What does wave nature of matter mean? Can a small particle be at multiple places at the same time? Do I have a wave nature? Why can’t I see it? Let’s try to answer these questions.

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## Wave Nature of Matter

In the earlier articles, we saw how light behaves both as a wave and particle. A particle is confined at a place. On the other hand, a wave is spread in space. We say that the nature of light depends on the nature of our observation. If you are observing phenomenon like the interference, diffraction or reflection, you will find that light is a wave. However, if you are looking at phenomena like the photoelectric effect, you will find that light has a particle character.

You might ask, which is it? Is light a wave or a particle? The answer is that it has a dual nature. You may also wonder whether it is a specific property of light! Does only light have a dual nature? What if other quantities had dual nature? How could we measure and prove that? Maybe these were the questions that led Louis Victor de Broglie to come up with one of the most revolutionary equations in Physics, the de Broglie equation.

**Browse more Topics under Dual Nature Of Radiation And Matter**

- Electron Emission
- Experimental Study of Photoelectric Effect
- Davisson and Germer Experiment
- Einstein’s Photoelectric Equation: Energy Quantum of Radiation

### Next up – A Few Lines of Math That Go A Long Way!

Let us recall the mass-energy equivalence of Einstein, E =mc^{2} …(1)

Also from Einstein-Plank relation, we have: E = hν …(2)

Furthermore, we see that equation (1) is applicable to particles with some “mass”. In other words equation (1) can be applied to particles and equation (2) is an equation for a wave of frequency ν. So the two were not equated until de Broglie had a breakthrough! We know that light can be a wave as well as a particle. In that case, we can say that equation (1) and (2) represent the same quantity. Consequently, we must have: hν = mc^{2 }. Since we know that ν = c/λ, we have:

h(c/λ) = mc^{2}

λ = h/mc; where ‘c’ is the velocity of light. If we have a wave of velocity, say ‘v’, we can write: λ = h/mv

or λ = h/p …(3)

where ‘p’ is the momentum of the wave-particle! See what we did here? We have mass – a particle property, in the same equation as wavelength – a wave property. Thus if matter exhibits wave properties, it must be given by equation (3). Equation (3) is the de Broglie equation and represents the wave-particle duality. Hence we say that everything in the Cosmos exhibits a dual nature. This is the wave nature of radiation and matter.

### Wavelength of Macroscopic Objects

“So you are telling me that I am not a particle but a wave? Where is it then?” First of all, you are both. Let us find out your wavelength. Suppose you have a mass of 55 kg. If you are at rest i.e. if the velocity = 0, then we see from equation (3), that λ is not defined. So not much help there! Let us say that you are moving at a velocity of 5 m/s. Using equation (3), we can see that

λ = h/(55)×5

λ = 6.63×10-34/275 ≈ 2.4×10-36 m

As you can see, you can’t “see” this small wavelength. Thus the wavelength of macroscopic objects is too small to have any observable effects on any property at normal velocities.

Learn more about Wave Optics.

### So de Broglie Guessed An Equation And Everyone Just Agreed?

Fortunately, there was a way to verify this equation. Let us see the equation again, λ = h/p

We know that K.E. = 1/2(mv^{2})

or K.E. = \( \frac{(mv)^2}{(2m)} \) = \( \frac{(p)^2}{(2m)} \)

Here, p is the momentum. Thus we have: p = \( \sqrt[]{2mE} \) …. (4)

Using (4) in (3), we have: λ = h/\( \sqrt[]{2mE} \) …(5)

Also for a charged particle, E = eV and we have: λ = h/\( \sqrt[]{2meV} \)

So for an electron e = 1.6×10^{-19} C and m = 9.10938356 × 10^{-31} kilograms, we have:

Hence we can verify the de Broglie equation if we observe the motion of an electron. This was done in the Davisson and Germer Experiment.

## Solved Examples For You

The de-Broglie wavelength of an electron (mass 1×10^{−30}kg, charge=1.6×10^{−19} C) with a kinetic energy of 200eV is: (Planck’s constant 6.6×10^{−34}J):

A) 9.60×10^{−11}m B) 8.25×10^{−11}m

C) 6.25×10^{−11}m C) 5.00×10^{−11}m

**Solution: **B) 8.25×10^{−11}m

We can directly use equation (5) i.e. λ = h/\( \sqrt[]{2mE} \). Substitution of the respective values gives the required result.

Eassy understand