Class 12 Physics Dual Nature Radiation Matter Wave Nature of Matter

## Wave Nature of Matter: De Broglie’s Hypothesis:

• De Broglie proposed that if the radiations could possess dual nature, matters could also possess dual nature.
• A particle of mass (m), moving with velocity (v) could behave like a wave under suitable conditions. And the corresponding wave related to that matter is called matter wave
• De Broglie’s wavelength for matter wave is given by: λ = h/p

Case-1- (Macroscopic object)

If we take an example of a car of mass= 900kg, moving with the velocity of 36km/hr (10m/s).

The wavelength associated with the car will be:

We can observe that the wavelength associated with the car is insignificant and can’t be detected experimentally.

Hence, for the macroscopic objects, the mass is so large that the matter wave associated with them becomes insignificant and negligible

Case-2- (Microscopic object)

If we take an electron, mass=9.1 X 10-31 kg, moving with the speed of light (3 X 108 m/s).

The wavelength associated with an electron will be:

Kinetic energy K = (1/2)mv2 = (mv)2/(2m) = p2/2m

Using De-Broglie’s Hypothesis:

Putting the value of V = 50V

The wavelength associated with electron is quite large and is experimentally observable. This is because the mass of a microscopic object is very small, so wavelength becomes sufficiently large andhence, observable.

Numerical Problems:

Question: Monochromatic light of wavelength 632.8nm is generated by a helium-neon laser having power of 9.42mW. Evaluate the following: a) energy and momentum of each photon, b) The number of photons emitted per second, and c) speed of a hydrogen atom to have momentum equal to that of an emitted photon by the laser.

Solution:

Given, λ = 632×10-9nm, P = 9.42×10-3W

1. Energy of a photon is given by: E = hv = hc/λ

Momentum of the photon is given by: p = hv/c = h/λ

b. Number of photons emitted per second (n) will be given by the equation:

Power = n×energy of a photon

9.42×10-3W = n×3.13×10-19 J

To find the speed of a hydrogen atom (v) to have momentum same as a photon:

v = p/m

Here, p = momentum of photon =1.043×10-27kgm/s, and

m=mass of a hydrogen atom= 1.67×10-24kg

Question: An electron has a kinetic energy of 120eV. Calculate: a) momentum, b) speed, and c) De Broglie wavelength of the electron

Solution:

Given, Kmax = 120×10-19J

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