Question

what do you understand by wave particale duality? Derive the de-Broglie reaction and give its experiment verification.
​

Answer 1

Explanation:

de Broglie equated the energy equations of Planck (wave) and Einstein (particle).

E=hv (planck's relation)

E=mc

2

(Einstein's mass - energy relation)

hv=mc

2

λ

hc

=mc

2

(or)

λ=

mc

h

For a particle moving with a velocity V,

λ=

mV

h

=

p

h

where p=mV is the momentum of the particle

Answer 2

Answer:

The fundamental property of matter that appears as a wave one instant and acts as a particle the next is referred to as wave-particle duality.

Explanation:

Wave-particle duality:

• In quantum physics, wave-particle duality states that every particle or quantum entity can be described as either a particle or a wave.

• It expresses the classical concepts of particle and wave's inability to completely explain the behaviour of quantum-scale objects.

• Light has the properties of both a wave and a particle, according to wave-particle theory.

De Broglie Equation:

Very low mass particles moving at speed less than that of light behave like a particle and wave. De Broglie derived an expression relating the mass of such smaller particles and its wavelength.

Step 1: Plank's quantum theory relates the energy of an electromagnetic wave to its wavelength or frequency.

$E=h \nu=\frac{h c}{\lambda} \ldots \ldots(1)$

Step 2: Einstein related the energy of particle matter to its mass and velocity, as $E=m c^2$.

As the smaller particle exhibits dual nature, and energy being the same, de Broglie equated both these relations for the particle moving with velocity ' v ' as,

$E=\frac{h c}{\lambda}=m v^2$

Step 3: Then,

$\frac{h}{\lambda}=m v$

or

$\lambda=\frac{h}{m v}=\frac{h}{\text { momentum }}:$

where ' $h$ ' is the Plank's constant.

This equation relating the momentum of a particle with its wavelength is de Broglie equation and the wavelength calculated using this relation is de Broglie wavelength.

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