of photon.

Q. Calculate the de-Broglie wavelength of an electron of kinetic energy of 1eV[ans. 12.22 A°] 

Q. The de Broglie wavelength of a beam of electron is 3.67 x 10^-10m. calculate the velocity of electron (ans. 1.99 x 10⁶