Question

De Broglie relation is true for

A. All particles

B. Charged particles only

C. Negatively charged particles only

D. Massless particles like photons only

Answer 1

Equate the energy terms from Einstein's energy equation and Planck's equation to derive a relationship between the momentum of the particle and its wavelength, also known as the De Broglie relation, to define the duality behaviour. Observe the relationship and comment for which particles does the relationship hold.

Formula used: Einstein’s energy equation:

$E=\text{ }m{{c}^{2}}$

Whereas, the Planck's equation is given as,

$E\text{ }=\text{ }h\nu$

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Complete step-by-step answer:

Dual behaviour of matter was proposed in the hypothesis presented by De Broglie. He proposed that just like light, matter also exhibits wave like and particle like properties. And this nature of matter was termed as dual behaviour of matter. Based on his observations, De Broglie derived a relationship between wavelength and momentum of matter which is known as the De Broglie relationship.

Einstein’s famous energy equation is expressed as:

$E=\text{ }m{{c}^{2}}$

Where, E is the energy of the object, m is its mass and c is the speed of light. This equation expresses properties of a particle.

Whereas, the Planck's equation is given as,

$E\text{ }=\text{ }h\nu$

$\text{Where, }h=6.62607015\times {{10}^{-34}}J\operatorname{s}$

Where, E again is the energy, h is the Planck's constant and $\nu$ is the frequency. This equation expresses properties of a wave.

When the expressions for energy are equated, we get:

$m{{c}^{2}}\text{ }=\text{ }h\nu$

Where frequency can be expressed in terms of wavelength $\lambda$ as:

$\nu \text{ }=\dfrac{c}{\lambda}$

For a particle, the speed of light is replaced with the object’s velocity $\text{v}$:

$m{{\text{v}}^{2}}\text{ }=\text{ }h\dfrac{\text{v}}{\lambda}$

$\Rightarrow \lambda =\dfrac{h}{m\text{v}}$

We know that mass times velocity for a particle is its momentum p. Therefore, the De Broglie relationship comes out to be:

$\lambda =\dfrac{h}{P}$

$\lambda$ is known as the de Broglie wavelength.

Since the De Broglie’s wavelength is applicable for any type of particle, the correct option is (A).

Note: In everyday objects it is tough to notice the dual nature of objects. Which is because the value of the De Broglie’s wavelength becomes very small. Consider the expression:

$\lambda =\dfrac{h}{p}$ Here, the momentum of the objects we observe is incomparably large to Planck's constant, which is of the order -34. Whereas, for smaller particles like electrons the small mass, in the order -27, wavelength is adequately small to observe a wave like nature.