 de Broglie Equation

de Broglie equation states that a 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, exhibit dual nature of a particle (having a momentum) and wave (expressed in frequency, 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 was 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.

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 the 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 de Broglie relation is that it is more useful to microscopic, fundamental particles like electron.

de Broglie Equation Derivation and de Broglie Wavelength

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

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

E  = hν  $=\frac{hc}{\lambda }$    …….(1)

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,

E = $=\frac{hc}{\lambda }=m{{v}^{2}}:$     Then, $\frac{h}{\lambda }=mv$ or $\lambda =\frac{h}{mv}=\frac{h}{\text{momentum}}:$ 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.

Relation Between de Broglie Equation and Bohr’s Hypothesis of Atom

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

Angular momentum of electron in orbit =  mvr $=\frac{nh}{2\pi }:$ 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 $\lambda =\frac{h}{mv}$   or    $mv=\frac{h}{\lambda }$

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

Circumference of an orbit = integral multiple of the wavelength.

$2\pi r=n\lambda$   or $\frac{1}{\lambda }=\frac{n}{2\pi r}$

Substituting for wave length, in the De Broglie equation,   $mv=\frac{nh}{2\pi r}$ or $mvr=n\times \frac{h}{2\pi }$

Hence the angular momentum of electron (mvr) is an integral multiple of a constant $\left( \frac{h}{2\pi } \right)$

Solved Problems on de Broglie Equation

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

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

Wavelength of $\lambda =\frac{h}{mv}=\frac{6.63\times {{10}^{-34}}}{100\times 100}=6.63\times {{10}^{-30}}m$

High-energy γ-radiations have wavelength 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 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 $\lambda =\frac{h}{mv}=\frac{6.63\times {{10}^{-34}}}{9.1\times {{10}^{-31}}1.6\times {{10}^{6}}}=4.55\times {{10}^{-10}}m$

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 are Eph and Pph respectively, then the correct statement from the following is –

4. a) $\frac{Ee}{Eph}=\frac{c}{2v}$  b) $\frac{Ee}{Eph}=\frac{v}{2c}$  c) $\frac{Pe}{Pph}=\frac{c}{2}$  d) $\frac{Pe}{Pph}=\frac{v}{2c}$

de Broglie wave length of  photon with velocity ‘c’ $\lambda ph=\frac{h}{{{m}_{ph}}c}=$

de Broglie wave length of electron with velocity ‘v’ $\lambda e=\frac{h}{{{m}_{e}}v}$ =

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

$\lambda e=\frac{h}{{{m}_{e}}v}=\lambda ph=\frac{h}{{{m}_{ph}}c}$

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

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

Or     $\frac{Ee}{Eph}=\frac{v}{2c}.$

The answer is ‘b’.

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

De Broglie wavelength $=\lambda =\frac{h}{mv}$

Assuming the mass of the electron to be the same in all orbits, $\lambda \,\alpha \frac{1}{v}$

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 = Vn = $=\frac{2\pi k{{e}^{2}}}{h}\times \frac{Z}{n}$

Atomic number and other constant being same,    ${{V}_{n}}\alpha \frac{1}{n}$ so,   $\lambda \,\alpha \,n$

Ratio of de Broglie wavelength $=\frac{{{\lambda }_{1}}}{{{\lambda }_{4}}}=\frac{v}{{{v}_{1}}}=\frac{1}{4}$

6. 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 $=mvr=\frac{nh}{2\pi }$

This relation can be rearranged as;   $2\pi r=\frac{nh}{mv}=n\times \frac{h}{mv}$

According to de Broglie, the wavelength of the electron particle $=\lambda =\frac{h}{mv}$

Substituting for the wavelength in the first equation,

$2\pi r=\frac{nh}{mv}=n\times \frac{h}{mv}=n\times \lambda$

Circumference of  ‘n’ th orbit $=2\pi \,\,{{r}_{n}}=n\times \lambda$

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