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# de Broglie Equation

The de Broglie equation states that matter can act as waves much like light and radiation, which also behave as waves and particles. The equation further explains that a beam of electrons can also be diffracted just like a beam of light. In essence, the de Broglie equation helps us understand the idea of matter having a wavelength.

Therefore, if we look at every moving particle, whether it is microscopic or macroscopic, it will have a wavelength.
In cases of macroscopic objects, the wave nature of matter can be detected, or it is visible.

## What Is de Broglie Equation?

The de Broglie equation is one of the equations that is commonly used to define the wave properties of matter. It basically describes the wave nature of the electron.

Electromagnetic radiation exhibits the dual nature of a particle (having a momentum) and wave (expressed in frequency and wavelength). Microscopic particle-like electrons also proved to possess this dual nature property.

Louis de Broglie, in his thesis, suggested that any moving particle, whether microscopic or macroscopic, will be associated with a wave character. It is called ‘Matter Waves’. He further proposed a relation between the velocity and momentum of a particle with the wavelength if the particle had to behave as a wave.

The particle and wave nature of matter, however, looked contradictory as it was not possible to prove the existence of both properties in any single experiment. This is because of the fact that every experiment is normally based on some principle, and results related to the principle are only reflected in that experiment and not the other.

Nonetheless, both properties are necessary to understand or describe the matter completely. Hence, particles and wave nature of matter are actually ‘complimentary’ to each other. It is not necessary for both to be present at the same time, though. The significance of the de Broglie relation is that it is more useful to microscopic, fundamental particles like electrons.

## de Broglie Equation Derivation and de Broglie Wavelength

Very low-mass particles moving at speeds less than that of light behave like particles and waves. De Broglie derived an expression relating the mass of such smaller particles and their wavelength.

Plank’s quantum theory relates the energy of an electromagnetic wave to its wavelength or frequency.

$$\begin{array}{l}E = hν =\frac{hc}{\lambda }…….(1)\end{array}$$

Einstein related the energy of particle matter to its mass and velocity as E = mc2……..(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,

$$\begin{array}{l}E =\frac{hc}{\lambda }=m{{v}^{2}}:\end{array}$$
Then,
$$\begin{array}{l}\frac{h}{\lambda }=mv\end{array}$$
or
$$\begin{array}{l}\lambda =\frac{h}{mv}=\frac{h}{\text{momentum}}:\end{array}$$
where ‘h’ is the Plank’s constant.

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

## Relation between de Broglie Equation and Bohr’s Hypothesis of Atom

Bohr postulated that the angular momentum of an electron revolving around the nucleus is quantized. Hence, the angular momentum will only be an integral multiple of a constant value and suggested the following expression.

Angular momentum of an electron in orbit

$$\begin{array}{l}= mvr =\frac{nh}{2\pi }:\end{array}$$
where ‘n’ is an integer with values of 1,2,3..

Bohr did not give any reason for such a proposal.

But, de Broglie equation gives a scientific validation for such an imaginative proposal.

By, de Broglie, equation

$$\begin{array}{l}\lambda =\frac{h}{mv}\end{array}$$
or
$$\begin{array}{l}mv=\frac{h}{\lambda }\end{array}$$

The electron wave in an orbit must be in phase, and so,

Circumference of an orbit = integral multiple of the wavelength.

$$\begin{array}{l}2\pi r=n\lambda\end{array}$$
or
$$\begin{array}{l}\frac{1}{\lambda }=\frac{n}{2\pi r}\end{array}$$

Substituting for wavelength, in the de Broglie equation,

$$\begin{array}{l}mv=\frac{nh}{2\pi r}\end{array}$$
or
$$\begin{array}{l}mvr=n\times \frac{h}{2\pi }\end{array}$$

Hence, the angular momentum of electron (mvr) is an integral multiple of a constant

$$\begin{array}{l}\left( \frac{h}{2\pi } \right)\end{array}$$

## Solved Problems on de Broglie Equation

1. Does de Broglie hypothesis have any relevance to macroscopic matter?

de Broglie relation can be applied to both microscopic and macroscopic. Taking, for example, a macro-sized 100 kg car moving at a speed of 100m/s will have a –

Wavelength of

$$\begin{array}{l}\lambda =\frac{h}{mv}=\frac{6.63\times {{10}^{-34}}}{100\times 100}=6.63\times {{10}^{-30}}m\end{array}$$

High-energy γ-radiations have wavelengths of only 10-12 m.

Very small wavelength corresponds to high frequencies. Waves below certain wavelength or beyond certain frequencies undergo particle-antiparticle annihilation to create mass. So, wave nature or de Broglie wavelength is not observable in the macroscopic matter.

2. Dual behaviour of matter proposed by de Broglie led to the discovery of the electron microscope often used for the highly magnified images of biological molecules and other types of material. If the velocity of the electron in this microscope is 1.6 × 106ms–1, calculate de Broglie wavelength associated with this electron.

de Broglie Wavelength of

$$\begin{array}{l}\lambda =\frac{h}{mv}=\frac{6.63\times {{10}^{-34}}}{9.1\times {{10}^{-31}}1.6\times {{10}^{6}}}=4.55\times {{10}^{-10}}m\end{array}$$

3. An electron and photon moving with speed ‘v’ and ‘c’, respectively, have the same de Broglie wavelength. If the kinetic energy and momentum of an electron are Ee and Pe and that of a proton is Eph and Pph, respectively, then the correct statement from the following is –

a)

$$\begin{array}{l}\frac{Ee}{Eph}=\frac{c}{2v}\end{array}$$
b)
$$\begin{array}{l}\frac{Ee}{Eph}=\frac{v}{2c}\end{array}$$
c)
$$\begin{array}{l}\frac{Pe}{Pph}=\frac{c}{2}\end{array}$$
d)
$$\begin{array}{l}\frac{Pe}{Pph}=\frac{v}{2c}\end{array}$$

de Broglie wavelength of a photon with velocity ‘c’

$$\begin{array}{l}\lambda ph=\frac{h}{{{m}_{ph}}c}\end{array}$$

de Broglie wave length of an electron with velocity ‘v’

$$\begin{array}{l}\lambda e=\frac{h}{{{m}_{e}}v}\end{array}$$

de Broglie wavelength of electron = de Broglie wavelength of a photon

$$\begin{array}{l}\lambda e=\frac{h}{{{m}_{e}}v}=\lambda ph=\frac{h}{{{m}_{ph}}c}\end{array}$$

$$\begin{array}{l}\frac{h}{{{m}_{e}}v}=\frac{h}{{{m}_{ph}}c}\,\,\,or\,\,{{m}_{e}}v={{m}_{ph}}c\end{array}$$
and
$$\begin{array}{l}\frac{{{m}_{e}}}{{{m}_{ph}}}=\frac{c}{v}\end{array}$$

$$\begin{array}{l}\frac{Ee}{Eph}=\frac{{{m}_{e}}{{v}^{2}}}{2{{m}_{ph}}{{c}^{2}}};\end{array}$$
Substituting for mass,
$$\begin{array}{l}\frac{Ee}{Eph}=\frac{c{{v}^{2}}}{2v{{c}^{2}}}\end{array}$$

Or

$$\begin{array}{l}\frac{Ee}{Eph}=\frac{v}{2c}.\end{array}$$

4. Is the wavelength of electrons on different orbits the same or different? If different what is the ratio of the wavelength in first and 4th orbits?

De Broglie wavelength

$$\begin{array}{l}=\lambda =\frac{h}{mv}\end{array}$$

Assuming the mass of the electron to be the same in all orbits,

$$\begin{array}{l}\lambda \,\alpha \frac{1}{v}\end{array}$$

The velocity of the electron varies in orbits, so the wavelength of the electron in different orbits will not be the same.

Velocity of the electron in an atom

$$\begin{array}{l}= V_n =\frac{2\pi k{{e}^{2}}}{h}\times \frac{Z}{n}\end{array}$$

Atomic number and other constant being same,

$$\begin{array}{l}{{V}_{n}}\alpha \frac{1}{n}\end{array}$$
so,
$$\begin{array}{l}\lambda \,\alpha \,n\end{array}$$

Ratio of de Broglie wavelength

$$\begin{array}{l}=\frac{{{\lambda }_{1}}}{{{\lambda }_{4}}}=\frac{v}{{{v}_{1}}}=\frac{1}{4}\end{array}$$

5. Show that the circumference of the Bohr orbit for the hydrogen atom is an integral multiple of the de Broglie wavelength associated with the electron revolving around the orbit.

According to Bohr’s postulate, angular momentum of electron

$$\begin{array}{l}=mvr=\frac{nh}{2\pi }\end{array}$$

This relation can be rearranged as;

$$\begin{array}{l}2\pi r=\frac{nh}{mv}=n\times \frac{h}{mv}\end{array}$$

According to de Broglie, the wavelength of the electron particle

$$\begin{array}{l}=\lambda =\frac{h}{mv}\end{array}$$

Substituting for the wavelength in the first equation,

$$\begin{array}{l}2\pi r=\frac{nh}{mv}=n\times \frac{h}{mv}=n\times \lambda\end{array}$$

Circumference of  ‘n’ th orbit

$$\begin{array}{l}=2\pi \,\,{{r}_{n}}=n\times \lambda\end{array}$$

Circumference of ‘n’ th orbit =  An integral number (n) × wavelength of the electron in the nth orbit.

## Frequently Asked Questions on the de Broglie Equation

Q1

### How does the de Broglie wavelength of the particle change when a particle is accelerated by applying an electric field?

When the particle is accelerated by applying an electric field, its velocity increases. Hence, its de Broglie wavelength decreases.

Q2

### How does the de Broglie wavelength of the proton change if its velocity is increased?

As the velocity of a proton increases, its de Broglie wavelength decreases.

Q3

### Why is it not possible to detect the de Broglie wavelength of a moving cricket ball?

The de Broglie wavelength of a moving cricket ball cannot be detected because its wavelength is too small to be detected.

Q4

### Which one will have a greater de Broglie wavelength if the electron and the proton move at the same speed?

The mass of the proton is greater than the electron, so the de Broglie wavelength of the proton is less than that of the electron.

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